Using Jquery Ajax to retrieve data from Mysql
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box;
}
29
list.php: A simple ajax code that I want to display only records of the Mysql table:
<html>
<head>
<script src="jquery-1.9.1.min.js">
</script>
<script>
$(document).ready(function() {
var response = '';
$.ajax({
type: "GET",
url: "Records.php",
async: false,
success: function(text) {
response = text;
}
});
alert(response);
});
</script>
</head>
<body>
<div id="div1">
<h2>Let jQuery AJAX Change This Text</h2>
</div>
<button>Get Records</button>
</body>
</html>
Records.php is the file to fetch records from Mysql.
In the Database are only two fields: 'Name', 'Address'.
<?php
//database name = "simple_ajax"
//table name = "users"
$con = mysql_connect("localhost","root","");
$dbs = mysql_select_db("simple_ajax",$con);
$result= mysql_query("select * from users");
$array = mysql_fetch_row($result);
?>
<tr>
<td>Name: </td>
<td>Address: </td>
</tr>
<?php
while ($row = mysql_fetch_array($result))
{
echo "<tr>";
echo "<td>$row[1]</td>";
echo "<td>$row[2]</td>";
echo "</tr>";
}
?>
This code is not working.
php mysql ajax jquery
edited Feb 19 at 7:07
Tushar Walzade
2,27931935
asked May 23 '13 at 7:06
RaviRavi
4021615
add a comment |
29
list.php: A simple ajax code that I want to display only records of the Mysql table:
<html>
<head>
<script src="jquery-1.9.1.min.js">
</script>
<script>
$(document).ready(function() {
var response = '';
$.ajax({
type: "GET",
url: "Records.php",
async: false,
success: function(text) {
response = text;
}
});
alert(response);
});
</script>
</head>
<body>
<div id="div1">
<h2>Let jQuery AJAX Change This Text</h2>
</div>
<button>Get Records</button>
</body>
</html>
Records.php is the file to fetch records from Mysql.
In the Database are only two fields: 'Name', 'Address'.
<?php
//database name = "simple_ajax"
//table name = "users"
$con = mysql_connect("localhost","root","");
$dbs = mysql_select_db("simple_ajax",$con);
$result= mysql_query("select * from users");
$array = mysql_fetch_row($result);
?>
<tr>
<td>Name: </td>
<td>Address: </td>
</tr>
<?php
while ($row = mysql_fetch_array($result))
{
echo "<tr>";
echo "<td>$row[1]</td>";
echo "<td>$row[2]</td>";
echo "</tr>";
}
?>
This code is not working.
php mysql ajax jquery
edited Feb 19 at 7:07
Tushar Walzade
2,27931935
asked May 23 '13 at 7:06
RaviRavi
4021615
2
What happens when you go toRecords.phpdirectly? Are there any error messages? You need to be more precise with whats not working.
– Phil Cross
May 23 '13 at 7:10
1
Are 1 and 2 names of columns in the table 'users'? If they are, try with echo "<td>".$row['1']."</td>"; In records.php
– V.Vachev
May 23 '13 at 7:12
@PhilCross: There is no any error is given it just put as a result for content on Respons.php , as file i have write as a output.
– Ravi
May 23 '13 at 7:55
1
What output do you exactly get ?
– Þaw
May 23 '13 at 8:20
add a comment |
29
29
29
26
list.php: A simple ajax code that I want to display only records of the Mysql table:
<html>
<head>
<script src="jquery-1.9.1.min.js">
</script>
<script>
$(document).ready(function() {
var response = '';
$.ajax({
type: "GET",
url: "Records.php",
async: false,
success: function(text) {
response = text;
}
});
alert(response);
});
</script>
</head>
<body>
<div id="div1">
<h2>Let jQuery AJAX Change This Text</h2>
</div>
<button>Get Records</button>
</body>
</html>
Records.php is the file to fetch records from Mysql.
In the Database are only two fields: 'Name', 'Address'.
<?php
//database name = "simple_ajax"
//table name = "users"
$con = mysql_connect("localhost","root","");
$dbs = mysql_select_db("simple_ajax",$con);
$result= mysql_query("select * from users");
$array = mysql_fetch_row($result);
?>
<tr>
<td>Name: </td>
<td>Address: </td>
</tr>
<?php
while ($row = mysql_fetch_array($result))
{
echo "<tr>";
echo "<td>$row[1]</td>";
echo "<td>$row[2]</td>";
echo "</tr>";
}
?>
This code is not working.
php mysql ajax jquery
edited Feb 19 at 7:07
Tushar Walzade
2,27931935
asked May 23 '13 at 7:06
RaviRavi
4021615
list.php: A simple ajax code that I want to display only records of the Mysql table:
<html>
<head>
<script src="jquery-1.9.1.min.js">
</script>
<script>
$(document).ready(function() {
var response = '';
$.ajax({
type: "GET",
url: "Records.php",
async: false,
success: function(text) {
response = text;
}
});
alert(response);
});
</script>
</head>
<body>
<div id="div1">
<h2>Let jQuery AJAX Change This Text</h2>
</div>
<button>Get Records</button>
</body>
</html>
Records.php is the file to fetch records from Mysql.
In the Database are only two fields: 'Name', 'Address'.
<?php
//database name = "simple_ajax"
//table name = "users"
$con = mysql_connect("localhost","root","");
$dbs = mysql_select_db("simple_ajax",$con);
$result= mysql_query("select * from users");
$array = mysql_fetch_row($result);
?>
<tr>
<td>Name: </td>
<td>Address: </td>
</tr>
<?php
while ($row = mysql_fetch_array($result))
{
echo "<tr>";
echo "<td>$row[1]</td>";
echo "<td>$row[2]</td>";
echo "</tr>";
}
?>
This code is not working.
php mysql ajax jquery
php mysql ajax jquery
edited Feb 19 at 7:07
Tushar Walzade
2,27931935
asked May 23 '13 at 7:06
RaviRavi
4021615
edited Feb 19 at 7:07
Tushar Walzade
2,27931935
asked May 23 '13 at 7:06
RaviRavi
4021615
edited Feb 19 at 7:07
Tushar Walzade
2,27931935
edited Feb 19 at 7:07
Tushar Walzade
2,27931935
edited Feb 19 at 7:07
Tushar Walzade
2,27931935
2,27931935
asked May 23 '13 at 7:06
RaviRavi
4021615
asked May 23 '13 at 7:06
RaviRavi
4021615
asked May 23 '13 at 7:06
RaviRavi
4021615
4021615
2
What happens when you go toRecords.phpdirectly? Are there any error messages? You need to be more precise with whats not working.
– Phil Cross
May 23 '13 at 7:10
1
Are 1 and 2 names of columns in the table 'users'? If they are, try with echo "<td>".$row['1']."</td>"; In records.php
– V.Vachev
May 23 '13 at 7:12
@PhilCross: There is no any error is given it just put as a result for content on Respons.php , as file i have write as a output.
– Ravi
May 23 '13 at 7:55
1
What output do you exactly get ?
– Þaw
May 23 '13 at 8:20
add a comment |
2
What happens when you go toRecords.phpdirectly? Are there any error messages? You need to be more precise with whats not working.
– Phil Cross
May 23 '13 at 7:10
1
Are 1 and 2 names of columns in the table 'users'? If they are, try with echo "<td>".$row['1']."</td>"; In records.php
– V.Vachev
May 23 '13 at 7:12
@PhilCross: There is no any error is given it just put as a result for content on Respons.php , as file i have write as a output.
– Ravi
May 23 '13 at 7:55
1
What output do you exactly get ?
– Þaw
May 23 '13 at 8:20
2
2
What happens when you go to
Records.php directly? Are there any error messages? You need to be more precise with whats not working.– Phil Cross
May 23 '13 at 7:10
What happens when you go to
Records.php directly? Are there any error messages? You need to be more precise with whats not working.– Phil Cross
May 23 '13 at 7:10
1
1
Are 1 and 2 names of columns in the table 'users'? If they are, try with echo "<td>".$row['1']."</td>"; In records.php
– V.Vachev
May 23 '13 at 7:12
Are 1 and 2 names of columns in the table 'users'? If they are, try with echo "<td>".$row['1']."</td>"; In records.php
– V.Vachev
May 23 '13 at 7:12
@PhilCross: There is no any error is given it just put as a result for content on Respons.php , as file i have write as a output.
– Ravi
May 23 '13 at 7:55
@PhilCross: There is no any error is given it just put as a result for content on Respons.php , as file i have write as a output.
– Ravi
May 23 '13 at 7:55
1
1
What output do you exactly get ?
– Þaw
May 23 '13 at 8:20
What output do you exactly get ?
– Þaw
May 23 '13 at 8:20
add a comment |
4 Answers
4
active
oldest
votes
53
For retrieving data using Ajax+jQuery, you must write the following code:
<html>
<script type="text/javascript" src="jquery-1.3.2.js"> </script>
<script type="text/javascript">
$(document).ready(function() {
$("#display").click(function() {
$.ajax({ //create an ajax request to display.php
type: "GET",
url: "display.php",
dataType: "html", //expect html to be returned
success: function(response){
$("#responsecontainer").html(response);
//alert(response);
}
});
});
});
</script>
<body>
<h3 align="center">Manage Student Details</h3>
<table border="1" align="center">
<tr>
<td> <input type="button" id="display" value="Display All Data" /> </td>
</tr>
</table>
<div id="responsecontainer" align="center">
</div>
</body>
</html>
For mysqli connection, write this:
<?php
$con=mysqli_connect("localhost","root","");
if (!$con)
{
die('Could not connect: ' . mysqli_error());
}
?>
For displaying the data from database, you must write this :
<?php
include("connection.php");
mysqli_select_db("samples",$con);
$result=mysqli_query("select * from student",$con);
echo "<table border='1' >
<tr>
<td align=center> <b>Roll No</b></td>
<td align=center><b>Name</b></td>
<td align=center><b>Address</b></td>
<td align=center><b>Stream</b></td></td>
<td align=center><b>Status</b></td>";
while($data = mysqli_fetch_row($result))
{
echo "<tr>";
echo "<td align=center>$data[0]</td>";
echo "<td align=center>$data[1]</td>";
echo "<td align=center>$data[2]</td>";
echo "<td align=center>$data[3]</td>";
echo "<td align=center>$data[4]</td>";
echo "</tr>";
}
echo "</table>";
?>
edited Nov 21 '17 at 11:21
Community♦
11
answered Jun 3 '13 at 10:24
Neha GandhiNeha Gandhi
69155
7
Just a note: There is no more support formysql_*functions, they are officially deprecated, no longer maintained and will be removed in the future. You should update your code with PDO or MySQLi to ensure the functionality of your project in the future.
– Amal Murali
Apr 23 '14 at 21:57
Hello Guys, I tried above code, but its returning full display file data along with php. Please can you let me know what can be issue. I am .net developer just working on php.
– user3816325
May 19 '16 at 3:50
@user3816325 check your server. I believe your server isnt running.
– prakashchhetri
Sep 10 '16 at 13:23
add a comment |
4
You can't return ajax return value. You stored global variable store your return values after return.
Or Change ur code like this one.
AjaxGet = function (url) {
var result = $.ajax({
type: "POST",
url: url,
param: '{}',
contentType: "application/json; charset=utf-8",
dataType: "json",
async: false,
success: function (data) {
// nothing needed here
}
}) .responseText ;
return result;
}
answered May 23 '13 at 7:25
Mariselvam PanneerselvamMariselvam Panneerselvam
10619
without Json i have to do.
– Ravi
May 23 '13 at 7:59
@RaviParmar change datatype "html".This is for example.
– Mariselvam Panneerselvam
May 23 '13 at 8:49
add a comment |
2
Please make sure your $row[1] , $row[2] contains correct value, we do assume here that 1 = Name , and 2 here is your Address field ?
Assuming you have correctly fetched your records from your Records.php, You can do something like this:
$(document).ready(function()
{
$('#getRecords').click(function()
{
var response = '';
$.ajax({ type: 'POST',
url: "Records.php",
async: false,
success : function(text){
$('#table1').html(text);
}
});
});
}
In your HTML
<table id="table1">
//Let jQuery AJAX Change This Text
</table>
<button id='getRecords'>Get Records</button>
A little note:
Try learing PDO http://php.net/manual/en/class.pdo.php since mysql_* functions are no longer encouraged..
answered May 23 '13 at 8:26
ÞawÞaw
1,55021535
add a comment |
1
$(document).ready(function(){
var response = '';
$.ajax({ type: "GET",
url: "Records.php",
async: false,
success : function(text)
{
response = text;
}
});
alert(response);
});
needs to be:
$(document).ready(function(){
$.ajax({ type: "GET",
url: "Records.php",
async: false,
success : function(text)
{
alert(text);
}
});
});
answered May 23 '13 at 8:15
gitaarikgitaarik
20.5k57176
add a comment |
protected by Community♦ Feb 19 '14 at 5:43
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
53
For retrieving data using Ajax+jQuery, you must write the following code:
<html>
<script type="text/javascript" src="jquery-1.3.2.js"> </script>
<script type="text/javascript">
$(document).ready(function() {
$("#display").click(function() {
$.ajax({ //create an ajax request to display.php
type: "GET",
url: "display.php",
dataType: "html", //expect html to be returned
success: function(response){
$("#responsecontainer").html(response);
//alert(response);
}
});
});
});
</script>
<body>
<h3 align="center">Manage Student Details</h3>
<table border="1" align="center">
<tr>
<td> <input type="button" id="display" value="Display All Data" /> </td>
</tr>
</table>
<div id="responsecontainer" align="center">
</div>
</body>
</html>
For mysqli connection, write this:
<?php
$con=mysqli_connect("localhost","root","");
if (!$con)
{
die('Could not connect: ' . mysqli_error());
}
?>
For displaying the data from database, you must write this :
<?php
include("connection.php");
mysqli_select_db("samples",$con);
$result=mysqli_query("select * from student",$con);
echo "<table border='1' >
<tr>
<td align=center> <b>Roll No</b></td>
<td align=center><b>Name</b></td>
<td align=center><b>Address</b></td>
<td align=center><b>Stream</b></td></td>
<td align=center><b>Status</b></td>";
while($data = mysqli_fetch_row($result))
{
echo "<tr>";
echo "<td align=center>$data[0]</td>";
echo "<td align=center>$data[1]</td>";
echo "<td align=center>$data[2]</td>";
echo "<td align=center>$data[3]</td>";
echo "<td align=center>$data[4]</td>";
echo "</tr>";
}
echo "</table>";
?>
edited Nov 21 '17 at 11:21
Community♦
11
answered Jun 3 '13 at 10:24
Neha GandhiNeha Gandhi
69155
7
Just a note: There is no more support formysql_*functions, they are officially deprecated, no longer maintained and will be removed in the future. You should update your code with PDO or MySQLi to ensure the functionality of your project in the future.
– Amal Murali
Apr 23 '14 at 21:57
Hello Guys, I tried above code, but its returning full display file data along with php. Please can you let me know what can be issue. I am .net developer just working on php.
– user3816325
May 19 '16 at 3:50
@user3816325 check your server. I believe your server isnt running.
– prakashchhetri
Sep 10 '16 at 13:23
add a comment |
53
For retrieving data using Ajax+jQuery, you must write the following code:
<html>
<script type="text/javascript" src="jquery-1.3.2.js"> </script>
<script type="text/javascript">
$(document).ready(function() {
$("#display").click(function() {
$.ajax({ //create an ajax request to display.php
type: "GET",
url: "display.php",
dataType: "html", //expect html to be returned
success: function(response){
$("#responsecontainer").html(response);
//alert(response);
}
});
});
});
</script>
<body>
<h3 align="center">Manage Student Details</h3>
<table border="1" align="center">
<tr>
<td> <input type="button" id="display" value="Display All Data" /> </td>
</tr>
</table>
<div id="responsecontainer" align="center">
</div>
</body>
</html>
For mysqli connection, write this:
<?php
$con=mysqli_connect("localhost","root","");
if (!$con)
{
die('Could not connect: ' . mysqli_error());
}
?>
For displaying the data from database, you must write this :
<?php
include("connection.php");
mysqli_select_db("samples",$con);
$result=mysqli_query("select * from student",$con);
echo "<table border='1' >
<tr>
<td align=center> <b>Roll No</b></td>
<td align=center><b>Name</b></td>
<td align=center><b>Address</b></td>
<td align=center><b>Stream</b></td></td>
<td align=center><b>Status</b></td>";
while($data = mysqli_fetch_row($result))
{
echo "<tr>";
echo "<td align=center>$data[0]</td>";
echo "<td align=center>$data[1]</td>";
echo "<td align=center>$data[2]</td>";
echo "<td align=center>$data[3]</td>";
echo "<td align=center>$data[4]</td>";
echo "</tr>";
}
echo "</table>";
?>
edited Nov 21 '17 at 11:21
Community♦
11
answered Jun 3 '13 at 10:24
Neha GandhiNeha Gandhi
69155
7
Just a note: There is no more support formysql_*functions, they are officially deprecated, no longer maintained and will be removed in the future. You should update your code with PDO or MySQLi to ensure the functionality of your project in the future.
– Amal Murali
Apr 23 '14 at 21:57
Hello Guys, I tried above code, but its returning full display file data along with php. Please can you let me know what can be issue. I am .net developer just working on php.
– user3816325
May 19 '16 at 3:50
@user3816325 check your server. I believe your server isnt running.
– prakashchhetri
Sep 10 '16 at 13:23
add a comment |
53
53
53
For retrieving data using Ajax+jQuery, you must write the following code:
<html>
<script type="text/javascript" src="jquery-1.3.2.js"> </script>
<script type="text/javascript">
$(document).ready(function() {
$("#display").click(function() {
$.ajax({ //create an ajax request to display.php
type: "GET",
url: "display.php",
dataType: "html", //expect html to be returned
success: function(response){
$("#responsecontainer").html(response);
//alert(response);
}
});
});
});
</script>
<body>
<h3 align="center">Manage Student Details</h3>
<table border="1" align="center">
<tr>
<td> <input type="button" id="display" value="Display All Data" /> </td>
</tr>
</table>
<div id="responsecontainer" align="center">
</div>
</body>
</html>
For mysqli connection, write this:
<?php
$con=mysqli_connect("localhost","root","");
if (!$con)
{
die('Could not connect: ' . mysqli_error());
}
?>
For displaying the data from database, you must write this :
<?php
include("connection.php");
mysqli_select_db("samples",$con);
$result=mysqli_query("select * from student",$con);
echo "<table border='1' >
<tr>
<td align=center> <b>Roll No</b></td>
<td align=center><b>Name</b></td>
<td align=center><b>Address</b></td>
<td align=center><b>Stream</b></td></td>
<td align=center><b>Status</b></td>";
while($data = mysqli_fetch_row($result))
{
echo "<tr>";
echo "<td align=center>$data[0]</td>";
echo "<td align=center>$data[1]</td>";
echo "<td align=center>$data[2]</td>";
echo "<td align=center>$data[3]</td>";
echo "<td align=center>$data[4]</td>";
echo "</tr>";
}
echo "</table>";
?>
edited Nov 21 '17 at 11:21
Community♦
11
answered Jun 3 '13 at 10:24
Neha GandhiNeha Gandhi
69155
For retrieving data using Ajax+jQuery, you must write the following code:
<html>
<script type="text/javascript" src="jquery-1.3.2.js"> </script>
<script type="text/javascript">
$(document).ready(function() {
$("#display").click(function() {
$.ajax({ //create an ajax request to display.php
type: "GET",
url: "display.php",
dataType: "html", //expect html to be returned
success: function(response){
$("#responsecontainer").html(response);
//alert(response);
}
});
});
});
</script>
<body>
<h3 align="center">Manage Student Details</h3>
<table border="1" align="center">
<tr>
<td> <input type="button" id="display" value="Display All Data" /> </td>
</tr>
</table>
<div id="responsecontainer" align="center">
</div>
</body>
</html>
For mysqli connection, write this:
<?php
$con=mysqli_connect("localhost","root","");
if (!$con)
{
die('Could not connect: ' . mysqli_error());
}
?>
For displaying the data from database, you must write this :
<?php
include("connection.php");
mysqli_select_db("samples",$con);
$result=mysqli_query("select * from student",$con);
echo "<table border='1' >
<tr>
<td align=center> <b>Roll No</b></td>
<td align=center><b>Name</b></td>
<td align=center><b>Address</b></td>
<td align=center><b>Stream</b></td></td>
<td align=center><b>Status</b></td>";
while($data = mysqli_fetch_row($result))
{
echo "<tr>";
echo "<td align=center>$data[0]</td>";
echo "<td align=center>$data[1]</td>";
echo "<td align=center>$data[2]</td>";
echo "<td align=center>$data[3]</td>";
echo "<td align=center>$data[4]</td>";
echo "</tr>";
}
echo "</table>";
?>
edited Nov 21 '17 at 11:21
Community♦
11
answered Jun 3 '13 at 10:24
Neha GandhiNeha Gandhi
69155
edited Nov 21 '17 at 11:21
Community♦
11
edited Nov 21 '17 at 11:21
Community♦
11
edited Nov 21 '17 at 11:21
Community♦
11
11
answered Jun 3 '13 at 10:24
Neha GandhiNeha Gandhi
69155
answered Jun 3 '13 at 10:24
Neha GandhiNeha Gandhi
69155
answered Jun 3 '13 at 10:24
Neha GandhiNeha Gandhi
69155
69155
7
Just a note: There is no more support formysql_*functions, they are officially deprecated, no longer maintained and will be removed in the future. You should update your code with PDO or MySQLi to ensure the functionality of your project in the future.
– Amal Murali
Apr 23 '14 at 21:57
Hello Guys, I tried above code, but its returning full display file data along with php. Please can you let me know what can be issue. I am .net developer just working on php.
– user3816325
May 19 '16 at 3:50
@user3816325 check your server. I believe your server isnt running.
– prakashchhetri
Sep 10 '16 at 13:23
add a comment |
7
Just a note: There is no more support formysql_*functions, they are officially deprecated, no longer maintained and will be removed in the future. You should update your code with PDO or MySQLi to ensure the functionality of your project in the future.
– Amal Murali
Apr 23 '14 at 21:57
Hello Guys, I tried above code, but its returning full display file data along with php. Please can you let me know what can be issue. I am .net developer just working on php.
– user3816325
May 19 '16 at 3:50
@user3816325 check your server. I believe your server isnt running.
– prakashchhetri
Sep 10 '16 at 13:23
7
7
Just a note: There is no more support for
mysql_* functions, they are officially deprecated, no longer maintained and will be removed in the future. You should update your code with PDO or MySQLi to ensure the functionality of your project in the future.– Amal Murali
Apr 23 '14 at 21:57
Just a note: There is no more support for
mysql_* functions, they are officially deprecated, no longer maintained and will be removed in the future. You should update your code with PDO or MySQLi to ensure the functionality of your project in the future.– Amal Murali
Apr 23 '14 at 21:57
Hello Guys, I tried above code, but its returning full display file data along with php. Please can you let me know what can be issue. I am .net developer just working on php.
– user3816325
May 19 '16 at 3:50
Hello Guys, I tried above code, but its returning full display file data along with php. Please can you let me know what can be issue. I am .net developer just working on php.
– user3816325
May 19 '16 at 3:50
@user3816325 check your server. I believe your server isnt running.
– prakashchhetri
Sep 10 '16 at 13:23
@user3816325 check your server. I believe your server isnt running.
– prakashchhetri
Sep 10 '16 at 13:23
add a comment |
4
You can't return ajax return value. You stored global variable store your return values after return.
Or Change ur code like this one.
AjaxGet = function (url) {
var result = $.ajax({
type: "POST",
url: url,
param: '{}',
contentType: "application/json; charset=utf-8",
dataType: "json",
async: false,
success: function (data) {
// nothing needed here
}
}) .responseText ;
return result;
}
answered May 23 '13 at 7:25
Mariselvam PanneerselvamMariselvam Panneerselvam
10619
without Json i have to do.
– Ravi
May 23 '13 at 7:59
@RaviParmar change datatype "html".This is for example.
– Mariselvam Panneerselvam
May 23 '13 at 8:49
add a comment |
4
You can't return ajax return value. You stored global variable store your return values after return.
Or Change ur code like this one.
AjaxGet = function (url) {
var result = $.ajax({
type: "POST",
url: url,
param: '{}',
contentType: "application/json; charset=utf-8",
dataType: "json",
async: false,
success: function (data) {
// nothing needed here
}
}) .responseText ;
return result;
}
answered May 23 '13 at 7:25
Mariselvam PanneerselvamMariselvam Panneerselvam
10619
without Json i have to do.
– Ravi
May 23 '13 at 7:59
@RaviParmar change datatype "html".This is for example.
– Mariselvam Panneerselvam
May 23 '13 at 8:49
add a comment |
4
4
4
You can't return ajax return value. You stored global variable store your return values after return.
Or Change ur code like this one.
AjaxGet = function (url) {
var result = $.ajax({
type: "POST",
url: url,
param: '{}',
contentType: "application/json; charset=utf-8",
dataType: "json",
async: false,
success: function (data) {
// nothing needed here
}
}) .responseText ;
return result;
}
answered May 23 '13 at 7:25
Mariselvam PanneerselvamMariselvam Panneerselvam
10619
You can't return ajax return value. You stored global variable store your return values after return.
Or Change ur code like this one.
AjaxGet = function (url) {
var result = $.ajax({
type: "POST",
url: url,
param: '{}',
contentType: "application/json; charset=utf-8",
dataType: "json",
async: false,
success: function (data) {
// nothing needed here
}
}) .responseText ;
return result;
}
answered May 23 '13 at 7:25
Mariselvam PanneerselvamMariselvam Panneerselvam
10619
answered May 23 '13 at 7:25
Mariselvam PanneerselvamMariselvam Panneerselvam
10619
answered May 23 '13 at 7:25
Mariselvam PanneerselvamMariselvam Panneerselvam
10619
answered May 23 '13 at 7:25
Mariselvam PanneerselvamMariselvam Panneerselvam
10619
10619
without Json i have to do.
– Ravi
May 23 '13 at 7:59
@RaviParmar change datatype "html".This is for example.
– Mariselvam Panneerselvam
May 23 '13 at 8:49
add a comment |
without Json i have to do.
– Ravi
May 23 '13 at 7:59
@RaviParmar change datatype "html".This is for example.
– Mariselvam Panneerselvam
May 23 '13 at 8:49
without Json i have to do.
– Ravi
May 23 '13 at 7:59
without Json i have to do.
– Ravi
May 23 '13 at 7:59
@RaviParmar change datatype "html".This is for example.
– Mariselvam Panneerselvam
May 23 '13 at 8:49
@RaviParmar change datatype "html".This is for example.
– Mariselvam Panneerselvam
May 23 '13 at 8:49
add a comment |
2
Please make sure your $row[1] , $row[2] contains correct value, we do assume here that 1 = Name , and 2 here is your Address field ?
Assuming you have correctly fetched your records from your Records.php, You can do something like this:
$(document).ready(function()
{
$('#getRecords').click(function()
{
var response = '';
$.ajax({ type: 'POST',
url: "Records.php",
async: false,
success : function(text){
$('#table1').html(text);
}
});
});
}
In your HTML
<table id="table1">
//Let jQuery AJAX Change This Text
</table>
<button id='getRecords'>Get Records</button>
A little note:
Try learing PDO http://php.net/manual/en/class.pdo.php since mysql_* functions are no longer encouraged..
answered May 23 '13 at 8:26
ÞawÞaw
1,55021535
add a comment |
2
Please make sure your $row[1] , $row[2] contains correct value, we do assume here that 1 = Name , and 2 here is your Address field ?
Assuming you have correctly fetched your records from your Records.php, You can do something like this:
$(document).ready(function()
{
$('#getRecords').click(function()
{
var response = '';
$.ajax({ type: 'POST',
url: "Records.php",
async: false,
success : function(text){
$('#table1').html(text);
}
});
});
}
In your HTML
<table id="table1">
//Let jQuery AJAX Change This Text
</table>
<button id='getRecords'>Get Records</button>
A little note:
Try learing PDO http://php.net/manual/en/class.pdo.php since mysql_* functions are no longer encouraged..
answered May 23 '13 at 8:26
ÞawÞaw
1,55021535
add a comment |
2
2
2
Please make sure your $row[1] , $row[2] contains correct value, we do assume here that 1 = Name , and 2 here is your Address field ?
Assuming you have correctly fetched your records from your Records.php, You can do something like this:
$(document).ready(function()
{
$('#getRecords').click(function()
{
var response = '';
$.ajax({ type: 'POST',
url: "Records.php",
async: false,
success : function(text){
$('#table1').html(text);
}
});
});
}
In your HTML
<table id="table1">
//Let jQuery AJAX Change This Text
</table>
<button id='getRecords'>Get Records</button>
A little note:
Try learing PDO http://php.net/manual/en/class.pdo.php since mysql_* functions are no longer encouraged..
answered May 23 '13 at 8:26
ÞawÞaw
1,55021535
Please make sure your $row[1] , $row[2] contains correct value, we do assume here that 1 = Name , and 2 here is your Address field ?
Assuming you have correctly fetched your records from your Records.php, You can do something like this:
$(document).ready(function()
{
$('#getRecords').click(function()
{
var response = '';
$.ajax({ type: 'POST',
url: "Records.php",
async: false,
success : function(text){
$('#table1').html(text);
}
});
});
}
In your HTML
<table id="table1">
//Let jQuery AJAX Change This Text
</table>
<button id='getRecords'>Get Records</button>
A little note:
Try learing PDO http://php.net/manual/en/class.pdo.php since mysql_* functions are no longer encouraged..
answered May 23 '13 at 8:26
ÞawÞaw
1,55021535
edited May 23 '13 at 8:32
answered May 23 '13 at 8:26
ÞawÞaw
1,55021535
answered May 23 '13 at 8:26
ÞawÞaw
1,55021535
answered May 23 '13 at 8:26
ÞawÞaw
1,55021535
1,55021535
add a comment |
add a comment |
1
$(document).ready(function(){
var response = '';
$.ajax({ type: "GET",
url: "Records.php",
async: false,
success : function(text)
{
response = text;
}
});
alert(response);
});
needs to be:
$(document).ready(function(){
$.ajax({ type: "GET",
url: "Records.php",
async: false,
success : function(text)
{
alert(text);
}
});
});
answered May 23 '13 at 8:15
gitaarikgitaarik
20.5k57176
add a comment |
1
$(document).ready(function(){
var response = '';
$.ajax({ type: "GET",
url: "Records.php",
async: false,
success : function(text)
{
response = text;
}
});
alert(response);
});
needs to be:
$(document).ready(function(){
$.ajax({ type: "GET",
url: "Records.php",
async: false,
success : function(text)
{
alert(text);
}
});
});
answered May 23 '13 at 8:15
gitaarikgitaarik
20.5k57176
add a comment |
1
1
1
$(document).ready(function(){
var response = '';
$.ajax({ type: "GET",
url: "Records.php",
async: false,
success : function(text)
{
response = text;
}
});
alert(response);
});
needs to be:
$(document).ready(function(){
$.ajax({ type: "GET",
url: "Records.php",
async: false,
success : function(text)
{
alert(text);
}
});
});
answered May 23 '13 at 8:15
gitaarikgitaarik
20.5k57176
$(document).ready(function(){
var response = '';
$.ajax({ type: "GET",
url: "Records.php",
async: false,
success : function(text)
{
response = text;
}
});
alert(response);
});
needs to be:
$(document).ready(function(){
$.ajax({ type: "GET",
url: "Records.php",
async: false,
success : function(text)
{
alert(text);
}
});
});
answered May 23 '13 at 8:15
gitaarikgitaarik
20.5k57176
edited May 23 '13 at 8:29
answered May 23 '13 at 8:15
gitaarikgitaarik
20.5k57176
answered May 23 '13 at 8:15
gitaarikgitaarik
20.5k57176
answered May 23 '13 at 8:15
gitaarikgitaarik
20.5k57176
20.5k57176
add a comment |
add a comment |
protected by Community♦ Feb 19 '14 at 5:43
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2
What happens when you go to
Records.phpdirectly? Are there any error messages? You need to be more precise with whats not working.– Phil Cross
May 23 '13 at 7:10
1
Are 1 and 2 names of columns in the table 'users'? If they are, try with echo "<td>".$row['1']."</td>"; In records.php
– V.Vachev
May 23 '13 at 7:12
@PhilCross: There is no any error is given it just put as a result for content on Respons.php , as file i have write as a output.
– Ravi
May 23 '13 at 7:55
1
What output do you exactly get ?
– Þaw
May 23 '13 at 8:20