“Notice: Undefined variable”, “Notice: Undefined index”, and “Notice: Undefined offset” using PHP
I'm running a PHP script, and continue to receive errors like:
Notice: Undefined variable: my_variable_name in C:wampwwwmypathindex.php on line 10
Notice: Undefined index: my_index C:wampwwwmypathindex.php on line 11
Line 10 and 11 looks like this:
echo "My variable value is: " . $my_variable_name;
echo "My index value is: " . $my_array["my_index"];
What is the meaning of these error messages?
Why do they appear all of a sudden? I used to use this script for years and I've never had any problem.
How do I fix them?
This is a General Reference question for people to link to as duplicate, instead of having to explain the issue over and over again. I feel this is necessary because most real-world answers on this issue are very specific.
Related Meta discussion:
- What can be done about repetitive questions?
- Do “reference questions” make sense?
php arrays variables warnings undefined-index
|
show 7 more comments
I'm running a PHP script, and continue to receive errors like:
Notice: Undefined variable: my_variable_name in C:wampwwwmypathindex.php on line 10
Notice: Undefined index: my_index C:wampwwwmypathindex.php on line 11
Line 10 and 11 looks like this:
echo "My variable value is: " . $my_variable_name;
echo "My index value is: " . $my_array["my_index"];
What is the meaning of these error messages?
Why do they appear all of a sudden? I used to use this script for years and I've never had any problem.
How do I fix them?
This is a General Reference question for people to link to as duplicate, instead of having to explain the issue over and over again. I feel this is necessary because most real-world answers on this issue are very specific.
Related Meta discussion:
- What can be done about repetitive questions?
- Do “reference questions” make sense?
php arrays variables warnings undefined-index
7
possible duplicate of Reference - What does this error mean in PHP?
– meagar♦
Oct 8 '13 at 15:25
2
the variable might not have been initialized. Are you initializing the variable from a post or get or any array? If that's the case you might not have an field in that array. That your accessing.
– Anish Silwal
Dec 15 '15 at 13:12
3
@Pekka웃 - I noticed the edit adding the "and "Notice: Undefined offset"" - Wouldn't it make more sense using "PHP: “Undefined variable”, “Undefined index”, “Undefined offset” notices" (even take out the PHP, since it is tagged as "php". Plus, the URL gets cut off atand-notice-undef
, just a suggestion so that the URL doesn't get cut off. Maybe even removing the (too many) quotes. OrPHP: “Undefined variable/index/offset” notices
– Funk Forty Niner
Jan 20 '17 at 15:24
2
@Fred I guess an argument can be made for both variations. THere's a chance that newbies will enter the entire line, including the "Notice:" into their search query, which I'm sure is the main traffic generator for this question. If the messages are there in full, that's likely to improve visibility in search engines
– Pekka 웃
Jan 20 '17 at 15:34
2
@Pekka웃 I understand. I only said that because the URL didn't get cut off before and now it does atand-notice-undef
. It was just a (few) suggestion(s). It just repeats itself also beingNotice: Undefined
.
– Funk Forty Niner
Jan 20 '17 at 15:35
|
show 7 more comments
I'm running a PHP script, and continue to receive errors like:
Notice: Undefined variable: my_variable_name in C:wampwwwmypathindex.php on line 10
Notice: Undefined index: my_index C:wampwwwmypathindex.php on line 11
Line 10 and 11 looks like this:
echo "My variable value is: " . $my_variable_name;
echo "My index value is: " . $my_array["my_index"];
What is the meaning of these error messages?
Why do they appear all of a sudden? I used to use this script for years and I've never had any problem.
How do I fix them?
This is a General Reference question for people to link to as duplicate, instead of having to explain the issue over and over again. I feel this is necessary because most real-world answers on this issue are very specific.
Related Meta discussion:
- What can be done about repetitive questions?
- Do “reference questions” make sense?
php arrays variables warnings undefined-index
I'm running a PHP script, and continue to receive errors like:
Notice: Undefined variable: my_variable_name in C:wampwwwmypathindex.php on line 10
Notice: Undefined index: my_index C:wampwwwmypathindex.php on line 11
Line 10 and 11 looks like this:
echo "My variable value is: " . $my_variable_name;
echo "My index value is: " . $my_array["my_index"];
What is the meaning of these error messages?
Why do they appear all of a sudden? I used to use this script for years and I've never had any problem.
How do I fix them?
This is a General Reference question for people to link to as duplicate, instead of having to explain the issue over and over again. I feel this is necessary because most real-world answers on this issue are very specific.
Related Meta discussion:
- What can be done about repetitive questions?
- Do “reference questions” make sense?
php arrays variables warnings undefined-index
php arrays variables warnings undefined-index
edited Oct 29 '18 at 20:02
community wiki
27 revs, 18 users 29%
Pekka 웃
7
possible duplicate of Reference - What does this error mean in PHP?
– meagar♦
Oct 8 '13 at 15:25
2
the variable might not have been initialized. Are you initializing the variable from a post or get or any array? If that's the case you might not have an field in that array. That your accessing.
– Anish Silwal
Dec 15 '15 at 13:12
3
@Pekka웃 - I noticed the edit adding the "and "Notice: Undefined offset"" - Wouldn't it make more sense using "PHP: “Undefined variable”, “Undefined index”, “Undefined offset” notices" (even take out the PHP, since it is tagged as "php". Plus, the URL gets cut off atand-notice-undef
, just a suggestion so that the URL doesn't get cut off. Maybe even removing the (too many) quotes. OrPHP: “Undefined variable/index/offset” notices
– Funk Forty Niner
Jan 20 '17 at 15:24
2
@Fred I guess an argument can be made for both variations. THere's a chance that newbies will enter the entire line, including the "Notice:" into their search query, which I'm sure is the main traffic generator for this question. If the messages are there in full, that's likely to improve visibility in search engines
– Pekka 웃
Jan 20 '17 at 15:34
2
@Pekka웃 I understand. I only said that because the URL didn't get cut off before and now it does atand-notice-undef
. It was just a (few) suggestion(s). It just repeats itself also beingNotice: Undefined
.
– Funk Forty Niner
Jan 20 '17 at 15:35
|
show 7 more comments
7
possible duplicate of Reference - What does this error mean in PHP?
– meagar♦
Oct 8 '13 at 15:25
2
the variable might not have been initialized. Are you initializing the variable from a post or get or any array? If that's the case you might not have an field in that array. That your accessing.
– Anish Silwal
Dec 15 '15 at 13:12
3
@Pekka웃 - I noticed the edit adding the "and "Notice: Undefined offset"" - Wouldn't it make more sense using "PHP: “Undefined variable”, “Undefined index”, “Undefined offset” notices" (even take out the PHP, since it is tagged as "php". Plus, the URL gets cut off atand-notice-undef
, just a suggestion so that the URL doesn't get cut off. Maybe even removing the (too many) quotes. OrPHP: “Undefined variable/index/offset” notices
– Funk Forty Niner
Jan 20 '17 at 15:24
2
@Fred I guess an argument can be made for both variations. THere's a chance that newbies will enter the entire line, including the "Notice:" into their search query, which I'm sure is the main traffic generator for this question. If the messages are there in full, that's likely to improve visibility in search engines
– Pekka 웃
Jan 20 '17 at 15:34
2
@Pekka웃 I understand. I only said that because the URL didn't get cut off before and now it does atand-notice-undef
. It was just a (few) suggestion(s). It just repeats itself also beingNotice: Undefined
.
– Funk Forty Niner
Jan 20 '17 at 15:35
7
7
possible duplicate of Reference - What does this error mean in PHP?
– meagar♦
Oct 8 '13 at 15:25
possible duplicate of Reference - What does this error mean in PHP?
– meagar♦
Oct 8 '13 at 15:25
2
2
the variable might not have been initialized. Are you initializing the variable from a post or get or any array? If that's the case you might not have an field in that array. That your accessing.
– Anish Silwal
Dec 15 '15 at 13:12
the variable might not have been initialized. Are you initializing the variable from a post or get or any array? If that's the case you might not have an field in that array. That your accessing.
– Anish Silwal
Dec 15 '15 at 13:12
3
3
@Pekka웃 - I noticed the edit adding the "and "Notice: Undefined offset"" - Wouldn't it make more sense using "PHP: “Undefined variable”, “Undefined index”, “Undefined offset” notices" (even take out the PHP, since it is tagged as "php". Plus, the URL gets cut off at
and-notice-undef
, just a suggestion so that the URL doesn't get cut off. Maybe even removing the (too many) quotes. Or PHP: “Undefined variable/index/offset” notices
– Funk Forty Niner
Jan 20 '17 at 15:24
@Pekka웃 - I noticed the edit adding the "and "Notice: Undefined offset"" - Wouldn't it make more sense using "PHP: “Undefined variable”, “Undefined index”, “Undefined offset” notices" (even take out the PHP, since it is tagged as "php". Plus, the URL gets cut off at
and-notice-undef
, just a suggestion so that the URL doesn't get cut off. Maybe even removing the (too many) quotes. Or PHP: “Undefined variable/index/offset” notices
– Funk Forty Niner
Jan 20 '17 at 15:24
2
2
@Fred I guess an argument can be made for both variations. THere's a chance that newbies will enter the entire line, including the "Notice:" into their search query, which I'm sure is the main traffic generator for this question. If the messages are there in full, that's likely to improve visibility in search engines
– Pekka 웃
Jan 20 '17 at 15:34
@Fred I guess an argument can be made for both variations. THere's a chance that newbies will enter the entire line, including the "Notice:" into their search query, which I'm sure is the main traffic generator for this question. If the messages are there in full, that's likely to improve visibility in search engines
– Pekka 웃
Jan 20 '17 at 15:34
2
2
@Pekka웃 I understand. I only said that because the URL didn't get cut off before and now it does at
and-notice-undef
. It was just a (few) suggestion(s). It just repeats itself also being Notice: Undefined
.– Funk Forty Niner
Jan 20 '17 at 15:35
@Pekka웃 I understand. I only said that because the URL didn't get cut off before and now it does at
and-notice-undef
. It was just a (few) suggestion(s). It just repeats itself also being Notice: Undefined
.– Funk Forty Niner
Jan 20 '17 at 15:35
|
show 7 more comments
27 Answers
27
active
oldest
votes
Notice: Undefined variable
From the vast wisdom of the PHP Manual:
Relying on the default value of an uninitialized variable is problematic in the case of including one file into another which uses the same variable name. It is also a major security risk with register_globals turned on. E_NOTICE level error is issued in case of working with uninitialized variables, however not in the case of appending elements to the uninitialized array. isset() language construct can be used to detect if a variable has been already initialized. Additionally and more ideal is the solution of empty() since it does not generate a warning or error message if the variable is not initialized.
From PHP documentation:
No warning is generated if the variable does not exist. That means
empty() is essentially the concise equivalent to !isset($var) || $var
== false.
This means that you could use only empty()
to determine if the variable is set, and in addition it checks the variable against the following, 0,"",null
.
Example:
$o = ;
@$var = ["",0,null,1,2,3,$foo,$o['myIndex']];
array_walk($var, function($v) {
echo (!isset($v) || $v == false) ? 'true ' : 'false';
echo ' ' . (empty($v) ? 'true' : 'false');
echo "n";
});
Test the above snippet in the 3v4l.org online PHP editor
Although PHP does not require a variable declaration, it does recommend it in order to avoid some security vulnerabilities or bugs where one would forget to give a value to a variable that will be used later in the script. What PHP does in the case of undeclared variables is issue a very low level error, E_NOTICE
, one that is not even reported by default, but the Manual advises to allow during development.
Ways to deal with the issue:
Recommended: Declare your variables, for example when you try to append a string to an undefined variable. Or use
isset()
/!empty()
to check if they are declared before referencing them, as in:
//Initializing variable
$value = ""; //Initialization value; Examples
//"" When you want to append stuff later
//0 When you want to add numbers later
//isset()
$value = isset($_POST['value']) ? $_POST['value'] : '';
//empty()
$value = !empty($_POST['value']) ? $_POST['value'] : '';
This has become much cleaner as of PHP 7.0, now you can use the null coalesce operator:
// Null coalesce operator - No need to explicitly initialize the variable.
$value = $_POST['value'] ?? '';
Set a custom error handler for E_NOTICE and redirect the messages away from the standard output (maybe to a log file):
set_error_handler('myHandlerForMinorErrors', E_NOTICE | E_STRICT)
Disable E_NOTICE from reporting. A quick way to exclude just
E_NOTICE
is:
error_reporting( error_reporting() & ~E_NOTICE )
Suppress the error with the @ operator.
Note: It's strongly recommended to implement just point 1.
Notice: Undefined index / Undefined offset
This notice appears when you (or PHP) try to access an undefined index of an array.
Ways to deal with the issue:
Check if the index exists before you access it. For this you can use
isset()
orarray_key_exists()
:
//isset()
$value = isset($array['my_index']) ? $array['my_index'] : '';
//array_key_exists()
$value = array_key_exists('my_index', $array) ? $array['my_index'] : '';
The language construct
list()
may generate this when it attempts to access an array index that does not exist:
list($a, $b) = array(0 => 'a');
//or
list($one, $two) = explode(',', 'test string');
Two variables are used to access two array elements, however there is only one array element, index 0
, so this will generate:
Notice: Undefined offset: 1
$_POST
/ $_GET
/ $_SESSION
variable
The notices above appear often when working with $_POST
, $_GET
or $_SESSION
. For $_POST
and $_GET
you just have to check if the index exists or not before you use them. For $_SESSION
you have to make sure you have the session started with session_start()
and that the index also exists.
Also note that all 3 variables are superglobals and are uppercase.
Related:
- Notice: Undefined variable
Notice: Undefined Index
7
@dieselpower44 A couple of thoughts: The "shut-up operator" (@
) has some performance issues. Also, since it suppresses all errors within a particular scope, using it without care might mask messages you wish you'd seen.
– IMSoP
Oct 24 '13 at 20:00
6
Hiding the issues is NOT the way to deal with issues. Items #2...#4 can be used only on production servers, not in general.
– Salman A
Sep 13 '14 at 8:09
1
Is it possible to shut-up the message inline (not in handler) when also a custom error handler is used?$var = @$_GET['nonexisting'];
still causes notice..
– Alph.Dev
Oct 11 '14 at 14:14
18
Why is it recommended to use 1.$value = isset($_POST['value']) ? $_POST['value'] : '';
instead of using 4.$value = @$_POST['value'];
?
– forsvunnet
Feb 9 '15 at 13:38
@twistedpixel Those 4 ways are independent, that's not a 4-step guide. So if you've chosen to use way 4, that means you didn't implement first 3 ways, so you you didn't supress any errors yet.
– Aycan Yaşıt
Aug 6 '15 at 9:58
|
show 5 more comments
Try these
Q1: this notice means $varname is not
defined at current scope of the
script.
Q2: Use of isset(), empty() conditions before using any suspicious variable works well.
// recommended solution for recent PHP versions
$user_name = $_SESSION['user_name'] ?? '';
// pre-7 PHP versions
$user_name = '';
if (!empty($_SESSION['user_name'])) {
$user_name = $_SESSION['user_name'];
}
Or, as a quick and dirty solution:
// not the best solution, but works
// in your php setting use, it helps hiding site wide notices
error_reporting(E_ALL ^ E_NOTICE);
Note about sessions:
When using sessions,
session_start();
is required to be placed inside all files using sessions.http://php.net/manual/en/features.sessions.php
If usingE_NOTICE
from thephp.ini
configuration file, doerror_reporting = (E_ALL & ~E_NOTICE)
– ahmd0
Oct 4 '14 at 5:39
1
This answer is being discussed on meta.
– Script47
Jun 25 '18 at 10:43
add a comment |
Error display @
operator
For undesired and redundant notices, one could use the dedicated @
operator to »hide« undefined variable/index messages.
$var = @($_GET["optional_param"]);
- This is usually discouraged. Newcomers tend to way overuse it.
- It's very inappropriate for code deep within the application logic (ignoring undeclared variables where you shouldn't), e.g. for function parameters, or in loops.
- There's one upside over the
isset?:
or??
super-supression however. Notices still can get logged. And one may resurrect@
-hidden notices with:set_error_handler("var_dump");
- Additonally you shouldn't habitually use/recommend
if (isset($_POST["shubmit"]))
in your initial code. - Newcomers won't spot such typos. It just deprives you of PHPs Notices for those very cases. Add
@
orisset
only after verifying functionality.
Fix the cause first. Not the notices.
- Additonally you shouldn't habitually use/recommend
@
is mainly acceptable for$_GET
/$_POST
input parameters, specifically if they're optional.
And since this covers the majority of such questions, let's expand on the most common causes:
$_GET
/ $_POST
/ $_REQUEST
undefined input
First thing you do when encountering an undefined index/offset, is check for typos:
$count = $_GET["whatnow?"];
- Is this an expected key name and present on each page request?
- Variable names and array indicies are case-sensitive in PHP.
Secondly, if the notice doesn't have an obvious cause, use
var_dump
orprint_r
to verify all input arrays for their curent content:
var_dump($_GET);
var_dump($_POST);
//print_r($_REQUEST);
Both will reveal if your script was invoked with the right or any parameters at all.
Alternativey or additionally use your browser devtools (F12) and inspect the network tab for requests and parameters:
POST parameters and GET input will be be shown separately.
For
$_GET
parameters you can also peek at theQUERY_STRING
in
print_r($_SERVER);
PHP has some rules to coalesce non-standard parameter names into the superglobals. Apache might do some rewriting as well.
You can also look at supplied raw$_COOKIES
and other HTTP request headers that way.
More obviously look at your browser address bar for GET parameters:
http://example.org/script.php?id=5&sort=desc
The
name=value
pairs after the?
question mark are your query (GET) parameters. Thus this URL could only possibly yield$_GET["id"]
and$_GET["sort"]
.
Finally check your
<form>
and<input>
declarations, if you expect a parameter but receive none.
- Ensure each required input has an
<input name=FOO>
- The
id=
ortitle=
attribute does not suffice. - A
method=POST
form ought to populate$_POST
. - Whereas a
method=GET
(or leaving it out) would yield$_GET
variables. - It's also possible for a form to supply
action=script.php?get=param
via $_GET and the remainingmethod=POST
fields in $_POST alongside. - With modern PHP configurations (≥ 5.6) it has become feasible (not fashionable) to use
$_REQUEST['vars']
again, which mashes GET and POST params.
- Ensure each required input has an
If you are employing mod_rewrite, then you should check both the
access.log
as well as enable theRewriteLog
to figure out absent parameters.
$_FILES
- The same sanity checks apply to file uploads and
$_FILES["formname"]
. - Moreover check for
enctype=multipart/form-data
- As well as
method=POST
in your<form>
declaration. - See also: PHP Undefined index error $_FILES?
$_COOKIE
- The
$_COOKIE
array is never populated right aftersetcookie()
, but only on any followup HTTP request. - Additionally their validity times out, they could be constraint to subdomains or individual paths, and user and browser can just reject or delete them.
2
If you are curious what is the performance impact, this article summarises it well, derickrethans.nl/….
– Gajus
Feb 11 '14 at 12:24
@GajusKuizinas There have been quoite a few changes since 2009, in particular php.net/ChangeLog-5.php#5.4.0 changes the outcome drastically (see "Zend Engine, performance" and "(silence) operator").
– mario
Feb 11 '14 at 12:37
Thanks @mario, interesting. Now, if someone was good enough to benchmark the two... 3v4l.org/CYVOn/perf#tabs 3v4l.org/FLp3D/perf#tabs According to this test, seem to be identical (notice that scale changes).
– Gajus
Feb 11 '14 at 16:30
I tested with PHP 5.4 and the performance still bad.
– Brynner Ferreira
Oct 3 '14 at 17:43
add a comment |
Generally because of "bad programming", and a possibility for mistakes now or later.
- If it's a mistake, make a proper assignment to the variable first: $varname=0;
- If it really is only defined sometimes, test for it:
if (isset($varname))
, before using it - If it's because you spelled it wrong, just correct that
- Maybe even turn of the warnings in you PHP-settings
4
Please don't turn warnings off. In stricter languages, they often mean "there might be a bug, you better check this line twice" - in a language as permissive as PHP, they often means "this code is crap and propably riddled with bugs; I'll try to make some sense of it but you better fix this ASAP".
– user395760
Nov 23 '10 at 21:37
4
Although I agree with the first three points, #4 is simply wrong. Hiding a problem won't make it go away, and it might even cause more problems down the road.
– Valentin Flachsel
Nov 23 '10 at 21:38
1
@Freek absolutely true, but in some scenarios (Bought script, zero technical knowledge, need it running by tomorrow...) it's the duct-tape solution - really bad, that always needs emphasizing, but an option
– Pekka 웃
Nov 23 '10 at 21:40
Duct-tape is good ... sometimes. Historically warnings have been turned of in standard PHP-settings, but defult settings have become more strict. Too bad many go back to the old settings, so as not to annoy the customers.
– Erik
Nov 23 '10 at 22:26
add a comment |
It means you are testing, evaluating, or printing a variable that you have not yet assigned anything to. It means you either have a typo, or you need to check that the variable was initialized to something first. Check your logic paths, it may be set in one path but not in another.
add a comment |
I didn't want to disable notice because it's helpful, but wanted to avoid too much typing.
My solution was this function:
function ifexists($varname)
{
return(isset($$varname)?$varname:null);
}
So if I want to reference to $name and echo if exists, I simply write:
<?=ifexists('name')?>
For array elements:
function ifexistsidx($var,$index)
{
return(isset($var[$index])?$var[$index]:null);
}
In page if I want to refer to $_REQUEST['name']:
<?=ifexistsidx($_REQUEST,'name')?>
2
Your ifexists() function doesn't work for me in PHP 5.3. The caller's variables are not available in the function's local scope (see Variable scope), unless they are Superglobals or you fiddle with $GLOBALS, so$foo = "BABAR"; ifexists('foo');
will in general return null. (Italics are php.net chapters.)
– skierpage
Jun 19 '12 at 22:14
now you will get "hello from"... what's the point? just check the valueif( !empty($user) and !empty($location) ) echo "hello $user ..."
– gcb
Jul 9 '13 at 5:31
add a comment |
The best way for getting input string is:
$value = filter_input(INPUT_POST, 'value');
This one-liner is almost equivalent to:
if (!isset($_POST['value'])) {
$value = null;
} elseif (is_array($_POST['value'])) {
$value = false;
} else {
$value = $_POST['value'];
}
If you absolutely want string value, just like:
$value = (string)filter_input(INPUT_POST, 'value');
add a comment |
Its because the variable '$user_location' is not getting defined. If you are using any if loop inside which you are declaring the '$user_location' variable then you must also have an else loop and define the same. For example:
$a=10;
if($a==5) { $user_location='Paris';} else { }
echo $user_location;
The above code will create error as The if loop is not satisfied and in the else loop '$user_location' was not defined. Still PHP was asked to echo out the variable. So to modify the code you must do the following:
$a=10;
if($a==5) { $user_location='Paris';} else { $user_location='SOMETHING OR BLANK'; }
echo $user_location;
add a comment |
In reply to ""Why do they appear all of a sudden? I used to use this script for years and I've never had any problem."
It is very common for most sites to operate under the "default" error reporting of "Show all errors, but not 'notices' and 'deprecated'". This will be set in php.ini and apply to all sites on the server. This means that those "notices" used in the examples will be suppressed (hidden) while other errors, considered more critical, will be shown/recorded.
The other critical setting is the errors can be hidden (i.e. display_errors
set to "off" or "syslog").
What will have happened in this case is that either the error_reporting
was changed to also show notices (as per examples) and/or that the settings were changed to display_errors
on screen (as opposed to suppressing them/logging them).
Why have they changed?
The obvious/simplest answer is that someone adjusted either of these settings in php.ini, or an upgraded version of PHP is now using a different php.ini from before. That's the first place to look.
However it is also possible to override these settings in
- .htconf (webserver configuration, including vhosts and sub-configurations)*
- .htaccess
- in php code itself
and any of these could also have been changed.
There is also the added complication that the web server configuration can enable/disable .htaccess directives, so if you have directives in .htaccess that suddenly start/stop working then you need to check for that.
(.htconf / .htaccess assume you're running as apache. If running command line this won't apply; if running IIS or other webserver then you'll need to check those configs accordingly)
Summary
- Check
error_reporting
anddisplay_errors
php directives in php.ini has not changed, or that you're not using a different php.ini from before. - Check
error_reporting
anddisplay_errors
php directives in .htconf (or vhosts etc) have not changed - Check
error_reporting
anddisplay_errors
php directives in .htaccess have not changed - If you have directive in .htaccess, check if they are still permitted in the .htconf file
- Finally check your code; possibly an unrelated library; to see if
error_reporting
anddisplay_errors
php directives have been set there.
add a comment |
the quick fix is to assign your variable to null at the top of your code
$user_location = null;
add a comment |
I used to curse this error, but it can be helpful to remind you to escape user input.
For instance, if you thought this was clever, shorthand code:
// Echo whatever the hell this is
<?=$_POST['something']?>
...Think again! A better solution is:
// If this is set, echo a filtered version
<?=isset($_POST['something']) ? html($_POST['something']) : ''?>
(I use a custom html()
function to escape characters, your mileage may vary)
add a comment |
I use all time own useful function exst() which automatically declare variables.
Your code will be -
$greeting = "Hello, ".exst($user_name, 'Visitor')." from ".exst($user_location);
/**
* Function exst() - Checks if the variable has been set
* (copy/paste it in any place of your code)
*
* If the variable is set and not empty returns the variable (no transformation)
* If the variable is not set or empty, returns the $default value
*
* @param mixed $var
* @param mixed $default
*
* @return mixed
*/
function exst( & $var, $default = "")
{
$t = "";
if ( !isset($var) || !$var ) {
if (isset($default) && $default != "") $t = $default;
}
else {
$t = $var;
}
if (is_string($t)) $t = trim($t);
return $t;
}
add a comment |
In a very Simple Language.
The mistake is you are using a variable $user_location
which is not defined by you earlier and it doesn't have any value So I recommend you to please declare this variable before using it, For Example:
$user_location = '';
Or
$user_location = 'Los Angles';
This is a very common error you can face.So don't worry just declare the variable and Enjoy Coding.
add a comment |
In PHP 7.0 it's now possible to use Null coalescing operator:
echo "My index value is: " . ($my_array["my_index"] ?? '');
Equals to:
echo "My index value is: " . (isset($my_array["my_index"]) ? $my_array["my_index"] : '');
PHP manual PHP 7.0
1
Wow, this answer needs to be voted up more now that this is available in PHP 7. Wonderful solution!!!
– OCDev
Sep 24 '17 at 13:17
This also works fine in if statements.if (is_array($my_array['idontexist'] ?? '')) { dosomething(); }
– a coder
Feb 23 '18 at 20:04
Your code is actually a nice overlooked bug: ?? - only checks forisset()
, if you passis_array()
- which is a boolean, unexpected behavior will follow.
– Lachezar Lechev
Feb 26 '18 at 1:20
add a comment |
why not keep things simple?
<?php
error_reporting(E_ALL); // making sure all notices are on
function idxVal(&$var, $default = null) {
return empty($var) ? $var = $default : $var;
}
echo idxVal($arr['test']); // returns null without any notice
echo idxVal($arr['hey ho'], 'yo'); // returns yo and assigns it to array index, nice
?>
add a comment |
WHY IS THIS HAPPENING?
Over time, PHP has become a more security-focused language. Settings which used to be turned off by default are now turned on by default. A perfect example of this is E_STRICT
, which became turned on by default as of PHP 5.4.0.
Furthermore, according to PHP documentation, by defualt, E_NOTICE
is disabled in php.ini. PHP docs recommend turning it on for debugging purposes. However, when I download PHP from the Ubuntu repository–and from BitNami's Windows stack–I see something else.
; Common Values:
; E_ALL (Show all errors, warnings and notices including coding standards.)
; E_ALL & ~E_NOTICE (Show all errors, except for notices)
; E_ALL & ~E_NOTICE & ~E_STRICT (Show all errors, except for notices and coding standards warnings.)
; E_COMPILE_ERROR|E_RECOVERABLE_ERROR|E_ERROR|E_CORE_ERROR (Show only errors)
; Default Value: E_ALL & ~E_NOTICE & ~E_STRICT & ~E_DEPRECATED
; Development Value: E_ALL
; Production Value: E_ALL & ~E_DEPRECATED & ~E_STRICT
; http://php.net/error-reporting
error_reporting = E_ALL & ~E_DEPRECATED & ~E_STRICT
Notice that error_reporting
is actually set to the production value by default, not to the "default" value by default. This is somewhat confusing and is not documented outside of php.ini, so I have not validated this on other distributions.
To answer your question, however, this error pops up now when it did not pop up before because:
You installed PHP and the new default settings are somewhat poorly documented, but do not exclude
E_NOTICE
.E_NOTICE
warnings like undefined variables and undefined indexes actually help to make your code cleaner and safer. I can tell you that, years ago, keepingE_NOTICE
enabled forced me to declare my variables. It made it a LOT easier to learn C, where not declaring variables is much bigger of a nuisance.
WHAT CAN I DO ABOUT IT?
Turn off
E_NOTICE
by copying the "Default value"E_ALL & ~E_NOTICE & ~E_STRICT & ~E_DEPRECATED
and replacing it with what is currently uncommented after the equals sign inerror_reporting =
. Restart Apache, or PHP if using CGI or FPM. Make sure you are editing the "right" php.ini. The correct one will be Apache if you are running PHP with Apache, fpm or php-fpm if running PHP-FPM, cgi if running PHP-CGI, etc. This is not the recommended method, but if you have legacy code that's going to be exceedingly difficult to edit, then it might be your best bet.Turn off
E_NOTICE
on the file or folder level. This might be preferable if you have some legacy code but want to do things the "right" way otherwise. To do this, you should consult Apache2, nginx, or whatever your server of choice is. In Apache, you would usephp_value
inside of<Directory>
.Rewrite your code to be cleaner. If you need to do this while moving to a production environment, or don't want someone to see your errors, make sure you are disabling any display of errors, and only logging your errors (see
display_errors
andlog_errors
in php.ini and your server settings).
To expand on option 3: This is the ideal. If you can go this route, you should. If you are not going this route initially, consider moving this route eventually by testing your code in a development environment. While you're at it, get rid of ~E_STRICT
and ~E_DEPRECATED
to see what might go wrong in the future. You're going to see a LOT of unfamiliar errors, but it's going to stop you from having any unpleasant problems when you need to upgrade PHP in the future.
WHAT DO THE ERRORS MEAN?
Undefined variable: my_variable_name
- This occurs when a variable has not been defined before use. When the PHP script is executed, it internally just assumes a null value. However, in which scenario would you need to check a variable before it was defined? Ultimately, this is an argument for "sloppy code". As a developer, I can tell you that I love it when I see an open source project where variables are defined as high up in their scopes as they can be defined. It makes it easier to tell what variables are going to pop up in the future, and makes it easier to read/learn the code.
function foo()
{
$my_variable_name = '';
//....
if ($my_variable_name) {
// perform some logic
}
}
Undefined index: my_index
- This occurs when you try to access a value in an array and it does not exist. To prevent this error, perform a conditional check.
// verbose way - generally better
if (isset($my_array['my_index'])) {
echo "My index value is: " . $my_array['my_index'];
}
// non-verbose ternary example - I use this sometimes for small rules.
$my_index_val = isset($my_array['my_index'])?$my_array['my_index']:'(undefined)';
echo "My index value is: " . $my_index_val;
Another option is to declare an empty array at the top of your function. This is not always possible.
$my_array = array(
'my_index' => ''
);
//...
$my_array['my_index'] = 'new string';
(additional tip)
- When I was encountering these and other issues, I used NetBeans IDE (free) and it gave me a host of warnings and notices. Some of them offer very helpful tips. This is not a requirement, and I don't use IDEs anymore except for large projects. I'm more of a
vim
person these days :).
add a comment |
undefined index means in an array you requested for unavailable array index for example
<?php
$newArray = {1,2,3,4,5};
print_r($newArray[5]);
?>
undefined variable means you have used completely not existing variable or which is not defined or initialized by that name for example
<?php print_r($myvar); ?>
undefined offset means in array you have asked for non existing key. And the
solution for this is to check before use
php> echo array_key_exists(1, $myarray);
add a comment |
Regarding this part of the question:
Why do they appear all of a sudden? I used to use this script for years and I've never had any problem.
No definite answers but here are a some possible explanations of why settings can 'suddenly' change:
You have upgraded PHP to a newer version which can have other defaults for error_reporting, display_errors or other relevant settings.
You have removed or introduced some code (possibly in a dependency) that sets relevant settings at runtime using
ini_set()
orerror_reporting()
(search for these in the code)You changed the webserver configuration (assuming apache here):
.htaccess
files and vhost configurations can also manipulate php settings.Usually notices don't get displayed / reported (see PHP manual)
so it is possible that when setting up the server, the php.ini file could not be loaded for some reason (file permissions??) and you were on the default settings. Later on, the 'bug' has been solved (by accident) and now it CAN load the correct php.ini file with the error_reporting set to show notices.
add a comment |
If working with classes you need to make sure you reference member variables using $this
:
class Person
{
protected $firstName;
protected $lastName;
public function setFullName($first, $last)
{
// Correct
$this->firstName = $first;
// Incorrect
$lastName = $last;
// Incorrect
$this->$lastName = $last;
}
}
add a comment |
Another reason why an undefined index notice will be thrown, would be that a column was omitted from a database query.
I.e.:
$query = "SELECT col1 FROM table WHERE col_x = ?";
Then trying to access more columns/rows inside a loop.
I.e.:
print_r($row['col1']);
print_r($row['col2']); // undefined index thrown
or in a while
loop:
while( $row = fetching_function($query) ) {
echo $row['col1'];
echo "<br>";
echo $row['col2']; // undefined index thrown
echo "<br>";
echo $row['col3']; // undefined index thrown
}
Something else that needs to be noted is that on a *NIX OS and Mac OS X, things are case-sensitive.
Consult the followning Q&A's on Stack:
Are table names in MySQL case sensitive?
mysql case sensitive table names in queries
MySql - Case Sensitive issue of tables in different server
I have to wonder why this was downvoted also.
– Funk Forty Niner
Oct 25 '18 at 0:26
add a comment |
Using a ternary is simple, readable, and clean:
Pre PHP 7
Assign a variable to the value of another variable if it's set, else assign null
(or whatever default value you need):
$newVariable = isset($thePotentialData) ? $thePotentialData : null;
PHP 7+
The same except using Null Coalescing Operator. There's no longer a need to call isset()
as this is built in, and no need to provide the variable to return as it's assumed to return the value of the variable being checked:
$newVariable = $thePotentialData ?? null;
Both will stop the Notices from the OP question, and both are the exact equivalent of:
if (isset($thePotentialData)) {
$newVariable = $thePotentialData;
} else {
$newVariable = null;
}
If you don't require setting a new variable then you can directly use the ternary's returned value, such as with echo
, function arguments, etc:
Echo:
echo 'Your name is: ' . isset($name) ? $name : 'You did not provide one';
Function:
$foreName = getForeName(isset($userId) ? $userId : null);
function getForeName($userId)
{
if ($userId === null) {
// Etc
}
}
The above will work just the same with arrays, including sessions etc, replacing the variable being checked with e.g.:$_SESSION['checkMe']
or however many levels deep you need, e.g.:$clients['personal']['address']['postcode']
Suppression:
It is possible to suppress the PHP Notices with @
or reduce your error reporting level, but it does not fix the problem, it simply stops it being reported in the error log. This means that your code still tried to use a variable that was not set, which may or may not mean something doesn't work as intended - depending on how crucial the missing value is.
You should really be checking for this issue and handling it appropriately, either serving a different message, or even just returning a null value for everything else to identify the precise state.
If you just care about the Notice not being in the error log, then as an option you could simply ignore the error log.
add a comment |
Probably you were using old PHP version until and now upgraded PHP thats the reason it was working without any error till now from years.
until PHP4 there was no error if you are using variable without defining it but as of PHP5 onwards it throws errors for codes like mentioned in question.
add a comment |
When dealing with files, a proper enctype and a POST method are required, which will trigger an undefined index notice if either are not included in the form.
The manual states the following basic syntax:
HTML
<!-- The data encoding type, enctype, MUST be specified as below -->
<form enctype="multipart/form-data" action="__URL__" method="POST">
<!-- MAX_FILE_SIZE must precede the file input field -->
<input type="hidden" name="MAX_FILE_SIZE" value="30000" />
<!-- Name of input element determines name in $_FILES array -->
Send this file: <input name="userfile" type="file" />
<input type="submit" value="Send File" />
</form>
PHP
<?php
// In PHP versions earlier than 4.1.0, $HTTP_POST_FILES should be used instead
// of $_FILES.
$uploaddir = '/var/www/uploads/';
$uploadfile = $uploaddir . basename($_FILES['userfile']['name']);
echo '<pre>';
if (move_uploaded_file($_FILES['userfile']['tmp_name'], $uploadfile)) {
echo "File is valid, and was successfully uploaded.n";
} else {
echo "Possible file upload attack!n";
}
echo 'Here is some more debugging info:';
print_r($_FILES);
print "</pre>";
?>
Reference:
- http://php.net/manual/en/features.file-upload.post-method.php
add a comment |
One common cause of a variable not existing after an HTML form has been submitted is the form element is not contained within a <form>
tag:
Example: Element not contained within the <form>
<form action="example.php" method="post">
<p>
<input type="text" name="name" />
<input type="submit" value="Submit" />
</p>
</form>
<select name="choice">
<option value="choice1">choice 1</option>
<option value="choice2">choice 2</option>
<option value="choice3">choice 3</option>
<option value="choice4">choice 4</option>
</select>
Example: Element now contained within the <form>
<form action="example.php" method="post">
<select name="choice">
<option value="choice1">choice 1</option>
<option value="choice2">choice 2</option>
<option value="choice3">choice 3</option>
<option value="choice4">choice 4</option>
</select>
<p>
<input type="text" name="name" />
<input type="submit" value="Submit" />
</p>
</form>
add a comment |
I asked a question about this and I was referred to this post with the message:
This question already has an answer here:
“Notice: Undefined variable”, “Notice: Undefined index”, and “Notice:
Undefined offset” using PHP
I am sharing my question and solution here:
This is the error:
Line 154 is the problem. This is what I have in line 154:
153 foreach($cities as $key => $city){
154 if(($city != 'London') && ($city != 'Madrid') && ($citiesCounterArray[$key] >= 1)){
I think the problem is that I am writing if conditions for the variable $city
, which is not the key but the value in $key => $city
. First, could you confirm if that is the cause of the warning? Second, if that is the problem, why is it that I cannot write a condition based on the value? Does it have to be with the key that I need to write the condition?
UPDATE 1: The problem is that when executing $citiesCounterArray[$key]
, sometimes the $key
corresponds to a key that does not exist in the $citiesCounterArray
array, but that is not always the case based on the data of my loop. What I need is to set a condition so that if $key
exists in the array, then run the code, otherwise, skip it.
UPDATE 2: This is how I fixed it by using array_key_exists()
:
foreach($cities as $key => $city){
if(array_key_exists($key, $citiesCounterArray)){
if(($city != 'London') && ($city != 'Madrid') && ($citiesCounterArray[$key] >= 1)){
add a comment |
Those notices are because you don't have the used variable defined
and my_index
key was not present into $my_array
variable.
Those notices were triggered every time, because your code
is not correct, but probably you didn't have the reporting of notices on.
Solve the bugs:
$my_variable_name = "Variable name"; // defining variable
echo "My variable value is: " . $my_variable_name;
if(isset($my_array["my_index"])){
echo "My index value is: " . $my_array["my_index"]; // check if my_index is set
}
Another way to get this out:
ini_set("error_reporting", false)
add a comment |
In PHP you need fist to define the variable after that you can use it.
We can check variable is defined or not in very efficient way!.
//If you only want to check variable has value and value has true and false value.
//But variable must be defined first.
if($my_variable_name){
}
//If you want to check variable is define or undefine
//Isset() does not check that variable has true or false value
//But it check null value of variable
if(isset($my_variable_name)){
}
Simple Explanation
//It will work with :- true,false,NULL
$defineVarialbe = false;
if($defineVarialbe){
echo "true";
}else{
echo "false";
}
//It will check variable is define or not and variable has null value.
if(isset($unDefineVarialbe)){
echo "true";
}else{
echo "false";
}
add a comment |
protected by Madara Uchiha♦ Dec 9 '12 at 12:00
Thank you for your interest in this question.
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Notice: Undefined variable
From the vast wisdom of the PHP Manual:
Relying on the default value of an uninitialized variable is problematic in the case of including one file into another which uses the same variable name. It is also a major security risk with register_globals turned on. E_NOTICE level error is issued in case of working with uninitialized variables, however not in the case of appending elements to the uninitialized array. isset() language construct can be used to detect if a variable has been already initialized. Additionally and more ideal is the solution of empty() since it does not generate a warning or error message if the variable is not initialized.
From PHP documentation:
No warning is generated if the variable does not exist. That means
empty() is essentially the concise equivalent to !isset($var) || $var
== false.
This means that you could use only empty()
to determine if the variable is set, and in addition it checks the variable against the following, 0,"",null
.
Example:
$o = ;
@$var = ["",0,null,1,2,3,$foo,$o['myIndex']];
array_walk($var, function($v) {
echo (!isset($v) || $v == false) ? 'true ' : 'false';
echo ' ' . (empty($v) ? 'true' : 'false');
echo "n";
});
Test the above snippet in the 3v4l.org online PHP editor
Although PHP does not require a variable declaration, it does recommend it in order to avoid some security vulnerabilities or bugs where one would forget to give a value to a variable that will be used later in the script. What PHP does in the case of undeclared variables is issue a very low level error, E_NOTICE
, one that is not even reported by default, but the Manual advises to allow during development.
Ways to deal with the issue:
Recommended: Declare your variables, for example when you try to append a string to an undefined variable. Or use
isset()
/!empty()
to check if they are declared before referencing them, as in:
//Initializing variable
$value = ""; //Initialization value; Examples
//"" When you want to append stuff later
//0 When you want to add numbers later
//isset()
$value = isset($_POST['value']) ? $_POST['value'] : '';
//empty()
$value = !empty($_POST['value']) ? $_POST['value'] : '';
This has become much cleaner as of PHP 7.0, now you can use the null coalesce operator:
// Null coalesce operator - No need to explicitly initialize the variable.
$value = $_POST['value'] ?? '';
Set a custom error handler for E_NOTICE and redirect the messages away from the standard output (maybe to a log file):
set_error_handler('myHandlerForMinorErrors', E_NOTICE | E_STRICT)
Disable E_NOTICE from reporting. A quick way to exclude just
E_NOTICE
is:
error_reporting( error_reporting() & ~E_NOTICE )
Suppress the error with the @ operator.
Note: It's strongly recommended to implement just point 1.
Notice: Undefined index / Undefined offset
This notice appears when you (or PHP) try to access an undefined index of an array.
Ways to deal with the issue:
Check if the index exists before you access it. For this you can use
isset()
orarray_key_exists()
:
//isset()
$value = isset($array['my_index']) ? $array['my_index'] : '';
//array_key_exists()
$value = array_key_exists('my_index', $array) ? $array['my_index'] : '';
The language construct
list()
may generate this when it attempts to access an array index that does not exist:
list($a, $b) = array(0 => 'a');
//or
list($one, $two) = explode(',', 'test string');
Two variables are used to access two array elements, however there is only one array element, index 0
, so this will generate:
Notice: Undefined offset: 1
$_POST
/ $_GET
/ $_SESSION
variable
The notices above appear often when working with $_POST
, $_GET
or $_SESSION
. For $_POST
and $_GET
you just have to check if the index exists or not before you use them. For $_SESSION
you have to make sure you have the session started with session_start()
and that the index also exists.
Also note that all 3 variables are superglobals and are uppercase.
Related:
- Notice: Undefined variable
Notice: Undefined Index
7
@dieselpower44 A couple of thoughts: The "shut-up operator" (@
) has some performance issues. Also, since it suppresses all errors within a particular scope, using it without care might mask messages you wish you'd seen.
– IMSoP
Oct 24 '13 at 20:00
6
Hiding the issues is NOT the way to deal with issues. Items #2...#4 can be used only on production servers, not in general.
– Salman A
Sep 13 '14 at 8:09
1
Is it possible to shut-up the message inline (not in handler) when also a custom error handler is used?$var = @$_GET['nonexisting'];
still causes notice..
– Alph.Dev
Oct 11 '14 at 14:14
18
Why is it recommended to use 1.$value = isset($_POST['value']) ? $_POST['value'] : '';
instead of using 4.$value = @$_POST['value'];
?
– forsvunnet
Feb 9 '15 at 13:38
@twistedpixel Those 4 ways are independent, that's not a 4-step guide. So if you've chosen to use way 4, that means you didn't implement first 3 ways, so you you didn't supress any errors yet.
– Aycan Yaşıt
Aug 6 '15 at 9:58
|
show 5 more comments
Notice: Undefined variable
From the vast wisdom of the PHP Manual:
Relying on the default value of an uninitialized variable is problematic in the case of including one file into another which uses the same variable name. It is also a major security risk with register_globals turned on. E_NOTICE level error is issued in case of working with uninitialized variables, however not in the case of appending elements to the uninitialized array. isset() language construct can be used to detect if a variable has been already initialized. Additionally and more ideal is the solution of empty() since it does not generate a warning or error message if the variable is not initialized.
From PHP documentation:
No warning is generated if the variable does not exist. That means
empty() is essentially the concise equivalent to !isset($var) || $var
== false.
This means that you could use only empty()
to determine if the variable is set, and in addition it checks the variable against the following, 0,"",null
.
Example:
$o = ;
@$var = ["",0,null,1,2,3,$foo,$o['myIndex']];
array_walk($var, function($v) {
echo (!isset($v) || $v == false) ? 'true ' : 'false';
echo ' ' . (empty($v) ? 'true' : 'false');
echo "n";
});
Test the above snippet in the 3v4l.org online PHP editor
Although PHP does not require a variable declaration, it does recommend it in order to avoid some security vulnerabilities or bugs where one would forget to give a value to a variable that will be used later in the script. What PHP does in the case of undeclared variables is issue a very low level error, E_NOTICE
, one that is not even reported by default, but the Manual advises to allow during development.
Ways to deal with the issue:
Recommended: Declare your variables, for example when you try to append a string to an undefined variable. Or use
isset()
/!empty()
to check if they are declared before referencing them, as in:
//Initializing variable
$value = ""; //Initialization value; Examples
//"" When you want to append stuff later
//0 When you want to add numbers later
//isset()
$value = isset($_POST['value']) ? $_POST['value'] : '';
//empty()
$value = !empty($_POST['value']) ? $_POST['value'] : '';
This has become much cleaner as of PHP 7.0, now you can use the null coalesce operator:
// Null coalesce operator - No need to explicitly initialize the variable.
$value = $_POST['value'] ?? '';
Set a custom error handler for E_NOTICE and redirect the messages away from the standard output (maybe to a log file):
set_error_handler('myHandlerForMinorErrors', E_NOTICE | E_STRICT)
Disable E_NOTICE from reporting. A quick way to exclude just
E_NOTICE
is:
error_reporting( error_reporting() & ~E_NOTICE )
Suppress the error with the @ operator.
Note: It's strongly recommended to implement just point 1.
Notice: Undefined index / Undefined offset
This notice appears when you (or PHP) try to access an undefined index of an array.
Ways to deal with the issue:
Check if the index exists before you access it. For this you can use
isset()
orarray_key_exists()
:
//isset()
$value = isset($array['my_index']) ? $array['my_index'] : '';
//array_key_exists()
$value = array_key_exists('my_index', $array) ? $array['my_index'] : '';
The language construct
list()
may generate this when it attempts to access an array index that does not exist:
list($a, $b) = array(0 => 'a');
//or
list($one, $two) = explode(',', 'test string');
Two variables are used to access two array elements, however there is only one array element, index 0
, so this will generate:
Notice: Undefined offset: 1
$_POST
/ $_GET
/ $_SESSION
variable
The notices above appear often when working with $_POST
, $_GET
or $_SESSION
. For $_POST
and $_GET
you just have to check if the index exists or not before you use them. For $_SESSION
you have to make sure you have the session started with session_start()
and that the index also exists.
Also note that all 3 variables are superglobals and are uppercase.
Related:
- Notice: Undefined variable
Notice: Undefined Index
7
@dieselpower44 A couple of thoughts: The "shut-up operator" (@
) has some performance issues. Also, since it suppresses all errors within a particular scope, using it without care might mask messages you wish you'd seen.
– IMSoP
Oct 24 '13 at 20:00
6
Hiding the issues is NOT the way to deal with issues. Items #2...#4 can be used only on production servers, not in general.
– Salman A
Sep 13 '14 at 8:09
1
Is it possible to shut-up the message inline (not in handler) when also a custom error handler is used?$var = @$_GET['nonexisting'];
still causes notice..
– Alph.Dev
Oct 11 '14 at 14:14
18
Why is it recommended to use 1.$value = isset($_POST['value']) ? $_POST['value'] : '';
instead of using 4.$value = @$_POST['value'];
?
– forsvunnet
Feb 9 '15 at 13:38
@twistedpixel Those 4 ways are independent, that's not a 4-step guide. So if you've chosen to use way 4, that means you didn't implement first 3 ways, so you you didn't supress any errors yet.
– Aycan Yaşıt
Aug 6 '15 at 9:58
|
show 5 more comments
Notice: Undefined variable
From the vast wisdom of the PHP Manual:
Relying on the default value of an uninitialized variable is problematic in the case of including one file into another which uses the same variable name. It is also a major security risk with register_globals turned on. E_NOTICE level error is issued in case of working with uninitialized variables, however not in the case of appending elements to the uninitialized array. isset() language construct can be used to detect if a variable has been already initialized. Additionally and more ideal is the solution of empty() since it does not generate a warning or error message if the variable is not initialized.
From PHP documentation:
No warning is generated if the variable does not exist. That means
empty() is essentially the concise equivalent to !isset($var) || $var
== false.
This means that you could use only empty()
to determine if the variable is set, and in addition it checks the variable against the following, 0,"",null
.
Example:
$o = ;
@$var = ["",0,null,1,2,3,$foo,$o['myIndex']];
array_walk($var, function($v) {
echo (!isset($v) || $v == false) ? 'true ' : 'false';
echo ' ' . (empty($v) ? 'true' : 'false');
echo "n";
});
Test the above snippet in the 3v4l.org online PHP editor
Although PHP does not require a variable declaration, it does recommend it in order to avoid some security vulnerabilities or bugs where one would forget to give a value to a variable that will be used later in the script. What PHP does in the case of undeclared variables is issue a very low level error, E_NOTICE
, one that is not even reported by default, but the Manual advises to allow during development.
Ways to deal with the issue:
Recommended: Declare your variables, for example when you try to append a string to an undefined variable. Or use
isset()
/!empty()
to check if they are declared before referencing them, as in:
//Initializing variable
$value = ""; //Initialization value; Examples
//"" When you want to append stuff later
//0 When you want to add numbers later
//isset()
$value = isset($_POST['value']) ? $_POST['value'] : '';
//empty()
$value = !empty($_POST['value']) ? $_POST['value'] : '';
This has become much cleaner as of PHP 7.0, now you can use the null coalesce operator:
// Null coalesce operator - No need to explicitly initialize the variable.
$value = $_POST['value'] ?? '';
Set a custom error handler for E_NOTICE and redirect the messages away from the standard output (maybe to a log file):
set_error_handler('myHandlerForMinorErrors', E_NOTICE | E_STRICT)
Disable E_NOTICE from reporting. A quick way to exclude just
E_NOTICE
is:
error_reporting( error_reporting() & ~E_NOTICE )
Suppress the error with the @ operator.
Note: It's strongly recommended to implement just point 1.
Notice: Undefined index / Undefined offset
This notice appears when you (or PHP) try to access an undefined index of an array.
Ways to deal with the issue:
Check if the index exists before you access it. For this you can use
isset()
orarray_key_exists()
:
//isset()
$value = isset($array['my_index']) ? $array['my_index'] : '';
//array_key_exists()
$value = array_key_exists('my_index', $array) ? $array['my_index'] : '';
The language construct
list()
may generate this when it attempts to access an array index that does not exist:
list($a, $b) = array(0 => 'a');
//or
list($one, $two) = explode(',', 'test string');
Two variables are used to access two array elements, however there is only one array element, index 0
, so this will generate:
Notice: Undefined offset: 1
$_POST
/ $_GET
/ $_SESSION
variable
The notices above appear often when working with $_POST
, $_GET
or $_SESSION
. For $_POST
and $_GET
you just have to check if the index exists or not before you use them. For $_SESSION
you have to make sure you have the session started with session_start()
and that the index also exists.
Also note that all 3 variables are superglobals and are uppercase.
Related:
- Notice: Undefined variable
Notice: Undefined Index
Notice: Undefined variable
From the vast wisdom of the PHP Manual:
Relying on the default value of an uninitialized variable is problematic in the case of including one file into another which uses the same variable name. It is also a major security risk with register_globals turned on. E_NOTICE level error is issued in case of working with uninitialized variables, however not in the case of appending elements to the uninitialized array. isset() language construct can be used to detect if a variable has been already initialized. Additionally and more ideal is the solution of empty() since it does not generate a warning or error message if the variable is not initialized.
From PHP documentation:
No warning is generated if the variable does not exist. That means
empty() is essentially the concise equivalent to !isset($var) || $var
== false.
This means that you could use only empty()
to determine if the variable is set, and in addition it checks the variable against the following, 0,"",null
.
Example:
$o = ;
@$var = ["",0,null,1,2,3,$foo,$o['myIndex']];
array_walk($var, function($v) {
echo (!isset($v) || $v == false) ? 'true ' : 'false';
echo ' ' . (empty($v) ? 'true' : 'false');
echo "n";
});
Test the above snippet in the 3v4l.org online PHP editor
Although PHP does not require a variable declaration, it does recommend it in order to avoid some security vulnerabilities or bugs where one would forget to give a value to a variable that will be used later in the script. What PHP does in the case of undeclared variables is issue a very low level error, E_NOTICE
, one that is not even reported by default, but the Manual advises to allow during development.
Ways to deal with the issue:
Recommended: Declare your variables, for example when you try to append a string to an undefined variable. Or use
isset()
/!empty()
to check if they are declared before referencing them, as in:
//Initializing variable
$value = ""; //Initialization value; Examples
//"" When you want to append stuff later
//0 When you want to add numbers later
//isset()
$value = isset($_POST['value']) ? $_POST['value'] : '';
//empty()
$value = !empty($_POST['value']) ? $_POST['value'] : '';
This has become much cleaner as of PHP 7.0, now you can use the null coalesce operator:
// Null coalesce operator - No need to explicitly initialize the variable.
$value = $_POST['value'] ?? '';
Set a custom error handler for E_NOTICE and redirect the messages away from the standard output (maybe to a log file):
set_error_handler('myHandlerForMinorErrors', E_NOTICE | E_STRICT)
Disable E_NOTICE from reporting. A quick way to exclude just
E_NOTICE
is:
error_reporting( error_reporting() & ~E_NOTICE )
Suppress the error with the @ operator.
Note: It's strongly recommended to implement just point 1.
Notice: Undefined index / Undefined offset
This notice appears when you (or PHP) try to access an undefined index of an array.
Ways to deal with the issue:
Check if the index exists before you access it. For this you can use
isset()
orarray_key_exists()
:
//isset()
$value = isset($array['my_index']) ? $array['my_index'] : '';
//array_key_exists()
$value = array_key_exists('my_index', $array) ? $array['my_index'] : '';
The language construct
list()
may generate this when it attempts to access an array index that does not exist:
list($a, $b) = array(0 => 'a');
//or
list($one, $two) = explode(',', 'test string');
Two variables are used to access two array elements, however there is only one array element, index 0
, so this will generate:
Notice: Undefined offset: 1
$_POST
/ $_GET
/ $_SESSION
variable
The notices above appear often when working with $_POST
, $_GET
or $_SESSION
. For $_POST
and $_GET
you just have to check if the index exists or not before you use them. For $_SESSION
you have to make sure you have the session started with session_start()
and that the index also exists.
Also note that all 3 variables are superglobals and are uppercase.
Related:
- Notice: Undefined variable
Notice: Undefined Index
edited May 25 '18 at 18:40
community wiki
23 revs, 11 users 27%
Rizier123
7
@dieselpower44 A couple of thoughts: The "shut-up operator" (@
) has some performance issues. Also, since it suppresses all errors within a particular scope, using it without care might mask messages you wish you'd seen.
– IMSoP
Oct 24 '13 at 20:00
6
Hiding the issues is NOT the way to deal with issues. Items #2...#4 can be used only on production servers, not in general.
– Salman A
Sep 13 '14 at 8:09
1
Is it possible to shut-up the message inline (not in handler) when also a custom error handler is used?$var = @$_GET['nonexisting'];
still causes notice..
– Alph.Dev
Oct 11 '14 at 14:14
18
Why is it recommended to use 1.$value = isset($_POST['value']) ? $_POST['value'] : '';
instead of using 4.$value = @$_POST['value'];
?
– forsvunnet
Feb 9 '15 at 13:38
@twistedpixel Those 4 ways are independent, that's not a 4-step guide. So if you've chosen to use way 4, that means you didn't implement first 3 ways, so you you didn't supress any errors yet.
– Aycan Yaşıt
Aug 6 '15 at 9:58
|
show 5 more comments
7
@dieselpower44 A couple of thoughts: The "shut-up operator" (@
) has some performance issues. Also, since it suppresses all errors within a particular scope, using it without care might mask messages you wish you'd seen.
– IMSoP
Oct 24 '13 at 20:00
6
Hiding the issues is NOT the way to deal with issues. Items #2...#4 can be used only on production servers, not in general.
– Salman A
Sep 13 '14 at 8:09
1
Is it possible to shut-up the message inline (not in handler) when also a custom error handler is used?$var = @$_GET['nonexisting'];
still causes notice..
– Alph.Dev
Oct 11 '14 at 14:14
18
Why is it recommended to use 1.$value = isset($_POST['value']) ? $_POST['value'] : '';
instead of using 4.$value = @$_POST['value'];
?
– forsvunnet
Feb 9 '15 at 13:38
@twistedpixel Those 4 ways are independent, that's not a 4-step guide. So if you've chosen to use way 4, that means you didn't implement first 3 ways, so you you didn't supress any errors yet.
– Aycan Yaşıt
Aug 6 '15 at 9:58
7
7
@dieselpower44 A couple of thoughts: The "shut-up operator" (
@
) has some performance issues. Also, since it suppresses all errors within a particular scope, using it without care might mask messages you wish you'd seen.– IMSoP
Oct 24 '13 at 20:00
@dieselpower44 A couple of thoughts: The "shut-up operator" (
@
) has some performance issues. Also, since it suppresses all errors within a particular scope, using it without care might mask messages you wish you'd seen.– IMSoP
Oct 24 '13 at 20:00
6
6
Hiding the issues is NOT the way to deal with issues. Items #2...#4 can be used only on production servers, not in general.
– Salman A
Sep 13 '14 at 8:09
Hiding the issues is NOT the way to deal with issues. Items #2...#4 can be used only on production servers, not in general.
– Salman A
Sep 13 '14 at 8:09
1
1
Is it possible to shut-up the message inline (not in handler) when also a custom error handler is used?
$var = @$_GET['nonexisting'];
still causes notice..– Alph.Dev
Oct 11 '14 at 14:14
Is it possible to shut-up the message inline (not in handler) when also a custom error handler is used?
$var = @$_GET['nonexisting'];
still causes notice..– Alph.Dev
Oct 11 '14 at 14:14
18
18
Why is it recommended to use 1.
$value = isset($_POST['value']) ? $_POST['value'] : '';
instead of using 4. $value = @$_POST['value'];
?– forsvunnet
Feb 9 '15 at 13:38
Why is it recommended to use 1.
$value = isset($_POST['value']) ? $_POST['value'] : '';
instead of using 4. $value = @$_POST['value'];
?– forsvunnet
Feb 9 '15 at 13:38
@twistedpixel Those 4 ways are independent, that's not a 4-step guide. So if you've chosen to use way 4, that means you didn't implement first 3 ways, so you you didn't supress any errors yet.
– Aycan Yaşıt
Aug 6 '15 at 9:58
@twistedpixel Those 4 ways are independent, that's not a 4-step guide. So if you've chosen to use way 4, that means you didn't implement first 3 ways, so you you didn't supress any errors yet.
– Aycan Yaşıt
Aug 6 '15 at 9:58
|
show 5 more comments
Try these
Q1: this notice means $varname is not
defined at current scope of the
script.
Q2: Use of isset(), empty() conditions before using any suspicious variable works well.
// recommended solution for recent PHP versions
$user_name = $_SESSION['user_name'] ?? '';
// pre-7 PHP versions
$user_name = '';
if (!empty($_SESSION['user_name'])) {
$user_name = $_SESSION['user_name'];
}
Or, as a quick and dirty solution:
// not the best solution, but works
// in your php setting use, it helps hiding site wide notices
error_reporting(E_ALL ^ E_NOTICE);
Note about sessions:
When using sessions,
session_start();
is required to be placed inside all files using sessions.http://php.net/manual/en/features.sessions.php
If usingE_NOTICE
from thephp.ini
configuration file, doerror_reporting = (E_ALL & ~E_NOTICE)
– ahmd0
Oct 4 '14 at 5:39
1
This answer is being discussed on meta.
– Script47
Jun 25 '18 at 10:43
add a comment |
Try these
Q1: this notice means $varname is not
defined at current scope of the
script.
Q2: Use of isset(), empty() conditions before using any suspicious variable works well.
// recommended solution for recent PHP versions
$user_name = $_SESSION['user_name'] ?? '';
// pre-7 PHP versions
$user_name = '';
if (!empty($_SESSION['user_name'])) {
$user_name = $_SESSION['user_name'];
}
Or, as a quick and dirty solution:
// not the best solution, but works
// in your php setting use, it helps hiding site wide notices
error_reporting(E_ALL ^ E_NOTICE);
Note about sessions:
When using sessions,
session_start();
is required to be placed inside all files using sessions.http://php.net/manual/en/features.sessions.php
If usingE_NOTICE
from thephp.ini
configuration file, doerror_reporting = (E_ALL & ~E_NOTICE)
– ahmd0
Oct 4 '14 at 5:39
1
This answer is being discussed on meta.
– Script47
Jun 25 '18 at 10:43
add a comment |
Try these
Q1: this notice means $varname is not
defined at current scope of the
script.
Q2: Use of isset(), empty() conditions before using any suspicious variable works well.
// recommended solution for recent PHP versions
$user_name = $_SESSION['user_name'] ?? '';
// pre-7 PHP versions
$user_name = '';
if (!empty($_SESSION['user_name'])) {
$user_name = $_SESSION['user_name'];
}
Or, as a quick and dirty solution:
// not the best solution, but works
// in your php setting use, it helps hiding site wide notices
error_reporting(E_ALL ^ E_NOTICE);
Note about sessions:
When using sessions,
session_start();
is required to be placed inside all files using sessions.http://php.net/manual/en/features.sessions.php
Try these
Q1: this notice means $varname is not
defined at current scope of the
script.
Q2: Use of isset(), empty() conditions before using any suspicious variable works well.
// recommended solution for recent PHP versions
$user_name = $_SESSION['user_name'] ?? '';
// pre-7 PHP versions
$user_name = '';
if (!empty($_SESSION['user_name'])) {
$user_name = $_SESSION['user_name'];
}
Or, as a quick and dirty solution:
// not the best solution, but works
// in your php setting use, it helps hiding site wide notices
error_reporting(E_ALL ^ E_NOTICE);
Note about sessions:
When using sessions,
session_start();
is required to be placed inside all files using sessions.http://php.net/manual/en/features.sessions.php
edited Jun 26 '18 at 14:58
community wiki
9 revs, 5 users 62%
Ish
If usingE_NOTICE
from thephp.ini
configuration file, doerror_reporting = (E_ALL & ~E_NOTICE)
– ahmd0
Oct 4 '14 at 5:39
1
This answer is being discussed on meta.
– Script47
Jun 25 '18 at 10:43
add a comment |
If usingE_NOTICE
from thephp.ini
configuration file, doerror_reporting = (E_ALL & ~E_NOTICE)
– ahmd0
Oct 4 '14 at 5:39
1
This answer is being discussed on meta.
– Script47
Jun 25 '18 at 10:43
If using
E_NOTICE
from the php.ini
configuration file, do error_reporting = (E_ALL & ~E_NOTICE)
– ahmd0
Oct 4 '14 at 5:39
If using
E_NOTICE
from the php.ini
configuration file, do error_reporting = (E_ALL & ~E_NOTICE)
– ahmd0
Oct 4 '14 at 5:39
1
1
This answer is being discussed on meta.
– Script47
Jun 25 '18 at 10:43
This answer is being discussed on meta.
– Script47
Jun 25 '18 at 10:43
add a comment |
Error display @
operator
For undesired and redundant notices, one could use the dedicated @
operator to »hide« undefined variable/index messages.
$var = @($_GET["optional_param"]);
- This is usually discouraged. Newcomers tend to way overuse it.
- It's very inappropriate for code deep within the application logic (ignoring undeclared variables where you shouldn't), e.g. for function parameters, or in loops.
- There's one upside over the
isset?:
or??
super-supression however. Notices still can get logged. And one may resurrect@
-hidden notices with:set_error_handler("var_dump");
- Additonally you shouldn't habitually use/recommend
if (isset($_POST["shubmit"]))
in your initial code. - Newcomers won't spot such typos. It just deprives you of PHPs Notices for those very cases. Add
@
orisset
only after verifying functionality.
Fix the cause first. Not the notices.
- Additonally you shouldn't habitually use/recommend
@
is mainly acceptable for$_GET
/$_POST
input parameters, specifically if they're optional.
And since this covers the majority of such questions, let's expand on the most common causes:
$_GET
/ $_POST
/ $_REQUEST
undefined input
First thing you do when encountering an undefined index/offset, is check for typos:
$count = $_GET["whatnow?"];
- Is this an expected key name and present on each page request?
- Variable names and array indicies are case-sensitive in PHP.
Secondly, if the notice doesn't have an obvious cause, use
var_dump
orprint_r
to verify all input arrays for their curent content:
var_dump($_GET);
var_dump($_POST);
//print_r($_REQUEST);
Both will reveal if your script was invoked with the right or any parameters at all.
Alternativey or additionally use your browser devtools (F12) and inspect the network tab for requests and parameters:
POST parameters and GET input will be be shown separately.
For
$_GET
parameters you can also peek at theQUERY_STRING
in
print_r($_SERVER);
PHP has some rules to coalesce non-standard parameter names into the superglobals. Apache might do some rewriting as well.
You can also look at supplied raw$_COOKIES
and other HTTP request headers that way.
More obviously look at your browser address bar for GET parameters:
http://example.org/script.php?id=5&sort=desc
The
name=value
pairs after the?
question mark are your query (GET) parameters. Thus this URL could only possibly yield$_GET["id"]
and$_GET["sort"]
.
Finally check your
<form>
and<input>
declarations, if you expect a parameter but receive none.
- Ensure each required input has an
<input name=FOO>
- The
id=
ortitle=
attribute does not suffice. - A
method=POST
form ought to populate$_POST
. - Whereas a
method=GET
(or leaving it out) would yield$_GET
variables. - It's also possible for a form to supply
action=script.php?get=param
via $_GET and the remainingmethod=POST
fields in $_POST alongside. - With modern PHP configurations (≥ 5.6) it has become feasible (not fashionable) to use
$_REQUEST['vars']
again, which mashes GET and POST params.
- Ensure each required input has an
If you are employing mod_rewrite, then you should check both the
access.log
as well as enable theRewriteLog
to figure out absent parameters.
$_FILES
- The same sanity checks apply to file uploads and
$_FILES["formname"]
. - Moreover check for
enctype=multipart/form-data
- As well as
method=POST
in your<form>
declaration. - See also: PHP Undefined index error $_FILES?
$_COOKIE
- The
$_COOKIE
array is never populated right aftersetcookie()
, but only on any followup HTTP request. - Additionally their validity times out, they could be constraint to subdomains or individual paths, and user and browser can just reject or delete them.
2
If you are curious what is the performance impact, this article summarises it well, derickrethans.nl/….
– Gajus
Feb 11 '14 at 12:24
@GajusKuizinas There have been quoite a few changes since 2009, in particular php.net/ChangeLog-5.php#5.4.0 changes the outcome drastically (see "Zend Engine, performance" and "(silence) operator").
– mario
Feb 11 '14 at 12:37
Thanks @mario, interesting. Now, if someone was good enough to benchmark the two... 3v4l.org/CYVOn/perf#tabs 3v4l.org/FLp3D/perf#tabs According to this test, seem to be identical (notice that scale changes).
– Gajus
Feb 11 '14 at 16:30
I tested with PHP 5.4 and the performance still bad.
– Brynner Ferreira
Oct 3 '14 at 17:43
add a comment |
Error display @
operator
For undesired and redundant notices, one could use the dedicated @
operator to »hide« undefined variable/index messages.
$var = @($_GET["optional_param"]);
- This is usually discouraged. Newcomers tend to way overuse it.
- It's very inappropriate for code deep within the application logic (ignoring undeclared variables where you shouldn't), e.g. for function parameters, or in loops.
- There's one upside over the
isset?:
or??
super-supression however. Notices still can get logged. And one may resurrect@
-hidden notices with:set_error_handler("var_dump");
- Additonally you shouldn't habitually use/recommend
if (isset($_POST["shubmit"]))
in your initial code. - Newcomers won't spot such typos. It just deprives you of PHPs Notices for those very cases. Add
@
orisset
only after verifying functionality.
Fix the cause first. Not the notices.
- Additonally you shouldn't habitually use/recommend
@
is mainly acceptable for$_GET
/$_POST
input parameters, specifically if they're optional.
And since this covers the majority of such questions, let's expand on the most common causes:
$_GET
/ $_POST
/ $_REQUEST
undefined input
First thing you do when encountering an undefined index/offset, is check for typos:
$count = $_GET["whatnow?"];
- Is this an expected key name and present on each page request?
- Variable names and array indicies are case-sensitive in PHP.
Secondly, if the notice doesn't have an obvious cause, use
var_dump
orprint_r
to verify all input arrays for their curent content:
var_dump($_GET);
var_dump($_POST);
//print_r($_REQUEST);
Both will reveal if your script was invoked with the right or any parameters at all.
Alternativey or additionally use your browser devtools (F12) and inspect the network tab for requests and parameters:
POST parameters and GET input will be be shown separately.
For
$_GET
parameters you can also peek at theQUERY_STRING
in
print_r($_SERVER);
PHP has some rules to coalesce non-standard parameter names into the superglobals. Apache might do some rewriting as well.
You can also look at supplied raw$_COOKIES
and other HTTP request headers that way.
More obviously look at your browser address bar for GET parameters:
http://example.org/script.php?id=5&sort=desc
The
name=value
pairs after the?
question mark are your query (GET) parameters. Thus this URL could only possibly yield$_GET["id"]
and$_GET["sort"]
.
Finally check your
<form>
and<input>
declarations, if you expect a parameter but receive none.
- Ensure each required input has an
<input name=FOO>
- The
id=
ortitle=
attribute does not suffice. - A
method=POST
form ought to populate$_POST
. - Whereas a
method=GET
(or leaving it out) would yield$_GET
variables. - It's also possible for a form to supply
action=script.php?get=param
via $_GET and the remainingmethod=POST
fields in $_POST alongside. - With modern PHP configurations (≥ 5.6) it has become feasible (not fashionable) to use
$_REQUEST['vars']
again, which mashes GET and POST params.
- Ensure each required input has an
If you are employing mod_rewrite, then you should check both the
access.log
as well as enable theRewriteLog
to figure out absent parameters.
$_FILES
- The same sanity checks apply to file uploads and
$_FILES["formname"]
. - Moreover check for
enctype=multipart/form-data
- As well as
method=POST
in your<form>
declaration. - See also: PHP Undefined index error $_FILES?
$_COOKIE
- The
$_COOKIE
array is never populated right aftersetcookie()
, but only on any followup HTTP request. - Additionally their validity times out, they could be constraint to subdomains or individual paths, and user and browser can just reject or delete them.
2
If you are curious what is the performance impact, this article summarises it well, derickrethans.nl/….
– Gajus
Feb 11 '14 at 12:24
@GajusKuizinas There have been quoite a few changes since 2009, in particular php.net/ChangeLog-5.php#5.4.0 changes the outcome drastically (see "Zend Engine, performance" and "(silence) operator").
– mario
Feb 11 '14 at 12:37
Thanks @mario, interesting. Now, if someone was good enough to benchmark the two... 3v4l.org/CYVOn/perf#tabs 3v4l.org/FLp3D/perf#tabs According to this test, seem to be identical (notice that scale changes).
– Gajus
Feb 11 '14 at 16:30
I tested with PHP 5.4 and the performance still bad.
– Brynner Ferreira
Oct 3 '14 at 17:43
add a comment |
Error display @
operator
For undesired and redundant notices, one could use the dedicated @
operator to »hide« undefined variable/index messages.
$var = @($_GET["optional_param"]);
- This is usually discouraged. Newcomers tend to way overuse it.
- It's very inappropriate for code deep within the application logic (ignoring undeclared variables where you shouldn't), e.g. for function parameters, or in loops.
- There's one upside over the
isset?:
or??
super-supression however. Notices still can get logged. And one may resurrect@
-hidden notices with:set_error_handler("var_dump");
- Additonally you shouldn't habitually use/recommend
if (isset($_POST["shubmit"]))
in your initial code. - Newcomers won't spot such typos. It just deprives you of PHPs Notices for those very cases. Add
@
orisset
only after verifying functionality.
Fix the cause first. Not the notices.
- Additonally you shouldn't habitually use/recommend
@
is mainly acceptable for$_GET
/$_POST
input parameters, specifically if they're optional.
And since this covers the majority of such questions, let's expand on the most common causes:
$_GET
/ $_POST
/ $_REQUEST
undefined input
First thing you do when encountering an undefined index/offset, is check for typos:
$count = $_GET["whatnow?"];
- Is this an expected key name and present on each page request?
- Variable names and array indicies are case-sensitive in PHP.
Secondly, if the notice doesn't have an obvious cause, use
var_dump
orprint_r
to verify all input arrays for their curent content:
var_dump($_GET);
var_dump($_POST);
//print_r($_REQUEST);
Both will reveal if your script was invoked with the right or any parameters at all.
Alternativey or additionally use your browser devtools (F12) and inspect the network tab for requests and parameters:
POST parameters and GET input will be be shown separately.
For
$_GET
parameters you can also peek at theQUERY_STRING
in
print_r($_SERVER);
PHP has some rules to coalesce non-standard parameter names into the superglobals. Apache might do some rewriting as well.
You can also look at supplied raw$_COOKIES
and other HTTP request headers that way.
More obviously look at your browser address bar for GET parameters:
http://example.org/script.php?id=5&sort=desc
The
name=value
pairs after the?
question mark are your query (GET) parameters. Thus this URL could only possibly yield$_GET["id"]
and$_GET["sort"]
.
Finally check your
<form>
and<input>
declarations, if you expect a parameter but receive none.
- Ensure each required input has an
<input name=FOO>
- The
id=
ortitle=
attribute does not suffice. - A
method=POST
form ought to populate$_POST
. - Whereas a
method=GET
(or leaving it out) would yield$_GET
variables. - It's also possible for a form to supply
action=script.php?get=param
via $_GET and the remainingmethod=POST
fields in $_POST alongside. - With modern PHP configurations (≥ 5.6) it has become feasible (not fashionable) to use
$_REQUEST['vars']
again, which mashes GET and POST params.
- Ensure each required input has an
If you are employing mod_rewrite, then you should check both the
access.log
as well as enable theRewriteLog
to figure out absent parameters.
$_FILES
- The same sanity checks apply to file uploads and
$_FILES["formname"]
. - Moreover check for
enctype=multipart/form-data
- As well as
method=POST
in your<form>
declaration. - See also: PHP Undefined index error $_FILES?
$_COOKIE
- The
$_COOKIE
array is never populated right aftersetcookie()
, but only on any followup HTTP request. - Additionally their validity times out, they could be constraint to subdomains or individual paths, and user and browser can just reject or delete them.
Error display @
operator
For undesired and redundant notices, one could use the dedicated @
operator to »hide« undefined variable/index messages.
$var = @($_GET["optional_param"]);
- This is usually discouraged. Newcomers tend to way overuse it.
- It's very inappropriate for code deep within the application logic (ignoring undeclared variables where you shouldn't), e.g. for function parameters, or in loops.
- There's one upside over the
isset?:
or??
super-supression however. Notices still can get logged. And one may resurrect@
-hidden notices with:set_error_handler("var_dump");
- Additonally you shouldn't habitually use/recommend
if (isset($_POST["shubmit"]))
in your initial code. - Newcomers won't spot such typos. It just deprives you of PHPs Notices for those very cases. Add
@
orisset
only after verifying functionality.
Fix the cause first. Not the notices.
- Additonally you shouldn't habitually use/recommend
@
is mainly acceptable for$_GET
/$_POST
input parameters, specifically if they're optional.
And since this covers the majority of such questions, let's expand on the most common causes:
$_GET
/ $_POST
/ $_REQUEST
undefined input
First thing you do when encountering an undefined index/offset, is check for typos:
$count = $_GET["whatnow?"];
- Is this an expected key name and present on each page request?
- Variable names and array indicies are case-sensitive in PHP.
Secondly, if the notice doesn't have an obvious cause, use
var_dump
orprint_r
to verify all input arrays for their curent content:
var_dump($_GET);
var_dump($_POST);
//print_r($_REQUEST);
Both will reveal if your script was invoked with the right or any parameters at all.
Alternativey or additionally use your browser devtools (F12) and inspect the network tab for requests and parameters:
POST parameters and GET input will be be shown separately.
For
$_GET
parameters you can also peek at theQUERY_STRING
in
print_r($_SERVER);
PHP has some rules to coalesce non-standard parameter names into the superglobals. Apache might do some rewriting as well.
You can also look at supplied raw$_COOKIES
and other HTTP request headers that way.
More obviously look at your browser address bar for GET parameters:
http://example.org/script.php?id=5&sort=desc
The
name=value
pairs after the?
question mark are your query (GET) parameters. Thus this URL could only possibly yield$_GET["id"]
and$_GET["sort"]
.
Finally check your
<form>
and<input>
declarations, if you expect a parameter but receive none.
- Ensure each required input has an
<input name=FOO>
- The
id=
ortitle=
attribute does not suffice. - A
method=POST
form ought to populate$_POST
. - Whereas a
method=GET
(or leaving it out) would yield$_GET
variables. - It's also possible for a form to supply
action=script.php?get=param
via $_GET and the remainingmethod=POST
fields in $_POST alongside. - With modern PHP configurations (≥ 5.6) it has become feasible (not fashionable) to use
$_REQUEST['vars']
again, which mashes GET and POST params.
- Ensure each required input has an
If you are employing mod_rewrite, then you should check both the
access.log
as well as enable theRewriteLog
to figure out absent parameters.
$_FILES
- The same sanity checks apply to file uploads and
$_FILES["formname"]
. - Moreover check for
enctype=multipart/form-data
- As well as
method=POST
in your<form>
declaration. - See also: PHP Undefined index error $_FILES?
$_COOKIE
- The
$_COOKIE
array is never populated right aftersetcookie()
, but only on any followup HTTP request. - Additionally their validity times out, they could be constraint to subdomains or individual paths, and user and browser can just reject or delete them.
edited Aug 22 '18 at 21:16
community wiki
3 revs
mario
2
If you are curious what is the performance impact, this article summarises it well, derickrethans.nl/….
– Gajus
Feb 11 '14 at 12:24
@GajusKuizinas There have been quoite a few changes since 2009, in particular php.net/ChangeLog-5.php#5.4.0 changes the outcome drastically (see "Zend Engine, performance" and "(silence) operator").
– mario
Feb 11 '14 at 12:37
Thanks @mario, interesting. Now, if someone was good enough to benchmark the two... 3v4l.org/CYVOn/perf#tabs 3v4l.org/FLp3D/perf#tabs According to this test, seem to be identical (notice that scale changes).
– Gajus
Feb 11 '14 at 16:30
I tested with PHP 5.4 and the performance still bad.
– Brynner Ferreira
Oct 3 '14 at 17:43
add a comment |
2
If you are curious what is the performance impact, this article summarises it well, derickrethans.nl/….
– Gajus
Feb 11 '14 at 12:24
@GajusKuizinas There have been quoite a few changes since 2009, in particular php.net/ChangeLog-5.php#5.4.0 changes the outcome drastically (see "Zend Engine, performance" and "(silence) operator").
– mario
Feb 11 '14 at 12:37
Thanks @mario, interesting. Now, if someone was good enough to benchmark the two... 3v4l.org/CYVOn/perf#tabs 3v4l.org/FLp3D/perf#tabs According to this test, seem to be identical (notice that scale changes).
– Gajus
Feb 11 '14 at 16:30
I tested with PHP 5.4 and the performance still bad.
– Brynner Ferreira
Oct 3 '14 at 17:43
2
2
If you are curious what is the performance impact, this article summarises it well, derickrethans.nl/….
– Gajus
Feb 11 '14 at 12:24
If you are curious what is the performance impact, this article summarises it well, derickrethans.nl/….
– Gajus
Feb 11 '14 at 12:24
@GajusKuizinas There have been quoite a few changes since 2009, in particular php.net/ChangeLog-5.php#5.4.0 changes the outcome drastically (see "Zend Engine, performance" and "(silence) operator").
– mario
Feb 11 '14 at 12:37
@GajusKuizinas There have been quoite a few changes since 2009, in particular php.net/ChangeLog-5.php#5.4.0 changes the outcome drastically (see "Zend Engine, performance" and "(silence) operator").
– mario
Feb 11 '14 at 12:37
Thanks @mario, interesting. Now, if someone was good enough to benchmark the two... 3v4l.org/CYVOn/perf#tabs 3v4l.org/FLp3D/perf#tabs According to this test, seem to be identical (notice that scale changes).
– Gajus
Feb 11 '14 at 16:30
Thanks @mario, interesting. Now, if someone was good enough to benchmark the two... 3v4l.org/CYVOn/perf#tabs 3v4l.org/FLp3D/perf#tabs According to this test, seem to be identical (notice that scale changes).
– Gajus
Feb 11 '14 at 16:30
I tested with PHP 5.4 and the performance still bad.
– Brynner Ferreira
Oct 3 '14 at 17:43
I tested with PHP 5.4 and the performance still bad.
– Brynner Ferreira
Oct 3 '14 at 17:43
add a comment |
Generally because of "bad programming", and a possibility for mistakes now or later.
- If it's a mistake, make a proper assignment to the variable first: $varname=0;
- If it really is only defined sometimes, test for it:
if (isset($varname))
, before using it - If it's because you spelled it wrong, just correct that
- Maybe even turn of the warnings in you PHP-settings
4
Please don't turn warnings off. In stricter languages, they often mean "there might be a bug, you better check this line twice" - in a language as permissive as PHP, they often means "this code is crap and propably riddled with bugs; I'll try to make some sense of it but you better fix this ASAP".
– user395760
Nov 23 '10 at 21:37
4
Although I agree with the first three points, #4 is simply wrong. Hiding a problem won't make it go away, and it might even cause more problems down the road.
– Valentin Flachsel
Nov 23 '10 at 21:38
1
@Freek absolutely true, but in some scenarios (Bought script, zero technical knowledge, need it running by tomorrow...) it's the duct-tape solution - really bad, that always needs emphasizing, but an option
– Pekka 웃
Nov 23 '10 at 21:40
Duct-tape is good ... sometimes. Historically warnings have been turned of in standard PHP-settings, but defult settings have become more strict. Too bad many go back to the old settings, so as not to annoy the customers.
– Erik
Nov 23 '10 at 22:26
add a comment |
Generally because of "bad programming", and a possibility for mistakes now or later.
- If it's a mistake, make a proper assignment to the variable first: $varname=0;
- If it really is only defined sometimes, test for it:
if (isset($varname))
, before using it - If it's because you spelled it wrong, just correct that
- Maybe even turn of the warnings in you PHP-settings
4
Please don't turn warnings off. In stricter languages, they often mean "there might be a bug, you better check this line twice" - in a language as permissive as PHP, they often means "this code is crap and propably riddled with bugs; I'll try to make some sense of it but you better fix this ASAP".
– user395760
Nov 23 '10 at 21:37
4
Although I agree with the first three points, #4 is simply wrong. Hiding a problem won't make it go away, and it might even cause more problems down the road.
– Valentin Flachsel
Nov 23 '10 at 21:38
1
@Freek absolutely true, but in some scenarios (Bought script, zero technical knowledge, need it running by tomorrow...) it's the duct-tape solution - really bad, that always needs emphasizing, but an option
– Pekka 웃
Nov 23 '10 at 21:40
Duct-tape is good ... sometimes. Historically warnings have been turned of in standard PHP-settings, but defult settings have become more strict. Too bad many go back to the old settings, so as not to annoy the customers.
– Erik
Nov 23 '10 at 22:26
add a comment |
Generally because of "bad programming", and a possibility for mistakes now or later.
- If it's a mistake, make a proper assignment to the variable first: $varname=0;
- If it really is only defined sometimes, test for it:
if (isset($varname))
, before using it - If it's because you spelled it wrong, just correct that
- Maybe even turn of the warnings in you PHP-settings
Generally because of "bad programming", and a possibility for mistakes now or later.
- If it's a mistake, make a proper assignment to the variable first: $varname=0;
- If it really is only defined sometimes, test for it:
if (isset($varname))
, before using it - If it's because you spelled it wrong, just correct that
- Maybe even turn of the warnings in you PHP-settings
edited Apr 11 '18 at 8:46
community wiki
2 revs, 2 users 83%
Erik
4
Please don't turn warnings off. In stricter languages, they often mean "there might be a bug, you better check this line twice" - in a language as permissive as PHP, they often means "this code is crap and propably riddled with bugs; I'll try to make some sense of it but you better fix this ASAP".
– user395760
Nov 23 '10 at 21:37
4
Although I agree with the first three points, #4 is simply wrong. Hiding a problem won't make it go away, and it might even cause more problems down the road.
– Valentin Flachsel
Nov 23 '10 at 21:38
1
@Freek absolutely true, but in some scenarios (Bought script, zero technical knowledge, need it running by tomorrow...) it's the duct-tape solution - really bad, that always needs emphasizing, but an option
– Pekka 웃
Nov 23 '10 at 21:40
Duct-tape is good ... sometimes. Historically warnings have been turned of in standard PHP-settings, but defult settings have become more strict. Too bad many go back to the old settings, so as not to annoy the customers.
– Erik
Nov 23 '10 at 22:26
add a comment |
4
Please don't turn warnings off. In stricter languages, they often mean "there might be a bug, you better check this line twice" - in a language as permissive as PHP, they often means "this code is crap and propably riddled with bugs; I'll try to make some sense of it but you better fix this ASAP".
– user395760
Nov 23 '10 at 21:37
4
Although I agree with the first three points, #4 is simply wrong. Hiding a problem won't make it go away, and it might even cause more problems down the road.
– Valentin Flachsel
Nov 23 '10 at 21:38
1
@Freek absolutely true, but in some scenarios (Bought script, zero technical knowledge, need it running by tomorrow...) it's the duct-tape solution - really bad, that always needs emphasizing, but an option
– Pekka 웃
Nov 23 '10 at 21:40
Duct-tape is good ... sometimes. Historically warnings have been turned of in standard PHP-settings, but defult settings have become more strict. Too bad many go back to the old settings, so as not to annoy the customers.
– Erik
Nov 23 '10 at 22:26
4
4
Please don't turn warnings off. In stricter languages, they often mean "there might be a bug, you better check this line twice" - in a language as permissive as PHP, they often means "this code is crap and propably riddled with bugs; I'll try to make some sense of it but you better fix this ASAP".
– user395760
Nov 23 '10 at 21:37
Please don't turn warnings off. In stricter languages, they often mean "there might be a bug, you better check this line twice" - in a language as permissive as PHP, they often means "this code is crap and propably riddled with bugs; I'll try to make some sense of it but you better fix this ASAP".
– user395760
Nov 23 '10 at 21:37
4
4
Although I agree with the first three points, #4 is simply wrong. Hiding a problem won't make it go away, and it might even cause more problems down the road.
– Valentin Flachsel
Nov 23 '10 at 21:38
Although I agree with the first three points, #4 is simply wrong. Hiding a problem won't make it go away, and it might even cause more problems down the road.
– Valentin Flachsel
Nov 23 '10 at 21:38
1
1
@Freek absolutely true, but in some scenarios (Bought script, zero technical knowledge, need it running by tomorrow...) it's the duct-tape solution - really bad, that always needs emphasizing, but an option
– Pekka 웃
Nov 23 '10 at 21:40
@Freek absolutely true, but in some scenarios (Bought script, zero technical knowledge, need it running by tomorrow...) it's the duct-tape solution - really bad, that always needs emphasizing, but an option
– Pekka 웃
Nov 23 '10 at 21:40
Duct-tape is good ... sometimes. Historically warnings have been turned of in standard PHP-settings, but defult settings have become more strict. Too bad many go back to the old settings, so as not to annoy the customers.
– Erik
Nov 23 '10 at 22:26
Duct-tape is good ... sometimes. Historically warnings have been turned of in standard PHP-settings, but defult settings have become more strict. Too bad many go back to the old settings, so as not to annoy the customers.
– Erik
Nov 23 '10 at 22:26
add a comment |
It means you are testing, evaluating, or printing a variable that you have not yet assigned anything to. It means you either have a typo, or you need to check that the variable was initialized to something first. Check your logic paths, it may be set in one path but not in another.
add a comment |
It means you are testing, evaluating, or printing a variable that you have not yet assigned anything to. It means you either have a typo, or you need to check that the variable was initialized to something first. Check your logic paths, it may be set in one path but not in another.
add a comment |
It means you are testing, evaluating, or printing a variable that you have not yet assigned anything to. It means you either have a typo, or you need to check that the variable was initialized to something first. Check your logic paths, it may be set in one path but not in another.
It means you are testing, evaluating, or printing a variable that you have not yet assigned anything to. It means you either have a typo, or you need to check that the variable was initialized to something first. Check your logic paths, it may be set in one path but not in another.
edited Nov 24 '10 at 2:43
community wiki
2 revs
DGM
add a comment |
add a comment |
I didn't want to disable notice because it's helpful, but wanted to avoid too much typing.
My solution was this function:
function ifexists($varname)
{
return(isset($$varname)?$varname:null);
}
So if I want to reference to $name and echo if exists, I simply write:
<?=ifexists('name')?>
For array elements:
function ifexistsidx($var,$index)
{
return(isset($var[$index])?$var[$index]:null);
}
In page if I want to refer to $_REQUEST['name']:
<?=ifexistsidx($_REQUEST,'name')?>
2
Your ifexists() function doesn't work for me in PHP 5.3. The caller's variables are not available in the function's local scope (see Variable scope), unless they are Superglobals or you fiddle with $GLOBALS, so$foo = "BABAR"; ifexists('foo');
will in general return null. (Italics are php.net chapters.)
– skierpage
Jun 19 '12 at 22:14
now you will get "hello from"... what's the point? just check the valueif( !empty($user) and !empty($location) ) echo "hello $user ..."
– gcb
Jul 9 '13 at 5:31
add a comment |
I didn't want to disable notice because it's helpful, but wanted to avoid too much typing.
My solution was this function:
function ifexists($varname)
{
return(isset($$varname)?$varname:null);
}
So if I want to reference to $name and echo if exists, I simply write:
<?=ifexists('name')?>
For array elements:
function ifexistsidx($var,$index)
{
return(isset($var[$index])?$var[$index]:null);
}
In page if I want to refer to $_REQUEST['name']:
<?=ifexistsidx($_REQUEST,'name')?>
2
Your ifexists() function doesn't work for me in PHP 5.3. The caller's variables are not available in the function's local scope (see Variable scope), unless they are Superglobals or you fiddle with $GLOBALS, so$foo = "BABAR"; ifexists('foo');
will in general return null. (Italics are php.net chapters.)
– skierpage
Jun 19 '12 at 22:14
now you will get "hello from"... what's the point? just check the valueif( !empty($user) and !empty($location) ) echo "hello $user ..."
– gcb
Jul 9 '13 at 5:31
add a comment |
I didn't want to disable notice because it's helpful, but wanted to avoid too much typing.
My solution was this function:
function ifexists($varname)
{
return(isset($$varname)?$varname:null);
}
So if I want to reference to $name and echo if exists, I simply write:
<?=ifexists('name')?>
For array elements:
function ifexistsidx($var,$index)
{
return(isset($var[$index])?$var[$index]:null);
}
In page if I want to refer to $_REQUEST['name']:
<?=ifexistsidx($_REQUEST,'name')?>
I didn't want to disable notice because it's helpful, but wanted to avoid too much typing.
My solution was this function:
function ifexists($varname)
{
return(isset($$varname)?$varname:null);
}
So if I want to reference to $name and echo if exists, I simply write:
<?=ifexists('name')?>
For array elements:
function ifexistsidx($var,$index)
{
return(isset($var[$index])?$var[$index]:null);
}
In page if I want to refer to $_REQUEST['name']:
<?=ifexistsidx($_REQUEST,'name')?>
answered Dec 30 '11 at 14:14
community wiki
Ferenci Zoltán László
2
Your ifexists() function doesn't work for me in PHP 5.3. The caller's variables are not available in the function's local scope (see Variable scope), unless they are Superglobals or you fiddle with $GLOBALS, so$foo = "BABAR"; ifexists('foo');
will in general return null. (Italics are php.net chapters.)
– skierpage
Jun 19 '12 at 22:14
now you will get "hello from"... what's the point? just check the valueif( !empty($user) and !empty($location) ) echo "hello $user ..."
– gcb
Jul 9 '13 at 5:31
add a comment |
2
Your ifexists() function doesn't work for me in PHP 5.3. The caller's variables are not available in the function's local scope (see Variable scope), unless they are Superglobals or you fiddle with $GLOBALS, so$foo = "BABAR"; ifexists('foo');
will in general return null. (Italics are php.net chapters.)
– skierpage
Jun 19 '12 at 22:14
now you will get "hello from"... what's the point? just check the valueif( !empty($user) and !empty($location) ) echo "hello $user ..."
– gcb
Jul 9 '13 at 5:31
2
2
Your ifexists() function doesn't work for me in PHP 5.3. The caller's variables are not available in the function's local scope (see Variable scope), unless they are Superglobals or you fiddle with $GLOBALS, so
$foo = "BABAR"; ifexists('foo');
will in general return null. (Italics are php.net chapters.)– skierpage
Jun 19 '12 at 22:14
Your ifexists() function doesn't work for me in PHP 5.3. The caller's variables are not available in the function's local scope (see Variable scope), unless they are Superglobals or you fiddle with $GLOBALS, so
$foo = "BABAR"; ifexists('foo');
will in general return null. (Italics are php.net chapters.)– skierpage
Jun 19 '12 at 22:14
now you will get "hello from"... what's the point? just check the value
if( !empty($user) and !empty($location) ) echo "hello $user ..."
– gcb
Jul 9 '13 at 5:31
now you will get "hello from"... what's the point? just check the value
if( !empty($user) and !empty($location) ) echo "hello $user ..."
– gcb
Jul 9 '13 at 5:31
add a comment |
The best way for getting input string is:
$value = filter_input(INPUT_POST, 'value');
This one-liner is almost equivalent to:
if (!isset($_POST['value'])) {
$value = null;
} elseif (is_array($_POST['value'])) {
$value = false;
} else {
$value = $_POST['value'];
}
If you absolutely want string value, just like:
$value = (string)filter_input(INPUT_POST, 'value');
add a comment |
The best way for getting input string is:
$value = filter_input(INPUT_POST, 'value');
This one-liner is almost equivalent to:
if (!isset($_POST['value'])) {
$value = null;
} elseif (is_array($_POST['value'])) {
$value = false;
} else {
$value = $_POST['value'];
}
If you absolutely want string value, just like:
$value = (string)filter_input(INPUT_POST, 'value');
add a comment |
The best way for getting input string is:
$value = filter_input(INPUT_POST, 'value');
This one-liner is almost equivalent to:
if (!isset($_POST['value'])) {
$value = null;
} elseif (is_array($_POST['value'])) {
$value = false;
} else {
$value = $_POST['value'];
}
If you absolutely want string value, just like:
$value = (string)filter_input(INPUT_POST, 'value');
The best way for getting input string is:
$value = filter_input(INPUT_POST, 'value');
This one-liner is almost equivalent to:
if (!isset($_POST['value'])) {
$value = null;
} elseif (is_array($_POST['value'])) {
$value = false;
} else {
$value = $_POST['value'];
}
If you absolutely want string value, just like:
$value = (string)filter_input(INPUT_POST, 'value');
answered Aug 26 '14 at 22:15
community wiki
mpyw
add a comment |
add a comment |
Its because the variable '$user_location' is not getting defined. If you are using any if loop inside which you are declaring the '$user_location' variable then you must also have an else loop and define the same. For example:
$a=10;
if($a==5) { $user_location='Paris';} else { }
echo $user_location;
The above code will create error as The if loop is not satisfied and in the else loop '$user_location' was not defined. Still PHP was asked to echo out the variable. So to modify the code you must do the following:
$a=10;
if($a==5) { $user_location='Paris';} else { $user_location='SOMETHING OR BLANK'; }
echo $user_location;
add a comment |
Its because the variable '$user_location' is not getting defined. If you are using any if loop inside which you are declaring the '$user_location' variable then you must also have an else loop and define the same. For example:
$a=10;
if($a==5) { $user_location='Paris';} else { }
echo $user_location;
The above code will create error as The if loop is not satisfied and in the else loop '$user_location' was not defined. Still PHP was asked to echo out the variable. So to modify the code you must do the following:
$a=10;
if($a==5) { $user_location='Paris';} else { $user_location='SOMETHING OR BLANK'; }
echo $user_location;
add a comment |
Its because the variable '$user_location' is not getting defined. If you are using any if loop inside which you are declaring the '$user_location' variable then you must also have an else loop and define the same. For example:
$a=10;
if($a==5) { $user_location='Paris';} else { }
echo $user_location;
The above code will create error as The if loop is not satisfied and in the else loop '$user_location' was not defined. Still PHP was asked to echo out the variable. So to modify the code you must do the following:
$a=10;
if($a==5) { $user_location='Paris';} else { $user_location='SOMETHING OR BLANK'; }
echo $user_location;
Its because the variable '$user_location' is not getting defined. If you are using any if loop inside which you are declaring the '$user_location' variable then you must also have an else loop and define the same. For example:
$a=10;
if($a==5) { $user_location='Paris';} else { }
echo $user_location;
The above code will create error as The if loop is not satisfied and in the else loop '$user_location' was not defined. Still PHP was asked to echo out the variable. So to modify the code you must do the following:
$a=10;
if($a==5) { $user_location='Paris';} else { $user_location='SOMETHING OR BLANK'; }
echo $user_location;
edited Sep 22 '13 at 12:55
community wiki
2 revs, 2 users 64%
Roger
add a comment |
add a comment |
In reply to ""Why do they appear all of a sudden? I used to use this script for years and I've never had any problem."
It is very common for most sites to operate under the "default" error reporting of "Show all errors, but not 'notices' and 'deprecated'". This will be set in php.ini and apply to all sites on the server. This means that those "notices" used in the examples will be suppressed (hidden) while other errors, considered more critical, will be shown/recorded.
The other critical setting is the errors can be hidden (i.e. display_errors
set to "off" or "syslog").
What will have happened in this case is that either the error_reporting
was changed to also show notices (as per examples) and/or that the settings were changed to display_errors
on screen (as opposed to suppressing them/logging them).
Why have they changed?
The obvious/simplest answer is that someone adjusted either of these settings in php.ini, or an upgraded version of PHP is now using a different php.ini from before. That's the first place to look.
However it is also possible to override these settings in
- .htconf (webserver configuration, including vhosts and sub-configurations)*
- .htaccess
- in php code itself
and any of these could also have been changed.
There is also the added complication that the web server configuration can enable/disable .htaccess directives, so if you have directives in .htaccess that suddenly start/stop working then you need to check for that.
(.htconf / .htaccess assume you're running as apache. If running command line this won't apply; if running IIS or other webserver then you'll need to check those configs accordingly)
Summary
- Check
error_reporting
anddisplay_errors
php directives in php.ini has not changed, or that you're not using a different php.ini from before. - Check
error_reporting
anddisplay_errors
php directives in .htconf (or vhosts etc) have not changed - Check
error_reporting
anddisplay_errors
php directives in .htaccess have not changed - If you have directive in .htaccess, check if they are still permitted in the .htconf file
- Finally check your code; possibly an unrelated library; to see if
error_reporting
anddisplay_errors
php directives have been set there.
add a comment |
In reply to ""Why do they appear all of a sudden? I used to use this script for years and I've never had any problem."
It is very common for most sites to operate under the "default" error reporting of "Show all errors, but not 'notices' and 'deprecated'". This will be set in php.ini and apply to all sites on the server. This means that those "notices" used in the examples will be suppressed (hidden) while other errors, considered more critical, will be shown/recorded.
The other critical setting is the errors can be hidden (i.e. display_errors
set to "off" or "syslog").
What will have happened in this case is that either the error_reporting
was changed to also show notices (as per examples) and/or that the settings were changed to display_errors
on screen (as opposed to suppressing them/logging them).
Why have they changed?
The obvious/simplest answer is that someone adjusted either of these settings in php.ini, or an upgraded version of PHP is now using a different php.ini from before. That's the first place to look.
However it is also possible to override these settings in
- .htconf (webserver configuration, including vhosts and sub-configurations)*
- .htaccess
- in php code itself
and any of these could also have been changed.
There is also the added complication that the web server configuration can enable/disable .htaccess directives, so if you have directives in .htaccess that suddenly start/stop working then you need to check for that.
(.htconf / .htaccess assume you're running as apache. If running command line this won't apply; if running IIS or other webserver then you'll need to check those configs accordingly)
Summary
- Check
error_reporting
anddisplay_errors
php directives in php.ini has not changed, or that you're not using a different php.ini from before. - Check
error_reporting
anddisplay_errors
php directives in .htconf (or vhosts etc) have not changed - Check
error_reporting
anddisplay_errors
php directives in .htaccess have not changed - If you have directive in .htaccess, check if they are still permitted in the .htconf file
- Finally check your code; possibly an unrelated library; to see if
error_reporting
anddisplay_errors
php directives have been set there.
add a comment |
In reply to ""Why do they appear all of a sudden? I used to use this script for years and I've never had any problem."
It is very common for most sites to operate under the "default" error reporting of "Show all errors, but not 'notices' and 'deprecated'". This will be set in php.ini and apply to all sites on the server. This means that those "notices" used in the examples will be suppressed (hidden) while other errors, considered more critical, will be shown/recorded.
The other critical setting is the errors can be hidden (i.e. display_errors
set to "off" or "syslog").
What will have happened in this case is that either the error_reporting
was changed to also show notices (as per examples) and/or that the settings were changed to display_errors
on screen (as opposed to suppressing them/logging them).
Why have they changed?
The obvious/simplest answer is that someone adjusted either of these settings in php.ini, or an upgraded version of PHP is now using a different php.ini from before. That's the first place to look.
However it is also possible to override these settings in
- .htconf (webserver configuration, including vhosts and sub-configurations)*
- .htaccess
- in php code itself
and any of these could also have been changed.
There is also the added complication that the web server configuration can enable/disable .htaccess directives, so if you have directives in .htaccess that suddenly start/stop working then you need to check for that.
(.htconf / .htaccess assume you're running as apache. If running command line this won't apply; if running IIS or other webserver then you'll need to check those configs accordingly)
Summary
- Check
error_reporting
anddisplay_errors
php directives in php.ini has not changed, or that you're not using a different php.ini from before. - Check
error_reporting
anddisplay_errors
php directives in .htconf (or vhosts etc) have not changed - Check
error_reporting
anddisplay_errors
php directives in .htaccess have not changed - If you have directive in .htaccess, check if they are still permitted in the .htconf file
- Finally check your code; possibly an unrelated library; to see if
error_reporting
anddisplay_errors
php directives have been set there.
In reply to ""Why do they appear all of a sudden? I used to use this script for years and I've never had any problem."
It is very common for most sites to operate under the "default" error reporting of "Show all errors, but not 'notices' and 'deprecated'". This will be set in php.ini and apply to all sites on the server. This means that those "notices" used in the examples will be suppressed (hidden) while other errors, considered more critical, will be shown/recorded.
The other critical setting is the errors can be hidden (i.e. display_errors
set to "off" or "syslog").
What will have happened in this case is that either the error_reporting
was changed to also show notices (as per examples) and/or that the settings were changed to display_errors
on screen (as opposed to suppressing them/logging them).
Why have they changed?
The obvious/simplest answer is that someone adjusted either of these settings in php.ini, or an upgraded version of PHP is now using a different php.ini from before. That's the first place to look.
However it is also possible to override these settings in
- .htconf (webserver configuration, including vhosts and sub-configurations)*
- .htaccess
- in php code itself
and any of these could also have been changed.
There is also the added complication that the web server configuration can enable/disable .htaccess directives, so if you have directives in .htaccess that suddenly start/stop working then you need to check for that.
(.htconf / .htaccess assume you're running as apache. If running command line this won't apply; if running IIS or other webserver then you'll need to check those configs accordingly)
Summary
- Check
error_reporting
anddisplay_errors
php directives in php.ini has not changed, or that you're not using a different php.ini from before. - Check
error_reporting
anddisplay_errors
php directives in .htconf (or vhosts etc) have not changed - Check
error_reporting
anddisplay_errors
php directives in .htaccess have not changed - If you have directive in .htaccess, check if they are still permitted in the .htconf file
- Finally check your code; possibly an unrelated library; to see if
error_reporting
anddisplay_errors
php directives have been set there.
answered Nov 23 '16 at 1:44
community wiki
Robbie
add a comment |
add a comment |
the quick fix is to assign your variable to null at the top of your code
$user_location = null;
add a comment |
the quick fix is to assign your variable to null at the top of your code
$user_location = null;
add a comment |
the quick fix is to assign your variable to null at the top of your code
$user_location = null;
the quick fix is to assign your variable to null at the top of your code
$user_location = null;
answered May 12 '14 at 19:02
community wiki
Shahin Mammadzada
add a comment |
add a comment |
I used to curse this error, but it can be helpful to remind you to escape user input.
For instance, if you thought this was clever, shorthand code:
// Echo whatever the hell this is
<?=$_POST['something']?>
...Think again! A better solution is:
// If this is set, echo a filtered version
<?=isset($_POST['something']) ? html($_POST['something']) : ''?>
(I use a custom html()
function to escape characters, your mileage may vary)
add a comment |
I used to curse this error, but it can be helpful to remind you to escape user input.
For instance, if you thought this was clever, shorthand code:
// Echo whatever the hell this is
<?=$_POST['something']?>
...Think again! A better solution is:
// If this is set, echo a filtered version
<?=isset($_POST['something']) ? html($_POST['something']) : ''?>
(I use a custom html()
function to escape characters, your mileage may vary)
add a comment |
I used to curse this error, but it can be helpful to remind you to escape user input.
For instance, if you thought this was clever, shorthand code:
// Echo whatever the hell this is
<?=$_POST['something']?>
...Think again! A better solution is:
// If this is set, echo a filtered version
<?=isset($_POST['something']) ? html($_POST['something']) : ''?>
(I use a custom html()
function to escape characters, your mileage may vary)
I used to curse this error, but it can be helpful to remind you to escape user input.
For instance, if you thought this was clever, shorthand code:
// Echo whatever the hell this is
<?=$_POST['something']?>
...Think again! A better solution is:
// If this is set, echo a filtered version
<?=isset($_POST['something']) ? html($_POST['something']) : ''?>
(I use a custom html()
function to escape characters, your mileage may vary)
answered Jul 28 '15 at 19:12
community wiki
rybo111
add a comment |
add a comment |
I use all time own useful function exst() which automatically declare variables.
Your code will be -
$greeting = "Hello, ".exst($user_name, 'Visitor')." from ".exst($user_location);
/**
* Function exst() - Checks if the variable has been set
* (copy/paste it in any place of your code)
*
* If the variable is set and not empty returns the variable (no transformation)
* If the variable is not set or empty, returns the $default value
*
* @param mixed $var
* @param mixed $default
*
* @return mixed
*/
function exst( & $var, $default = "")
{
$t = "";
if ( !isset($var) || !$var ) {
if (isset($default) && $default != "") $t = $default;
}
else {
$t = $var;
}
if (is_string($t)) $t = trim($t);
return $t;
}
add a comment |
I use all time own useful function exst() which automatically declare variables.
Your code will be -
$greeting = "Hello, ".exst($user_name, 'Visitor')." from ".exst($user_location);
/**
* Function exst() - Checks if the variable has been set
* (copy/paste it in any place of your code)
*
* If the variable is set and not empty returns the variable (no transformation)
* If the variable is not set or empty, returns the $default value
*
* @param mixed $var
* @param mixed $default
*
* @return mixed
*/
function exst( & $var, $default = "")
{
$t = "";
if ( !isset($var) || !$var ) {
if (isset($default) && $default != "") $t = $default;
}
else {
$t = $var;
}
if (is_string($t)) $t = trim($t);
return $t;
}
add a comment |
I use all time own useful function exst() which automatically declare variables.
Your code will be -
$greeting = "Hello, ".exst($user_name, 'Visitor')." from ".exst($user_location);
/**
* Function exst() - Checks if the variable has been set
* (copy/paste it in any place of your code)
*
* If the variable is set and not empty returns the variable (no transformation)
* If the variable is not set or empty, returns the $default value
*
* @param mixed $var
* @param mixed $default
*
* @return mixed
*/
function exst( & $var, $default = "")
{
$t = "";
if ( !isset($var) || !$var ) {
if (isset($default) && $default != "") $t = $default;
}
else {
$t = $var;
}
if (is_string($t)) $t = trim($t);
return $t;
}
I use all time own useful function exst() which automatically declare variables.
Your code will be -
$greeting = "Hello, ".exst($user_name, 'Visitor')." from ".exst($user_location);
/**
* Function exst() - Checks if the variable has been set
* (copy/paste it in any place of your code)
*
* If the variable is set and not empty returns the variable (no transformation)
* If the variable is not set or empty, returns the $default value
*
* @param mixed $var
* @param mixed $default
*
* @return mixed
*/
function exst( & $var, $default = "")
{
$t = "";
if ( !isset($var) || !$var ) {
if (isset($default) && $default != "") $t = $default;
}
else {
$t = $var;
}
if (is_string($t)) $t = trim($t);
return $t;
}
edited Apr 9 '13 at 14:07
community wiki
2 revs
user2253362
add a comment |
add a comment |
In a very Simple Language.
The mistake is you are using a variable $user_location
which is not defined by you earlier and it doesn't have any value So I recommend you to please declare this variable before using it, For Example:
$user_location = '';
Or
$user_location = 'Los Angles';
This is a very common error you can face.So don't worry just declare the variable and Enjoy Coding.
add a comment |
In a very Simple Language.
The mistake is you are using a variable $user_location
which is not defined by you earlier and it doesn't have any value So I recommend you to please declare this variable before using it, For Example:
$user_location = '';
Or
$user_location = 'Los Angles';
This is a very common error you can face.So don't worry just declare the variable and Enjoy Coding.
add a comment |
In a very Simple Language.
The mistake is you are using a variable $user_location
which is not defined by you earlier and it doesn't have any value So I recommend you to please declare this variable before using it, For Example:
$user_location = '';
Or
$user_location = 'Los Angles';
This is a very common error you can face.So don't worry just declare the variable and Enjoy Coding.
In a very Simple Language.
The mistake is you are using a variable $user_location
which is not defined by you earlier and it doesn't have any value So I recommend you to please declare this variable before using it, For Example:
$user_location = '';
Or
$user_location = 'Los Angles';
This is a very common error you can face.So don't worry just declare the variable and Enjoy Coding.
answered Dec 18 '15 at 11:06
community wiki
Rishabh Seth
add a comment |
add a comment |
In PHP 7.0 it's now possible to use Null coalescing operator:
echo "My index value is: " . ($my_array["my_index"] ?? '');
Equals to:
echo "My index value is: " . (isset($my_array["my_index"]) ? $my_array["my_index"] : '');
PHP manual PHP 7.0
1
Wow, this answer needs to be voted up more now that this is available in PHP 7. Wonderful solution!!!
– OCDev
Sep 24 '17 at 13:17
This also works fine in if statements.if (is_array($my_array['idontexist'] ?? '')) { dosomething(); }
– a coder
Feb 23 '18 at 20:04
Your code is actually a nice overlooked bug: ?? - only checks forisset()
, if you passis_array()
- which is a boolean, unexpected behavior will follow.
– Lachezar Lechev
Feb 26 '18 at 1:20
add a comment |
In PHP 7.0 it's now possible to use Null coalescing operator:
echo "My index value is: " . ($my_array["my_index"] ?? '');
Equals to:
echo "My index value is: " . (isset($my_array["my_index"]) ? $my_array["my_index"] : '');
PHP manual PHP 7.0
1
Wow, this answer needs to be voted up more now that this is available in PHP 7. Wonderful solution!!!
– OCDev
Sep 24 '17 at 13:17
This also works fine in if statements.if (is_array($my_array['idontexist'] ?? '')) { dosomething(); }
– a coder
Feb 23 '18 at 20:04
Your code is actually a nice overlooked bug: ?? - only checks forisset()
, if you passis_array()
- which is a boolean, unexpected behavior will follow.
– Lachezar Lechev
Feb 26 '18 at 1:20
add a comment |
In PHP 7.0 it's now possible to use Null coalescing operator:
echo "My index value is: " . ($my_array["my_index"] ?? '');
Equals to:
echo "My index value is: " . (isset($my_array["my_index"]) ? $my_array["my_index"] : '');
PHP manual PHP 7.0
In PHP 7.0 it's now possible to use Null coalescing operator:
echo "My index value is: " . ($my_array["my_index"] ?? '');
Equals to:
echo "My index value is: " . (isset($my_array["my_index"]) ? $my_array["my_index"] : '');
PHP manual PHP 7.0
answered Dec 13 '16 at 13:34
community wiki
Lachezar Lechev
1
Wow, this answer needs to be voted up more now that this is available in PHP 7. Wonderful solution!!!
– OCDev
Sep 24 '17 at 13:17
This also works fine in if statements.if (is_array($my_array['idontexist'] ?? '')) { dosomething(); }
– a coder
Feb 23 '18 at 20:04
Your code is actually a nice overlooked bug: ?? - only checks forisset()
, if you passis_array()
- which is a boolean, unexpected behavior will follow.
– Lachezar Lechev
Feb 26 '18 at 1:20
add a comment |
1
Wow, this answer needs to be voted up more now that this is available in PHP 7. Wonderful solution!!!
– OCDev
Sep 24 '17 at 13:17
This also works fine in if statements.if (is_array($my_array['idontexist'] ?? '')) { dosomething(); }
– a coder
Feb 23 '18 at 20:04
Your code is actually a nice overlooked bug: ?? - only checks forisset()
, if you passis_array()
- which is a boolean, unexpected behavior will follow.
– Lachezar Lechev
Feb 26 '18 at 1:20
1
1
Wow, this answer needs to be voted up more now that this is available in PHP 7. Wonderful solution!!!
– OCDev
Sep 24 '17 at 13:17
Wow, this answer needs to be voted up more now that this is available in PHP 7. Wonderful solution!!!
– OCDev
Sep 24 '17 at 13:17
This also works fine in if statements.
if (is_array($my_array['idontexist'] ?? '')) { dosomething(); }
– a coder
Feb 23 '18 at 20:04
This also works fine in if statements.
if (is_array($my_array['idontexist'] ?? '')) { dosomething(); }
– a coder
Feb 23 '18 at 20:04
Your code is actually a nice overlooked bug: ?? - only checks for
isset()
, if you pass is_array()
- which is a boolean, unexpected behavior will follow.– Lachezar Lechev
Feb 26 '18 at 1:20
Your code is actually a nice overlooked bug: ?? - only checks for
isset()
, if you pass is_array()
- which is a boolean, unexpected behavior will follow.– Lachezar Lechev
Feb 26 '18 at 1:20
add a comment |
why not keep things simple?
<?php
error_reporting(E_ALL); // making sure all notices are on
function idxVal(&$var, $default = null) {
return empty($var) ? $var = $default : $var;
}
echo idxVal($arr['test']); // returns null without any notice
echo idxVal($arr['hey ho'], 'yo'); // returns yo and assigns it to array index, nice
?>
add a comment |
why not keep things simple?
<?php
error_reporting(E_ALL); // making sure all notices are on
function idxVal(&$var, $default = null) {
return empty($var) ? $var = $default : $var;
}
echo idxVal($arr['test']); // returns null without any notice
echo idxVal($arr['hey ho'], 'yo'); // returns yo and assigns it to array index, nice
?>
add a comment |
why not keep things simple?
<?php
error_reporting(E_ALL); // making sure all notices are on
function idxVal(&$var, $default = null) {
return empty($var) ? $var = $default : $var;
}
echo idxVal($arr['test']); // returns null without any notice
echo idxVal($arr['hey ho'], 'yo'); // returns yo and assigns it to array index, nice
?>
why not keep things simple?
<?php
error_reporting(E_ALL); // making sure all notices are on
function idxVal(&$var, $default = null) {
return empty($var) ? $var = $default : $var;
}
echo idxVal($arr['test']); // returns null without any notice
echo idxVal($arr['hey ho'], 'yo'); // returns yo and assigns it to array index, nice
?>
answered Jun 6 '13 at 20:37
community wiki
gts
add a comment |
add a comment |
WHY IS THIS HAPPENING?
Over time, PHP has become a more security-focused language. Settings which used to be turned off by default are now turned on by default. A perfect example of this is E_STRICT
, which became turned on by default as of PHP 5.4.0.
Furthermore, according to PHP documentation, by defualt, E_NOTICE
is disabled in php.ini. PHP docs recommend turning it on for debugging purposes. However, when I download PHP from the Ubuntu repository–and from BitNami's Windows stack–I see something else.
; Common Values:
; E_ALL (Show all errors, warnings and notices including coding standards.)
; E_ALL & ~E_NOTICE (Show all errors, except for notices)
; E_ALL & ~E_NOTICE & ~E_STRICT (Show all errors, except for notices and coding standards warnings.)
; E_COMPILE_ERROR|E_RECOVERABLE_ERROR|E_ERROR|E_CORE_ERROR (Show only errors)
; Default Value: E_ALL & ~E_NOTICE & ~E_STRICT & ~E_DEPRECATED
; Development Value: E_ALL
; Production Value: E_ALL & ~E_DEPRECATED & ~E_STRICT
; http://php.net/error-reporting
error_reporting = E_ALL & ~E_DEPRECATED & ~E_STRICT
Notice that error_reporting
is actually set to the production value by default, not to the "default" value by default. This is somewhat confusing and is not documented outside of php.ini, so I have not validated this on other distributions.
To answer your question, however, this error pops up now when it did not pop up before because:
You installed PHP and the new default settings are somewhat poorly documented, but do not exclude
E_NOTICE
.E_NOTICE
warnings like undefined variables and undefined indexes actually help to make your code cleaner and safer. I can tell you that, years ago, keepingE_NOTICE
enabled forced me to declare my variables. It made it a LOT easier to learn C, where not declaring variables is much bigger of a nuisance.
WHAT CAN I DO ABOUT IT?
Turn off
E_NOTICE
by copying the "Default value"E_ALL & ~E_NOTICE & ~E_STRICT & ~E_DEPRECATED
and replacing it with what is currently uncommented after the equals sign inerror_reporting =
. Restart Apache, or PHP if using CGI or FPM. Make sure you are editing the "right" php.ini. The correct one will be Apache if you are running PHP with Apache, fpm or php-fpm if running PHP-FPM, cgi if running PHP-CGI, etc. This is not the recommended method, but if you have legacy code that's going to be exceedingly difficult to edit, then it might be your best bet.Turn off
E_NOTICE
on the file or folder level. This might be preferable if you have some legacy code but want to do things the "right" way otherwise. To do this, you should consult Apache2, nginx, or whatever your server of choice is. In Apache, you would usephp_value
inside of<Directory>
.Rewrite your code to be cleaner. If you need to do this while moving to a production environment, or don't want someone to see your errors, make sure you are disabling any display of errors, and only logging your errors (see
display_errors
andlog_errors
in php.ini and your server settings).
To expand on option 3: This is the ideal. If you can go this route, you should. If you are not going this route initially, consider moving this route eventually by testing your code in a development environment. While you're at it, get rid of ~E_STRICT
and ~E_DEPRECATED
to see what might go wrong in the future. You're going to see a LOT of unfamiliar errors, but it's going to stop you from having any unpleasant problems when you need to upgrade PHP in the future.
WHAT DO THE ERRORS MEAN?
Undefined variable: my_variable_name
- This occurs when a variable has not been defined before use. When the PHP script is executed, it internally just assumes a null value. However, in which scenario would you need to check a variable before it was defined? Ultimately, this is an argument for "sloppy code". As a developer, I can tell you that I love it when I see an open source project where variables are defined as high up in their scopes as they can be defined. It makes it easier to tell what variables are going to pop up in the future, and makes it easier to read/learn the code.
function foo()
{
$my_variable_name = '';
//....
if ($my_variable_name) {
// perform some logic
}
}
Undefined index: my_index
- This occurs when you try to access a value in an array and it does not exist. To prevent this error, perform a conditional check.
// verbose way - generally better
if (isset($my_array['my_index'])) {
echo "My index value is: " . $my_array['my_index'];
}
// non-verbose ternary example - I use this sometimes for small rules.
$my_index_val = isset($my_array['my_index'])?$my_array['my_index']:'(undefined)';
echo "My index value is: " . $my_index_val;
Another option is to declare an empty array at the top of your function. This is not always possible.
$my_array = array(
'my_index' => ''
);
//...
$my_array['my_index'] = 'new string';
(additional tip)
- When I was encountering these and other issues, I used NetBeans IDE (free) and it gave me a host of warnings and notices. Some of them offer very helpful tips. This is not a requirement, and I don't use IDEs anymore except for large projects. I'm more of a
vim
person these days :).
add a comment |
WHY IS THIS HAPPENING?
Over time, PHP has become a more security-focused language. Settings which used to be turned off by default are now turned on by default. A perfect example of this is E_STRICT
, which became turned on by default as of PHP 5.4.0.
Furthermore, according to PHP documentation, by defualt, E_NOTICE
is disabled in php.ini. PHP docs recommend turning it on for debugging purposes. However, when I download PHP from the Ubuntu repository–and from BitNami's Windows stack–I see something else.
; Common Values:
; E_ALL (Show all errors, warnings and notices including coding standards.)
; E_ALL & ~E_NOTICE (Show all errors, except for notices)
; E_ALL & ~E_NOTICE & ~E_STRICT (Show all errors, except for notices and coding standards warnings.)
; E_COMPILE_ERROR|E_RECOVERABLE_ERROR|E_ERROR|E_CORE_ERROR (Show only errors)
; Default Value: E_ALL & ~E_NOTICE & ~E_STRICT & ~E_DEPRECATED
; Development Value: E_ALL
; Production Value: E_ALL & ~E_DEPRECATED & ~E_STRICT
; http://php.net/error-reporting
error_reporting = E_ALL & ~E_DEPRECATED & ~E_STRICT
Notice that error_reporting
is actually set to the production value by default, not to the "default" value by default. This is somewhat confusing and is not documented outside of php.ini, so I have not validated this on other distributions.
To answer your question, however, this error pops up now when it did not pop up before because:
You installed PHP and the new default settings are somewhat poorly documented, but do not exclude
E_NOTICE
.E_NOTICE
warnings like undefined variables and undefined indexes actually help to make your code cleaner and safer. I can tell you that, years ago, keepingE_NOTICE
enabled forced me to declare my variables. It made it a LOT easier to learn C, where not declaring variables is much bigger of a nuisance.
WHAT CAN I DO ABOUT IT?
Turn off
E_NOTICE
by copying the "Default value"E_ALL & ~E_NOTICE & ~E_STRICT & ~E_DEPRECATED
and replacing it with what is currently uncommented after the equals sign inerror_reporting =
. Restart Apache, or PHP if using CGI or FPM. Make sure you are editing the "right" php.ini. The correct one will be Apache if you are running PHP with Apache, fpm or php-fpm if running PHP-FPM, cgi if running PHP-CGI, etc. This is not the recommended method, but if you have legacy code that's going to be exceedingly difficult to edit, then it might be your best bet.Turn off
E_NOTICE
on the file or folder level. This might be preferable if you have some legacy code but want to do things the "right" way otherwise. To do this, you should consult Apache2, nginx, or whatever your server of choice is. In Apache, you would usephp_value
inside of<Directory>
.Rewrite your code to be cleaner. If you need to do this while moving to a production environment, or don't want someone to see your errors, make sure you are disabling any display of errors, and only logging your errors (see
display_errors
andlog_errors
in php.ini and your server settings).
To expand on option 3: This is the ideal. If you can go this route, you should. If you are not going this route initially, consider moving this route eventually by testing your code in a development environment. While you're at it, get rid of ~E_STRICT
and ~E_DEPRECATED
to see what might go wrong in the future. You're going to see a LOT of unfamiliar errors, but it's going to stop you from having any unpleasant problems when you need to upgrade PHP in the future.
WHAT DO THE ERRORS MEAN?
Undefined variable: my_variable_name
- This occurs when a variable has not been defined before use. When the PHP script is executed, it internally just assumes a null value. However, in which scenario would you need to check a variable before it was defined? Ultimately, this is an argument for "sloppy code". As a developer, I can tell you that I love it when I see an open source project where variables are defined as high up in their scopes as they can be defined. It makes it easier to tell what variables are going to pop up in the future, and makes it easier to read/learn the code.
function foo()
{
$my_variable_name = '';
//....
if ($my_variable_name) {
// perform some logic
}
}
Undefined index: my_index
- This occurs when you try to access a value in an array and it does not exist. To prevent this error, perform a conditional check.
// verbose way - generally better
if (isset($my_array['my_index'])) {
echo "My index value is: " . $my_array['my_index'];
}
// non-verbose ternary example - I use this sometimes for small rules.
$my_index_val = isset($my_array['my_index'])?$my_array['my_index']:'(undefined)';
echo "My index value is: " . $my_index_val;
Another option is to declare an empty array at the top of your function. This is not always possible.
$my_array = array(
'my_index' => ''
);
//...
$my_array['my_index'] = 'new string';
(additional tip)
- When I was encountering these and other issues, I used NetBeans IDE (free) and it gave me a host of warnings and notices. Some of them offer very helpful tips. This is not a requirement, and I don't use IDEs anymore except for large projects. I'm more of a
vim
person these days :).
add a comment |
WHY IS THIS HAPPENING?
Over time, PHP has become a more security-focused language. Settings which used to be turned off by default are now turned on by default. A perfect example of this is E_STRICT
, which became turned on by default as of PHP 5.4.0.
Furthermore, according to PHP documentation, by defualt, E_NOTICE
is disabled in php.ini. PHP docs recommend turning it on for debugging purposes. However, when I download PHP from the Ubuntu repository–and from BitNami's Windows stack–I see something else.
; Common Values:
; E_ALL (Show all errors, warnings and notices including coding standards.)
; E_ALL & ~E_NOTICE (Show all errors, except for notices)
; E_ALL & ~E_NOTICE & ~E_STRICT (Show all errors, except for notices and coding standards warnings.)
; E_COMPILE_ERROR|E_RECOVERABLE_ERROR|E_ERROR|E_CORE_ERROR (Show only errors)
; Default Value: E_ALL & ~E_NOTICE & ~E_STRICT & ~E_DEPRECATED
; Development Value: E_ALL
; Production Value: E_ALL & ~E_DEPRECATED & ~E_STRICT
; http://php.net/error-reporting
error_reporting = E_ALL & ~E_DEPRECATED & ~E_STRICT
Notice that error_reporting
is actually set to the production value by default, not to the "default" value by default. This is somewhat confusing and is not documented outside of php.ini, so I have not validated this on other distributions.
To answer your question, however, this error pops up now when it did not pop up before because:
You installed PHP and the new default settings are somewhat poorly documented, but do not exclude
E_NOTICE
.E_NOTICE
warnings like undefined variables and undefined indexes actually help to make your code cleaner and safer. I can tell you that, years ago, keepingE_NOTICE
enabled forced me to declare my variables. It made it a LOT easier to learn C, where not declaring variables is much bigger of a nuisance.
WHAT CAN I DO ABOUT IT?
Turn off
E_NOTICE
by copying the "Default value"E_ALL & ~E_NOTICE & ~E_STRICT & ~E_DEPRECATED
and replacing it with what is currently uncommented after the equals sign inerror_reporting =
. Restart Apache, or PHP if using CGI or FPM. Make sure you are editing the "right" php.ini. The correct one will be Apache if you are running PHP with Apache, fpm or php-fpm if running PHP-FPM, cgi if running PHP-CGI, etc. This is not the recommended method, but if you have legacy code that's going to be exceedingly difficult to edit, then it might be your best bet.Turn off
E_NOTICE
on the file or folder level. This might be preferable if you have some legacy code but want to do things the "right" way otherwise. To do this, you should consult Apache2, nginx, or whatever your server of choice is. In Apache, you would usephp_value
inside of<Directory>
.Rewrite your code to be cleaner. If you need to do this while moving to a production environment, or don't want someone to see your errors, make sure you are disabling any display of errors, and only logging your errors (see
display_errors
andlog_errors
in php.ini and your server settings).
To expand on option 3: This is the ideal. If you can go this route, you should. If you are not going this route initially, consider moving this route eventually by testing your code in a development environment. While you're at it, get rid of ~E_STRICT
and ~E_DEPRECATED
to see what might go wrong in the future. You're going to see a LOT of unfamiliar errors, but it's going to stop you from having any unpleasant problems when you need to upgrade PHP in the future.
WHAT DO THE ERRORS MEAN?
Undefined variable: my_variable_name
- This occurs when a variable has not been defined before use. When the PHP script is executed, it internally just assumes a null value. However, in which scenario would you need to check a variable before it was defined? Ultimately, this is an argument for "sloppy code". As a developer, I can tell you that I love it when I see an open source project where variables are defined as high up in their scopes as they can be defined. It makes it easier to tell what variables are going to pop up in the future, and makes it easier to read/learn the code.
function foo()
{
$my_variable_name = '';
//....
if ($my_variable_name) {
// perform some logic
}
}
Undefined index: my_index
- This occurs when you try to access a value in an array and it does not exist. To prevent this error, perform a conditional check.
// verbose way - generally better
if (isset($my_array['my_index'])) {
echo "My index value is: " . $my_array['my_index'];
}
// non-verbose ternary example - I use this sometimes for small rules.
$my_index_val = isset($my_array['my_index'])?$my_array['my_index']:'(undefined)';
echo "My index value is: " . $my_index_val;
Another option is to declare an empty array at the top of your function. This is not always possible.
$my_array = array(
'my_index' => ''
);
//...
$my_array['my_index'] = 'new string';
(additional tip)
- When I was encountering these and other issues, I used NetBeans IDE (free) and it gave me a host of warnings and notices. Some of them offer very helpful tips. This is not a requirement, and I don't use IDEs anymore except for large projects. I'm more of a
vim
person these days :).
WHY IS THIS HAPPENING?
Over time, PHP has become a more security-focused language. Settings which used to be turned off by default are now turned on by default. A perfect example of this is E_STRICT
, which became turned on by default as of PHP 5.4.0.
Furthermore, according to PHP documentation, by defualt, E_NOTICE
is disabled in php.ini. PHP docs recommend turning it on for debugging purposes. However, when I download PHP from the Ubuntu repository–and from BitNami's Windows stack–I see something else.
; Common Values:
; E_ALL (Show all errors, warnings and notices including coding standards.)
; E_ALL & ~E_NOTICE (Show all errors, except for notices)
; E_ALL & ~E_NOTICE & ~E_STRICT (Show all errors, except for notices and coding standards warnings.)
; E_COMPILE_ERROR|E_RECOVERABLE_ERROR|E_ERROR|E_CORE_ERROR (Show only errors)
; Default Value: E_ALL & ~E_NOTICE & ~E_STRICT & ~E_DEPRECATED
; Development Value: E_ALL
; Production Value: E_ALL & ~E_DEPRECATED & ~E_STRICT
; http://php.net/error-reporting
error_reporting = E_ALL & ~E_DEPRECATED & ~E_STRICT
Notice that error_reporting
is actually set to the production value by default, not to the "default" value by default. This is somewhat confusing and is not documented outside of php.ini, so I have not validated this on other distributions.
To answer your question, however, this error pops up now when it did not pop up before because:
You installed PHP and the new default settings are somewhat poorly documented, but do not exclude
E_NOTICE
.E_NOTICE
warnings like undefined variables and undefined indexes actually help to make your code cleaner and safer. I can tell you that, years ago, keepingE_NOTICE
enabled forced me to declare my variables. It made it a LOT easier to learn C, where not declaring variables is much bigger of a nuisance.
WHAT CAN I DO ABOUT IT?
Turn off
E_NOTICE
by copying the "Default value"E_ALL & ~E_NOTICE & ~E_STRICT & ~E_DEPRECATED
and replacing it with what is currently uncommented after the equals sign inerror_reporting =
. Restart Apache, or PHP if using CGI or FPM. Make sure you are editing the "right" php.ini. The correct one will be Apache if you are running PHP with Apache, fpm or php-fpm if running PHP-FPM, cgi if running PHP-CGI, etc. This is not the recommended method, but if you have legacy code that's going to be exceedingly difficult to edit, then it might be your best bet.Turn off
E_NOTICE
on the file or folder level. This might be preferable if you have some legacy code but want to do things the "right" way otherwise. To do this, you should consult Apache2, nginx, or whatever your server of choice is. In Apache, you would usephp_value
inside of<Directory>
.Rewrite your code to be cleaner. If you need to do this while moving to a production environment, or don't want someone to see your errors, make sure you are disabling any display of errors, and only logging your errors (see
display_errors
andlog_errors
in php.ini and your server settings).
To expand on option 3: This is the ideal. If you can go this route, you should. If you are not going this route initially, consider moving this route eventually by testing your code in a development environment. While you're at it, get rid of ~E_STRICT
and ~E_DEPRECATED
to see what might go wrong in the future. You're going to see a LOT of unfamiliar errors, but it's going to stop you from having any unpleasant problems when you need to upgrade PHP in the future.
WHAT DO THE ERRORS MEAN?
Undefined variable: my_variable_name
- This occurs when a variable has not been defined before use. When the PHP script is executed, it internally just assumes a null value. However, in which scenario would you need to check a variable before it was defined? Ultimately, this is an argument for "sloppy code". As a developer, I can tell you that I love it when I see an open source project where variables are defined as high up in their scopes as they can be defined. It makes it easier to tell what variables are going to pop up in the future, and makes it easier to read/learn the code.
function foo()
{
$my_variable_name = '';
//....
if ($my_variable_name) {
// perform some logic
}
}
Undefined index: my_index
- This occurs when you try to access a value in an array and it does not exist. To prevent this error, perform a conditional check.
// verbose way - generally better
if (isset($my_array['my_index'])) {
echo "My index value is: " . $my_array['my_index'];
}
// non-verbose ternary example - I use this sometimes for small rules.
$my_index_val = isset($my_array['my_index'])?$my_array['my_index']:'(undefined)';
echo "My index value is: " . $my_index_val;
Another option is to declare an empty array at the top of your function. This is not always possible.
$my_array = array(
'my_index' => ''
);
//...
$my_array['my_index'] = 'new string';
(additional tip)
- When I was encountering these and other issues, I used NetBeans IDE (free) and it gave me a host of warnings and notices. Some of them offer very helpful tips. This is not a requirement, and I don't use IDEs anymore except for large projects. I'm more of a
vim
person these days :).
answered Nov 28 '16 at 23:01
community wiki
smcjones
add a comment |
add a comment |
undefined index means in an array you requested for unavailable array index for example
<?php
$newArray = {1,2,3,4,5};
print_r($newArray[5]);
?>
undefined variable means you have used completely not existing variable or which is not defined or initialized by that name for example
<?php print_r($myvar); ?>
undefined offset means in array you have asked for non existing key. And the
solution for this is to check before use
php> echo array_key_exists(1, $myarray);
add a comment |
undefined index means in an array you requested for unavailable array index for example
<?php
$newArray = {1,2,3,4,5};
print_r($newArray[5]);
?>
undefined variable means you have used completely not existing variable or which is not defined or initialized by that name for example
<?php print_r($myvar); ?>
undefined offset means in array you have asked for non existing key. And the
solution for this is to check before use
php> echo array_key_exists(1, $myarray);
add a comment |
undefined index means in an array you requested for unavailable array index for example
<?php
$newArray = {1,2,3,4,5};
print_r($newArray[5]);
?>
undefined variable means you have used completely not existing variable or which is not defined or initialized by that name for example
<?php print_r($myvar); ?>
undefined offset means in array you have asked for non existing key. And the
solution for this is to check before use
php> echo array_key_exists(1, $myarray);
undefined index means in an array you requested for unavailable array index for example
<?php
$newArray = {1,2,3,4,5};
print_r($newArray[5]);
?>
undefined variable means you have used completely not existing variable or which is not defined or initialized by that name for example
<?php print_r($myvar); ?>
undefined offset means in array you have asked for non existing key. And the
solution for this is to check before use
php> echo array_key_exists(1, $myarray);
edited Jun 30 '17 at 1:02
community wiki
2 revs, 2 users 84%
Sintayehu Abaynhe
add a comment |
add a comment |
Regarding this part of the question:
Why do they appear all of a sudden? I used to use this script for years and I've never had any problem.
No definite answers but here are a some possible explanations of why settings can 'suddenly' change:
You have upgraded PHP to a newer version which can have other defaults for error_reporting, display_errors or other relevant settings.
You have removed or introduced some code (possibly in a dependency) that sets relevant settings at runtime using
ini_set()
orerror_reporting()
(search for these in the code)You changed the webserver configuration (assuming apache here):
.htaccess
files and vhost configurations can also manipulate php settings.Usually notices don't get displayed / reported (see PHP manual)
so it is possible that when setting up the server, the php.ini file could not be loaded for some reason (file permissions??) and you were on the default settings. Later on, the 'bug' has been solved (by accident) and now it CAN load the correct php.ini file with the error_reporting set to show notices.
add a comment |
Regarding this part of the question:
Why do they appear all of a sudden? I used to use this script for years and I've never had any problem.
No definite answers but here are a some possible explanations of why settings can 'suddenly' change:
You have upgraded PHP to a newer version which can have other defaults for error_reporting, display_errors or other relevant settings.
You have removed or introduced some code (possibly in a dependency) that sets relevant settings at runtime using
ini_set()
orerror_reporting()
(search for these in the code)You changed the webserver configuration (assuming apache here):
.htaccess
files and vhost configurations can also manipulate php settings.Usually notices don't get displayed / reported (see PHP manual)
so it is possible that when setting up the server, the php.ini file could not be loaded for some reason (file permissions??) and you were on the default settings. Later on, the 'bug' has been solved (by accident) and now it CAN load the correct php.ini file with the error_reporting set to show notices.
add a comment |
Regarding this part of the question:
Why do they appear all of a sudden? I used to use this script for years and I've never had any problem.
No definite answers but here are a some possible explanations of why settings can 'suddenly' change:
You have upgraded PHP to a newer version which can have other defaults for error_reporting, display_errors or other relevant settings.
You have removed or introduced some code (possibly in a dependency) that sets relevant settings at runtime using
ini_set()
orerror_reporting()
(search for these in the code)You changed the webserver configuration (assuming apache here):
.htaccess
files and vhost configurations can also manipulate php settings.Usually notices don't get displayed / reported (see PHP manual)
so it is possible that when setting up the server, the php.ini file could not be loaded for some reason (file permissions??) and you were on the default settings. Later on, the 'bug' has been solved (by accident) and now it CAN load the correct php.ini file with the error_reporting set to show notices.
Regarding this part of the question:
Why do they appear all of a sudden? I used to use this script for years and I've never had any problem.
No definite answers but here are a some possible explanations of why settings can 'suddenly' change:
You have upgraded PHP to a newer version which can have other defaults for error_reporting, display_errors or other relevant settings.
You have removed or introduced some code (possibly in a dependency) that sets relevant settings at runtime using
ini_set()
orerror_reporting()
(search for these in the code)You changed the webserver configuration (assuming apache here):
.htaccess
files and vhost configurations can also manipulate php settings.Usually notices don't get displayed / reported (see PHP manual)
so it is possible that when setting up the server, the php.ini file could not be loaded for some reason (file permissions??) and you were on the default settings. Later on, the 'bug' has been solved (by accident) and now it CAN load the correct php.ini file with the error_reporting set to show notices.
answered Nov 28 '16 at 19:31
community wiki
Wouter de Winter
add a comment |
add a comment |
If working with classes you need to make sure you reference member variables using $this
:
class Person
{
protected $firstName;
protected $lastName;
public function setFullName($first, $last)
{
// Correct
$this->firstName = $first;
// Incorrect
$lastName = $last;
// Incorrect
$this->$lastName = $last;
}
}
add a comment |
If working with classes you need to make sure you reference member variables using $this
:
class Person
{
protected $firstName;
protected $lastName;
public function setFullName($first, $last)
{
// Correct
$this->firstName = $first;
// Incorrect
$lastName = $last;
// Incorrect
$this->$lastName = $last;
}
}
add a comment |
If working with classes you need to make sure you reference member variables using $this
:
class Person
{
protected $firstName;
protected $lastName;
public function setFullName($first, $last)
{
// Correct
$this->firstName = $first;
// Incorrect
$lastName = $last;
// Incorrect
$this->$lastName = $last;
}
}
If working with classes you need to make sure you reference member variables using $this
:
class Person
{
protected $firstName;
protected $lastName;
public function setFullName($first, $last)
{
// Correct
$this->firstName = $first;
// Incorrect
$lastName = $last;
// Incorrect
$this->$lastName = $last;
}
}
edited Jul 20 '17 at 11:44
community wiki
2 revs
John Conde
add a comment |
add a comment |
Another reason why an undefined index notice will be thrown, would be that a column was omitted from a database query.
I.e.:
$query = "SELECT col1 FROM table WHERE col_x = ?";
Then trying to access more columns/rows inside a loop.
I.e.:
print_r($row['col1']);
print_r($row['col2']); // undefined index thrown
or in a while
loop:
while( $row = fetching_function($query) ) {
echo $row['col1'];
echo "<br>";
echo $row['col2']; // undefined index thrown
echo "<br>";
echo $row['col3']; // undefined index thrown
}
Something else that needs to be noted is that on a *NIX OS and Mac OS X, things are case-sensitive.
Consult the followning Q&A's on Stack:
Are table names in MySQL case sensitive?
mysql case sensitive table names in queries
MySql - Case Sensitive issue of tables in different server
I have to wonder why this was downvoted also.
– Funk Forty Niner
Oct 25 '18 at 0:26
add a comment |
Another reason why an undefined index notice will be thrown, would be that a column was omitted from a database query.
I.e.:
$query = "SELECT col1 FROM table WHERE col_x = ?";
Then trying to access more columns/rows inside a loop.
I.e.:
print_r($row['col1']);
print_r($row['col2']); // undefined index thrown
or in a while
loop:
while( $row = fetching_function($query) ) {
echo $row['col1'];
echo "<br>";
echo $row['col2']; // undefined index thrown
echo "<br>";
echo $row['col3']; // undefined index thrown
}
Something else that needs to be noted is that on a *NIX OS and Mac OS X, things are case-sensitive.
Consult the followning Q&A's on Stack:
Are table names in MySQL case sensitive?
mysql case sensitive table names in queries
MySql - Case Sensitive issue of tables in different server
I have to wonder why this was downvoted also.
– Funk Forty Niner
Oct 25 '18 at 0:26
add a comment |
Another reason why an undefined index notice will be thrown, would be that a column was omitted from a database query.
I.e.:
$query = "SELECT col1 FROM table WHERE col_x = ?";
Then trying to access more columns/rows inside a loop.
I.e.:
print_r($row['col1']);
print_r($row['col2']); // undefined index thrown
or in a while
loop:
while( $row = fetching_function($query) ) {
echo $row['col1'];
echo "<br>";
echo $row['col2']; // undefined index thrown
echo "<br>";
echo $row['col3']; // undefined index thrown
}
Something else that needs to be noted is that on a *NIX OS and Mac OS X, things are case-sensitive.
Consult the followning Q&A's on Stack:
Are table names in MySQL case sensitive?
mysql case sensitive table names in queries
MySql - Case Sensitive issue of tables in different server
Another reason why an undefined index notice will be thrown, would be that a column was omitted from a database query.
I.e.:
$query = "SELECT col1 FROM table WHERE col_x = ?";
Then trying to access more columns/rows inside a loop.
I.e.:
print_r($row['col1']);
print_r($row['col2']); // undefined index thrown
or in a while
loop:
while( $row = fetching_function($query) ) {
echo $row['col1'];
echo "<br>";
echo $row['col2']; // undefined index thrown
echo "<br>";
echo $row['col3']; // undefined index thrown
}
Something else that needs to be noted is that on a *NIX OS and Mac OS X, things are case-sensitive.
Consult the followning Q&A's on Stack:
Are table names in MySQL case sensitive?
mysql case sensitive table names in queries
MySql - Case Sensitive issue of tables in different server
edited Oct 25 '17 at 10:15
community wiki
3 revs
Fred -ii-
I have to wonder why this was downvoted also.
– Funk Forty Niner
Oct 25 '18 at 0:26
add a comment |
I have to wonder why this was downvoted also.
– Funk Forty Niner
Oct 25 '18 at 0:26
I have to wonder why this was downvoted also.
– Funk Forty Niner
Oct 25 '18 at 0:26
I have to wonder why this was downvoted also.
– Funk Forty Niner
Oct 25 '18 at 0:26
add a comment |
Using a ternary is simple, readable, and clean:
Pre PHP 7
Assign a variable to the value of another variable if it's set, else assign null
(or whatever default value you need):
$newVariable = isset($thePotentialData) ? $thePotentialData : null;
PHP 7+
The same except using Null Coalescing Operator. There's no longer a need to call isset()
as this is built in, and no need to provide the variable to return as it's assumed to return the value of the variable being checked:
$newVariable = $thePotentialData ?? null;
Both will stop the Notices from the OP question, and both are the exact equivalent of:
if (isset($thePotentialData)) {
$newVariable = $thePotentialData;
} else {
$newVariable = null;
}
If you don't require setting a new variable then you can directly use the ternary's returned value, such as with echo
, function arguments, etc:
Echo:
echo 'Your name is: ' . isset($name) ? $name : 'You did not provide one';
Function:
$foreName = getForeName(isset($userId) ? $userId : null);
function getForeName($userId)
{
if ($userId === null) {
// Etc
}
}
The above will work just the same with arrays, including sessions etc, replacing the variable being checked with e.g.:$_SESSION['checkMe']
or however many levels deep you need, e.g.:$clients['personal']['address']['postcode']
Suppression:
It is possible to suppress the PHP Notices with @
or reduce your error reporting level, but it does not fix the problem, it simply stops it being reported in the error log. This means that your code still tried to use a variable that was not set, which may or may not mean something doesn't work as intended - depending on how crucial the missing value is.
You should really be checking for this issue and handling it appropriately, either serving a different message, or even just returning a null value for everything else to identify the precise state.
If you just care about the Notice not being in the error log, then as an option you could simply ignore the error log.
add a comment |
Using a ternary is simple, readable, and clean:
Pre PHP 7
Assign a variable to the value of another variable if it's set, else assign null
(or whatever default value you need):
$newVariable = isset($thePotentialData) ? $thePotentialData : null;
PHP 7+
The same except using Null Coalescing Operator. There's no longer a need to call isset()
as this is built in, and no need to provide the variable to return as it's assumed to return the value of the variable being checked:
$newVariable = $thePotentialData ?? null;
Both will stop the Notices from the OP question, and both are the exact equivalent of:
if (isset($thePotentialData)) {
$newVariable = $thePotentialData;
} else {
$newVariable = null;
}
If you don't require setting a new variable then you can directly use the ternary's returned value, such as with echo
, function arguments, etc:
Echo:
echo 'Your name is: ' . isset($name) ? $name : 'You did not provide one';
Function:
$foreName = getForeName(isset($userId) ? $userId : null);
function getForeName($userId)
{
if ($userId === null) {
// Etc
}
}
The above will work just the same with arrays, including sessions etc, replacing the variable being checked with e.g.:$_SESSION['checkMe']
or however many levels deep you need, e.g.:$clients['personal']['address']['postcode']
Suppression:
It is possible to suppress the PHP Notices with @
or reduce your error reporting level, but it does not fix the problem, it simply stops it being reported in the error log. This means that your code still tried to use a variable that was not set, which may or may not mean something doesn't work as intended - depending on how crucial the missing value is.
You should really be checking for this issue and handling it appropriately, either serving a different message, or even just returning a null value for everything else to identify the precise state.
If you just care about the Notice not being in the error log, then as an option you could simply ignore the error log.
add a comment |
Using a ternary is simple, readable, and clean:
Pre PHP 7
Assign a variable to the value of another variable if it's set, else assign null
(or whatever default value you need):
$newVariable = isset($thePotentialData) ? $thePotentialData : null;
PHP 7+
The same except using Null Coalescing Operator. There's no longer a need to call isset()
as this is built in, and no need to provide the variable to return as it's assumed to return the value of the variable being checked:
$newVariable = $thePotentialData ?? null;
Both will stop the Notices from the OP question, and both are the exact equivalent of:
if (isset($thePotentialData)) {
$newVariable = $thePotentialData;
} else {
$newVariable = null;
}
If you don't require setting a new variable then you can directly use the ternary's returned value, such as with echo
, function arguments, etc:
Echo:
echo 'Your name is: ' . isset($name) ? $name : 'You did not provide one';
Function:
$foreName = getForeName(isset($userId) ? $userId : null);
function getForeName($userId)
{
if ($userId === null) {
// Etc
}
}
The above will work just the same with arrays, including sessions etc, replacing the variable being checked with e.g.:$_SESSION['checkMe']
or however many levels deep you need, e.g.:$clients['personal']['address']['postcode']
Suppression:
It is possible to suppress the PHP Notices with @
or reduce your error reporting level, but it does not fix the problem, it simply stops it being reported in the error log. This means that your code still tried to use a variable that was not set, which may or may not mean something doesn't work as intended - depending on how crucial the missing value is.
You should really be checking for this issue and handling it appropriately, either serving a different message, or even just returning a null value for everything else to identify the precise state.
If you just care about the Notice not being in the error log, then as an option you could simply ignore the error log.
Using a ternary is simple, readable, and clean:
Pre PHP 7
Assign a variable to the value of another variable if it's set, else assign null
(or whatever default value you need):
$newVariable = isset($thePotentialData) ? $thePotentialData : null;
PHP 7+
The same except using Null Coalescing Operator. There's no longer a need to call isset()
as this is built in, and no need to provide the variable to return as it's assumed to return the value of the variable being checked:
$newVariable = $thePotentialData ?? null;
Both will stop the Notices from the OP question, and both are the exact equivalent of:
if (isset($thePotentialData)) {
$newVariable = $thePotentialData;
} else {
$newVariable = null;
}
If you don't require setting a new variable then you can directly use the ternary's returned value, such as with echo
, function arguments, etc:
Echo:
echo 'Your name is: ' . isset($name) ? $name : 'You did not provide one';
Function:
$foreName = getForeName(isset($userId) ? $userId : null);
function getForeName($userId)
{
if ($userId === null) {
// Etc
}
}
The above will work just the same with arrays, including sessions etc, replacing the variable being checked with e.g.:$_SESSION['checkMe']
or however many levels deep you need, e.g.:$clients['personal']['address']['postcode']
Suppression:
It is possible to suppress the PHP Notices with @
or reduce your error reporting level, but it does not fix the problem, it simply stops it being reported in the error log. This means that your code still tried to use a variable that was not set, which may or may not mean something doesn't work as intended - depending on how crucial the missing value is.
You should really be checking for this issue and handling it appropriately, either serving a different message, or even just returning a null value for everything else to identify the precise state.
If you just care about the Notice not being in the error log, then as an option you could simply ignore the error log.
answered Jul 6 '18 at 18:56
community wiki
James
add a comment |
add a comment |
Probably you were using old PHP version until and now upgraded PHP thats the reason it was working without any error till now from years.
until PHP4 there was no error if you are using variable without defining it but as of PHP5 onwards it throws errors for codes like mentioned in question.
add a comment |
Probably you were using old PHP version until and now upgraded PHP thats the reason it was working without any error till now from years.
until PHP4 there was no error if you are using variable without defining it but as of PHP5 onwards it throws errors for codes like mentioned in question.
add a comment |
Probably you were using old PHP version until and now upgraded PHP thats the reason it was working without any error till now from years.
until PHP4 there was no error if you are using variable without defining it but as of PHP5 onwards it throws errors for codes like mentioned in question.
Probably you were using old PHP version until and now upgraded PHP thats the reason it was working without any error till now from years.
until PHP4 there was no error if you are using variable without defining it but as of PHP5 onwards it throws errors for codes like mentioned in question.
answered Aug 30 '17 at 9:08
community wiki
phvish
add a comment |
add a comment |
When dealing with files, a proper enctype and a POST method are required, which will trigger an undefined index notice if either are not included in the form.
The manual states the following basic syntax:
HTML
<!-- The data encoding type, enctype, MUST be specified as below -->
<form enctype="multipart/form-data" action="__URL__" method="POST">
<!-- MAX_FILE_SIZE must precede the file input field -->
<input type="hidden" name="MAX_FILE_SIZE" value="30000" />
<!-- Name of input element determines name in $_FILES array -->
Send this file: <input name="userfile" type="file" />
<input type="submit" value="Send File" />
</form>
PHP
<?php
// In PHP versions earlier than 4.1.0, $HTTP_POST_FILES should be used instead
// of $_FILES.
$uploaddir = '/var/www/uploads/';
$uploadfile = $uploaddir . basename($_FILES['userfile']['name']);
echo '<pre>';
if (move_uploaded_file($_FILES['userfile']['tmp_name'], $uploadfile)) {
echo "File is valid, and was successfully uploaded.n";
} else {
echo "Possible file upload attack!n";
}
echo 'Here is some more debugging info:';
print_r($_FILES);
print "</pre>";
?>
Reference:
- http://php.net/manual/en/features.file-upload.post-method.php
add a comment |
When dealing with files, a proper enctype and a POST method are required, which will trigger an undefined index notice if either are not included in the form.
The manual states the following basic syntax:
HTML
<!-- The data encoding type, enctype, MUST be specified as below -->
<form enctype="multipart/form-data" action="__URL__" method="POST">
<!-- MAX_FILE_SIZE must precede the file input field -->
<input type="hidden" name="MAX_FILE_SIZE" value="30000" />
<!-- Name of input element determines name in $_FILES array -->
Send this file: <input name="userfile" type="file" />
<input type="submit" value="Send File" />
</form>
PHP
<?php
// In PHP versions earlier than 4.1.0, $HTTP_POST_FILES should be used instead
// of $_FILES.
$uploaddir = '/var/www/uploads/';
$uploadfile = $uploaddir . basename($_FILES['userfile']['name']);
echo '<pre>';
if (move_uploaded_file($_FILES['userfile']['tmp_name'], $uploadfile)) {
echo "File is valid, and was successfully uploaded.n";
} else {
echo "Possible file upload attack!n";
}
echo 'Here is some more debugging info:';
print_r($_FILES);
print "</pre>";
?>
Reference:
- http://php.net/manual/en/features.file-upload.post-method.php
add a comment |
When dealing with files, a proper enctype and a POST method are required, which will trigger an undefined index notice if either are not included in the form.
The manual states the following basic syntax:
HTML
<!-- The data encoding type, enctype, MUST be specified as below -->
<form enctype="multipart/form-data" action="__URL__" method="POST">
<!-- MAX_FILE_SIZE must precede the file input field -->
<input type="hidden" name="MAX_FILE_SIZE" value="30000" />
<!-- Name of input element determines name in $_FILES array -->
Send this file: <input name="userfile" type="file" />
<input type="submit" value="Send File" />
</form>
PHP
<?php
// In PHP versions earlier than 4.1.0, $HTTP_POST_FILES should be used instead
// of $_FILES.
$uploaddir = '/var/www/uploads/';
$uploadfile = $uploaddir . basename($_FILES['userfile']['name']);
echo '<pre>';
if (move_uploaded_file($_FILES['userfile']['tmp_name'], $uploadfile)) {
echo "File is valid, and was successfully uploaded.n";
} else {
echo "Possible file upload attack!n";
}
echo 'Here is some more debugging info:';
print_r($_FILES);
print "</pre>";
?>
Reference:
- http://php.net/manual/en/features.file-upload.post-method.php
When dealing with files, a proper enctype and a POST method are required, which will trigger an undefined index notice if either are not included in the form.
The manual states the following basic syntax:
HTML
<!-- The data encoding type, enctype, MUST be specified as below -->
<form enctype="multipart/form-data" action="__URL__" method="POST">
<!-- MAX_FILE_SIZE must precede the file input field -->
<input type="hidden" name="MAX_FILE_SIZE" value="30000" />
<!-- Name of input element determines name in $_FILES array -->
Send this file: <input name="userfile" type="file" />
<input type="submit" value="Send File" />
</form>
PHP
<?php
// In PHP versions earlier than 4.1.0, $HTTP_POST_FILES should be used instead
// of $_FILES.
$uploaddir = '/var/www/uploads/';
$uploadfile = $uploaddir . basename($_FILES['userfile']['name']);
echo '<pre>';
if (move_uploaded_file($_FILES['userfile']['tmp_name'], $uploadfile)) {
echo "File is valid, and was successfully uploaded.n";
} else {
echo "Possible file upload attack!n";
}
echo 'Here is some more debugging info:';
print_r($_FILES);
print "</pre>";
?>
Reference:
- http://php.net/manual/en/features.file-upload.post-method.php
answered Oct 7 '17 at 12:21
community wiki
Funk Forty Niner
add a comment |
add a comment |
One common cause of a variable not existing after an HTML form has been submitted is the form element is not contained within a <form>
tag:
Example: Element not contained within the <form>
<form action="example.php" method="post">
<p>
<input type="text" name="name" />
<input type="submit" value="Submit" />
</p>
</form>
<select name="choice">
<option value="choice1">choice 1</option>
<option value="choice2">choice 2</option>
<option value="choice3">choice 3</option>
<option value="choice4">choice 4</option>
</select>
Example: Element now contained within the <form>
<form action="example.php" method="post">
<select name="choice">
<option value="choice1">choice 1</option>
<option value="choice2">choice 2</option>
<option value="choice3">choice 3</option>
<option value="choice4">choice 4</option>
</select>
<p>
<input type="text" name="name" />
<input type="submit" value="Submit" />
</p>
</form>
add a comment |
One common cause of a variable not existing after an HTML form has been submitted is the form element is not contained within a <form>
tag:
Example: Element not contained within the <form>
<form action="example.php" method="post">
<p>
<input type="text" name="name" />
<input type="submit" value="Submit" />
</p>
</form>
<select name="choice">
<option value="choice1">choice 1</option>
<option value="choice2">choice 2</option>
<option value="choice3">choice 3</option>
<option value="choice4">choice 4</option>
</select>
Example: Element now contained within the <form>
<form action="example.php" method="post">
<select name="choice">
<option value="choice1">choice 1</option>
<option value="choice2">choice 2</option>
<option value="choice3">choice 3</option>
<option value="choice4">choice 4</option>
</select>
<p>
<input type="text" name="name" />
<input type="submit" value="Submit" />
</p>
</form>
add a comment |
One common cause of a variable not existing after an HTML form has been submitted is the form element is not contained within a <form>
tag:
Example: Element not contained within the <form>
<form action="example.php" method="post">
<p>
<input type="text" name="name" />
<input type="submit" value="Submit" />
</p>
</form>
<select name="choice">
<option value="choice1">choice 1</option>
<option value="choice2">choice 2</option>
<option value="choice3">choice 3</option>
<option value="choice4">choice 4</option>
</select>
Example: Element now contained within the <form>
<form action="example.php" method="post">
<select name="choice">
<option value="choice1">choice 1</option>
<option value="choice2">choice 2</option>
<option value="choice3">choice 3</option>
<option value="choice4">choice 4</option>
</select>
<p>
<input type="text" name="name" />
<input type="submit" value="Submit" />
</p>
</form>
One common cause of a variable not existing after an HTML form has been submitted is the form element is not contained within a <form>
tag:
Example: Element not contained within the <form>
<form action="example.php" method="post">
<p>
<input type="text" name="name" />
<input type="submit" value="Submit" />
</p>
</form>
<select name="choice">
<option value="choice1">choice 1</option>
<option value="choice2">choice 2</option>
<option value="choice3">choice 3</option>
<option value="choice4">choice 4</option>
</select>
Example: Element now contained within the <form>
<form action="example.php" method="post">
<select name="choice">
<option value="choice1">choice 1</option>
<option value="choice2">choice 2</option>
<option value="choice3">choice 3</option>
<option value="choice4">choice 4</option>
</select>
<p>
<input type="text" name="name" />
<input type="submit" value="Submit" />
</p>
</form>
answered Jul 5 '18 at 20:46
community wiki
John Conde
add a comment |
add a comment |
I asked a question about this and I was referred to this post with the message:
This question already has an answer here:
“Notice: Undefined variable”, “Notice: Undefined index”, and “Notice:
Undefined offset” using PHP
I am sharing my question and solution here:
This is the error:
Line 154 is the problem. This is what I have in line 154:
153 foreach($cities as $key => $city){
154 if(($city != 'London') && ($city != 'Madrid') && ($citiesCounterArray[$key] >= 1)){
I think the problem is that I am writing if conditions for the variable $city
, which is not the key but the value in $key => $city
. First, could you confirm if that is the cause of the warning? Second, if that is the problem, why is it that I cannot write a condition based on the value? Does it have to be with the key that I need to write the condition?
UPDATE 1: The problem is that when executing $citiesCounterArray[$key]
, sometimes the $key
corresponds to a key that does not exist in the $citiesCounterArray
array, but that is not always the case based on the data of my loop. What I need is to set a condition so that if $key
exists in the array, then run the code, otherwise, skip it.
UPDATE 2: This is how I fixed it by using array_key_exists()
:
foreach($cities as $key => $city){
if(array_key_exists($key, $citiesCounterArray)){
if(($city != 'London') && ($city != 'Madrid') && ($citiesCounterArray[$key] >= 1)){
add a comment |
I asked a question about this and I was referred to this post with the message:
This question already has an answer here:
“Notice: Undefined variable”, “Notice: Undefined index”, and “Notice:
Undefined offset” using PHP
I am sharing my question and solution here:
This is the error:
Line 154 is the problem. This is what I have in line 154:
153 foreach($cities as $key => $city){
154 if(($city != 'London') && ($city != 'Madrid') && ($citiesCounterArray[$key] >= 1)){
I think the problem is that I am writing if conditions for the variable $city
, which is not the key but the value in $key => $city
. First, could you confirm if that is the cause of the warning? Second, if that is the problem, why is it that I cannot write a condition based on the value? Does it have to be with the key that I need to write the condition?
UPDATE 1: The problem is that when executing $citiesCounterArray[$key]
, sometimes the $key
corresponds to a key that does not exist in the $citiesCounterArray
array, but that is not always the case based on the data of my loop. What I need is to set a condition so that if $key
exists in the array, then run the code, otherwise, skip it.
UPDATE 2: This is how I fixed it by using array_key_exists()
:
foreach($cities as $key => $city){
if(array_key_exists($key, $citiesCounterArray)){
if(($city != 'London') && ($city != 'Madrid') && ($citiesCounterArray[$key] >= 1)){
add a comment |
I asked a question about this and I was referred to this post with the message:
This question already has an answer here:
“Notice: Undefined variable”, “Notice: Undefined index”, and “Notice:
Undefined offset” using PHP
I am sharing my question and solution here:
This is the error:
Line 154 is the problem. This is what I have in line 154:
153 foreach($cities as $key => $city){
154 if(($city != 'London') && ($city != 'Madrid') && ($citiesCounterArray[$key] >= 1)){
I think the problem is that I am writing if conditions for the variable $city
, which is not the key but the value in $key => $city
. First, could you confirm if that is the cause of the warning? Second, if that is the problem, why is it that I cannot write a condition based on the value? Does it have to be with the key that I need to write the condition?
UPDATE 1: The problem is that when executing $citiesCounterArray[$key]
, sometimes the $key
corresponds to a key that does not exist in the $citiesCounterArray
array, but that is not always the case based on the data of my loop. What I need is to set a condition so that if $key
exists in the array, then run the code, otherwise, skip it.
UPDATE 2: This is how I fixed it by using array_key_exists()
:
foreach($cities as $key => $city){
if(array_key_exists($key, $citiesCounterArray)){
if(($city != 'London') && ($city != 'Madrid') && ($citiesCounterArray[$key] >= 1)){
I asked a question about this and I was referred to this post with the message:
This question already has an answer here:
“Notice: Undefined variable”, “Notice: Undefined index”, and “Notice:
Undefined offset” using PHP
I am sharing my question and solution here:
This is the error:
Line 154 is the problem. This is what I have in line 154:
153 foreach($cities as $key => $city){
154 if(($city != 'London') && ($city != 'Madrid') && ($citiesCounterArray[$key] >= 1)){
I think the problem is that I am writing if conditions for the variable $city
, which is not the key but the value in $key => $city
. First, could you confirm if that is the cause of the warning? Second, if that is the problem, why is it that I cannot write a condition based on the value? Does it have to be with the key that I need to write the condition?
UPDATE 1: The problem is that when executing $citiesCounterArray[$key]
, sometimes the $key
corresponds to a key that does not exist in the $citiesCounterArray
array, but that is not always the case based on the data of my loop. What I need is to set a condition so that if $key
exists in the array, then run the code, otherwise, skip it.
UPDATE 2: This is how I fixed it by using array_key_exists()
:
foreach($cities as $key => $city){
if(array_key_exists($key, $citiesCounterArray)){
if(($city != 'London') && ($city != 'Madrid') && ($citiesCounterArray[$key] >= 1)){
answered Sep 1 '18 at 18:31
community wiki
Jaime Montoya
add a comment |
add a comment |
Those notices are because you don't have the used variable defined
and my_index
key was not present into $my_array
variable.
Those notices were triggered every time, because your code
is not correct, but probably you didn't have the reporting of notices on.
Solve the bugs:
$my_variable_name = "Variable name"; // defining variable
echo "My variable value is: " . $my_variable_name;
if(isset($my_array["my_index"])){
echo "My index value is: " . $my_array["my_index"]; // check if my_index is set
}
Another way to get this out:
ini_set("error_reporting", false)
add a comment |
Those notices are because you don't have the used variable defined
and my_index
key was not present into $my_array
variable.
Those notices were triggered every time, because your code
is not correct, but probably you didn't have the reporting of notices on.
Solve the bugs:
$my_variable_name = "Variable name"; // defining variable
echo "My variable value is: " . $my_variable_name;
if(isset($my_array["my_index"])){
echo "My index value is: " . $my_array["my_index"]; // check if my_index is set
}
Another way to get this out:
ini_set("error_reporting", false)
add a comment |
Those notices are because you don't have the used variable defined
and my_index
key was not present into $my_array
variable.
Those notices were triggered every time, because your code
is not correct, but probably you didn't have the reporting of notices on.
Solve the bugs:
$my_variable_name = "Variable name"; // defining variable
echo "My variable value is: " . $my_variable_name;
if(isset($my_array["my_index"])){
echo "My index value is: " . $my_array["my_index"]; // check if my_index is set
}
Another way to get this out:
ini_set("error_reporting", false)
Those notices are because you don't have the used variable defined
and my_index
key was not present into $my_array
variable.
Those notices were triggered every time, because your code
is not correct, but probably you didn't have the reporting of notices on.
Solve the bugs:
$my_variable_name = "Variable name"; // defining variable
echo "My variable value is: " . $my_variable_name;
if(isset($my_array["my_index"])){
echo "My index value is: " . $my_array["my_index"]; // check if my_index is set
}
Another way to get this out:
ini_set("error_reporting", false)
answered Jan 25 '18 at 13:09
community wiki
Andrei Todorut
add a comment |
add a comment |
In PHP you need fist to define the variable after that you can use it.
We can check variable is defined or not in very efficient way!.
//If you only want to check variable has value and value has true and false value.
//But variable must be defined first.
if($my_variable_name){
}
//If you want to check variable is define or undefine
//Isset() does not check that variable has true or false value
//But it check null value of variable
if(isset($my_variable_name)){
}
Simple Explanation
//It will work with :- true,false,NULL
$defineVarialbe = false;
if($defineVarialbe){
echo "true";
}else{
echo "false";
}
//It will check variable is define or not and variable has null value.
if(isset($unDefineVarialbe)){
echo "true";
}else{
echo "false";
}
add a comment |
In PHP you need fist to define the variable after that you can use it.
We can check variable is defined or not in very efficient way!.
//If you only want to check variable has value and value has true and false value.
//But variable must be defined first.
if($my_variable_name){
}
//If you want to check variable is define or undefine
//Isset() does not check that variable has true or false value
//But it check null value of variable
if(isset($my_variable_name)){
}
Simple Explanation
//It will work with :- true,false,NULL
$defineVarialbe = false;
if($defineVarialbe){
echo "true";
}else{
echo "false";
}
//It will check variable is define or not and variable has null value.
if(isset($unDefineVarialbe)){
echo "true";
}else{
echo "false";
}
add a comment |
In PHP you need fist to define the variable after that you can use it.
We can check variable is defined or not in very efficient way!.
//If you only want to check variable has value and value has true and false value.
//But variable must be defined first.
if($my_variable_name){
}
//If you want to check variable is define or undefine
//Isset() does not check that variable has true or false value
//But it check null value of variable
if(isset($my_variable_name)){
}
Simple Explanation
//It will work with :- true,false,NULL
$defineVarialbe = false;
if($defineVarialbe){
echo "true";
}else{
echo "false";
}
//It will check variable is define or not and variable has null value.
if(isset($unDefineVarialbe)){
echo "true";
}else{
echo "false";
}
In PHP you need fist to define the variable after that you can use it.
We can check variable is defined or not in very efficient way!.
//If you only want to check variable has value and value has true and false value.
//But variable must be defined first.
if($my_variable_name){
}
//If you want to check variable is define or undefine
//Isset() does not check that variable has true or false value
//But it check null value of variable
if(isset($my_variable_name)){
}
Simple Explanation
//It will work with :- true,false,NULL
$defineVarialbe = false;
if($defineVarialbe){
echo "true";
}else{
echo "false";
}
//It will check variable is define or not and variable has null value.
if(isset($unDefineVarialbe)){
echo "true";
}else{
echo "false";
}
answered May 17 '18 at 2:43
community wiki
Manish Goswami
add a comment |
add a comment |
protected by Madara Uchiha♦ Dec 9 '12 at 12:00
Thank you for your interest in this question.
Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).
Would you like to answer one of these unanswered questions instead?
7
possible duplicate of Reference - What does this error mean in PHP?
– meagar♦
Oct 8 '13 at 15:25
2
the variable might not have been initialized. Are you initializing the variable from a post or get or any array? If that's the case you might not have an field in that array. That your accessing.
– Anish Silwal
Dec 15 '15 at 13:12
3
@Pekka웃 - I noticed the edit adding the "and "Notice: Undefined offset"" - Wouldn't it make more sense using "PHP: “Undefined variable”, “Undefined index”, “Undefined offset” notices" (even take out the PHP, since it is tagged as "php". Plus, the URL gets cut off at
and-notice-undef
, just a suggestion so that the URL doesn't get cut off. Maybe even removing the (too many) quotes. OrPHP: “Undefined variable/index/offset” notices
– Funk Forty Niner
Jan 20 '17 at 15:24
2
@Fred I guess an argument can be made for both variations. THere's a chance that newbies will enter the entire line, including the "Notice:" into their search query, which I'm sure is the main traffic generator for this question. If the messages are there in full, that's likely to improve visibility in search engines
– Pekka 웃
Jan 20 '17 at 15:34
2
@Pekka웃 I understand. I only said that because the URL didn't get cut off before and now it does at
and-notice-undef
. It was just a (few) suggestion(s). It just repeats itself also beingNotice: Undefined
.– Funk Forty Niner
Jan 20 '17 at 15:35