How to handle two separate responses from the server to a JQuery AJAX request?
I need to use AJAX to read and write some data from my server, in one operation. I don't want the write to hold up the read, so of course on the server I could read first and respond with that data, but then how do I alert the front end to an error with the write?
Front end is Javascript and JQuery, backend is PHP - or Laravel to be precise (tho I think this question is language agnostic).
My standard JQuery AJAX request:
$.ajax({
type: 'POST',
url: 'savedata',
dataType: 'json',
data: { data_to_save: JSON.stringify(mydata) }, // JSON encoded data
success: function (data_from_server)
{
// Update UI with data_from_server
},
error: function (data)
{
console.log('Error:', data.responseText);
}
});
}
Server side (psuedocode):
1. Read requested data from database (using a key in the POST params)
2. Reply to front end with requested data
3. Write data_to_save to the database
4. If there was an error in saving, how to notify the frontend as response has already been sent?
Or I could do two separate AJAX requests on the front end, one to read, one to write, but do I just list them one after the other in the code, or do I only fire the second in the success block of the first?
Two AJAX requests also seems much more inefficient on both front and back end.
javascript php jquery ajax laravel
edited Nov 22 at 19:53
Billal Begueradj
5,631132637
asked Nov 22 at 18:27
Alex Kerr
223118
add a comment |
I need to use AJAX to read and write some data from my server, in one operation. I don't want the write to hold up the read, so of course on the server I could read first and respond with that data, but then how do I alert the front end to an error with the write?
Front end is Javascript and JQuery, backend is PHP - or Laravel to be precise (tho I think this question is language agnostic).
My standard JQuery AJAX request:
$.ajax({
type: 'POST',
url: 'savedata',
dataType: 'json',
data: { data_to_save: JSON.stringify(mydata) }, // JSON encoded data
success: function (data_from_server)
{
// Update UI with data_from_server
},
error: function (data)
{
console.log('Error:', data.responseText);
}
});
}
Server side (psuedocode):
1. Read requested data from database (using a key in the POST params)
2. Reply to front end with requested data
3. Write data_to_save to the database
4. If there was an error in saving, how to notify the frontend as response has already been sent?
Or I could do two separate AJAX requests on the front end, one to read, one to write, but do I just list them one after the other in the code, or do I only fire the second in the success block of the first?
Two AJAX requests also seems much more inefficient on both front and back end.
javascript php jquery ajax laravel
edited Nov 22 at 19:53
Billal Begueradj
5,631132637
asked Nov 22 at 18:27
Alex Kerr
223118
Most db writes are very fast. Do you have some long process running before you can write to db?
– charlietfl
Nov 22 at 18:33
After you send a response you usually can't send another response, however you may be able to broadcast something to a client through a socket. However they can be tricky to set up properly
– apokryfos
Nov 22 at 18:46
Could you not wait until after the write to send the response?
– Ghostrydr
Nov 22 at 19:41
You can send short messages in either direction through sockets, laravel is set up nicely for this, I highly recommend checkout out the related videos on laracasts for this.
– wheelmaker
Nov 22 at 21:02
How did u go???
– adis
Nov 24 at 8:31
add a comment |
I need to use AJAX to read and write some data from my server, in one operation. I don't want the write to hold up the read, so of course on the server I could read first and respond with that data, but then how do I alert the front end to an error with the write?
Front end is Javascript and JQuery, backend is PHP - or Laravel to be precise (tho I think this question is language agnostic).
My standard JQuery AJAX request:
$.ajax({
type: 'POST',
url: 'savedata',
dataType: 'json',
data: { data_to_save: JSON.stringify(mydata) }, // JSON encoded data
success: function (data_from_server)
{
// Update UI with data_from_server
},
error: function (data)
{
console.log('Error:', data.responseText);
}
});
}
Server side (psuedocode):
1. Read requested data from database (using a key in the POST params)
2. Reply to front end with requested data
3. Write data_to_save to the database
4. If there was an error in saving, how to notify the frontend as response has already been sent?
Or I could do two separate AJAX requests on the front end, one to read, one to write, but do I just list them one after the other in the code, or do I only fire the second in the success block of the first?
Two AJAX requests also seems much more inefficient on both front and back end.
javascript php jquery ajax laravel
edited Nov 22 at 19:53
Billal Begueradj
5,631132637
asked Nov 22 at 18:27
Alex Kerr
223118
I need to use AJAX to read and write some data from my server, in one operation. I don't want the write to hold up the read, so of course on the server I could read first and respond with that data, but then how do I alert the front end to an error with the write?
Front end is Javascript and JQuery, backend is PHP - or Laravel to be precise (tho I think this question is language agnostic).
My standard JQuery AJAX request:
$.ajax({
type: 'POST',
url: 'savedata',
dataType: 'json',
data: { data_to_save: JSON.stringify(mydata) }, // JSON encoded data
success: function (data_from_server)
{
// Update UI with data_from_server
},
error: function (data)
{
console.log('Error:', data.responseText);
}
});
}
Server side (psuedocode):
1. Read requested data from database (using a key in the POST params)
2. Reply to front end with requested data
3. Write data_to_save to the database
4. If there was an error in saving, how to notify the frontend as response has already been sent?
Or I could do two separate AJAX requests on the front end, one to read, one to write, but do I just list them one after the other in the code, or do I only fire the second in the success block of the first?
Two AJAX requests also seems much more inefficient on both front and back end.
javascript php jquery ajax laravel
javascript php jquery ajax laravel
edited Nov 22 at 19:53
Billal Begueradj
5,631132637
asked Nov 22 at 18:27
Alex Kerr
223118
edited Nov 22 at 19:53
Billal Begueradj
5,631132637
asked Nov 22 at 18:27
Alex Kerr
223118
edited Nov 22 at 19:53
Billal Begueradj
5,631132637
edited Nov 22 at 19:53
Billal Begueradj
5,631132637
edited Nov 22 at 19:53
Billal Begueradj
5,631132637
5,631132637
asked Nov 22 at 18:27
Alex Kerr
223118
asked Nov 22 at 18:27
Alex Kerr
223118
asked Nov 22 at 18:27
Alex Kerr
223118
223118
Most db writes are very fast. Do you have some long process running before you can write to db?
– charlietfl
Nov 22 at 18:33
After you send a response you usually can't send another response, however you may be able to broadcast something to a client through a socket. However they can be tricky to set up properly
– apokryfos
Nov 22 at 18:46
Could you not wait until after the write to send the response?
– Ghostrydr
Nov 22 at 19:41
You can send short messages in either direction through sockets, laravel is set up nicely for this, I highly recommend checkout out the related videos on laracasts for this.
– wheelmaker
Nov 22 at 21:02
How did u go???
– adis
Nov 24 at 8:31
add a comment |
Most db writes are very fast. Do you have some long process running before you can write to db?
– charlietfl
Nov 22 at 18:33
After you send a response you usually can't send another response, however you may be able to broadcast something to a client through a socket. However they can be tricky to set up properly
– apokryfos
Nov 22 at 18:46
Could you not wait until after the write to send the response?
– Ghostrydr
Nov 22 at 19:41
You can send short messages in either direction through sockets, laravel is set up nicely for this, I highly recommend checkout out the related videos on laracasts for this.
– wheelmaker
Nov 22 at 21:02
How did u go???
– adis
Nov 24 at 8:31
Most db writes are very fast. Do you have some long process running before you can write to db?
– charlietfl
Nov 22 at 18:33
Most db writes are very fast. Do you have some long process running before you can write to db?
– charlietfl
Nov 22 at 18:33
After you send a response you usually can't send another response, however you may be able to broadcast something to a client through a socket. However they can be tricky to set up properly
– apokryfos
Nov 22 at 18:46
After you send a response you usually can't send another response, however you may be able to broadcast something to a client through a socket. However they can be tricky to set up properly
– apokryfos
Nov 22 at 18:46
Could you not wait until after the write to send the response?
– Ghostrydr
Nov 22 at 19:41
Could you not wait until after the write to send the response?
– Ghostrydr
Nov 22 at 19:41
You can send short messages in either direction through sockets, laravel is set up nicely for this, I highly recommend checkout out the related videos on laracasts for this.
– wheelmaker
Nov 22 at 21:02
You can send short messages in either direction through sockets, laravel is set up nicely for this, I highly recommend checkout out the related videos on laracasts for this.
– wheelmaker
Nov 22 at 21:02
How did u go???
– adis
Nov 24 at 8:31
How did u go???
– adis
Nov 24 at 8:31
add a comment |
2 Answers
2
active
oldest
votes
Well..... You can do it with 2 methods.
1. Splitting process.
I means first Ask for data. If error, Show error. If success request to Write data, get response If error show error.
2. send response after both operation
first read data, then write data. Send your response like,
{
"read":{
"status":1
},
"write":{
"status":0,
"error":"the error message"
}
}
answered Nov 22 at 18:47
Banujan Balendrakumar
512210
add a comment |
What you can do is, in the controller action of saving the data, use a try catch block.
PHP File
try
{
save_data()
}
catch(Exception $e)
{
return response->json(['message' => $e->getMessage() ], 500);
}
500 means server error status. You can define it yourself or get it from the Exception object. Using a try catch block ensures the ajax call to know that a success/ error has occured.
JS File
In the error function of ajax, you can add data argument, and console.log(data), it will catch the error from server side.
Goodluck!
edited Nov 22 at 19:44
Marius
867
answered Nov 22 at 19:06
adis
157114
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53436458%2fhow-to-handle-two-separate-responses-from-the-server-to-a-jquery-ajax-request%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Well..... You can do it with 2 methods.
1. Splitting process.
I means first Ask for data. If error, Show error. If success request to Write data, get response If error show error.
2. send response after both operation
first read data, then write data. Send your response like,
{
"read":{
"status":1
},
"write":{
"status":0,
"error":"the error message"
}
}
answered Nov 22 at 18:47
Banujan Balendrakumar
512210
add a comment |
Well..... You can do it with 2 methods.
1. Splitting process.
I means first Ask for data. If error, Show error. If success request to Write data, get response If error show error.
2. send response after both operation
first read data, then write data. Send your response like,
{
"read":{
"status":1
},
"write":{
"status":0,
"error":"the error message"
}
}
answered Nov 22 at 18:47
Banujan Balendrakumar
512210
add a comment |
Well..... You can do it with 2 methods.
1. Splitting process.
I means first Ask for data. If error, Show error. If success request to Write data, get response If error show error.
2. send response after both operation
first read data, then write data. Send your response like,
{
"read":{
"status":1
},
"write":{
"status":0,
"error":"the error message"
}
}
answered Nov 22 at 18:47
Banujan Balendrakumar
512210
Well..... You can do it with 2 methods.
1. Splitting process.
I means first Ask for data. If error, Show error. If success request to Write data, get response If error show error.
2. send response after both operation
first read data, then write data. Send your response like,
{
"read":{
"status":1
},
"write":{
"status":0,
"error":"the error message"
}
}
answered Nov 22 at 18:47
Banujan Balendrakumar
512210
answered Nov 22 at 18:47
Banujan Balendrakumar
512210
answered Nov 22 at 18:47
Banujan Balendrakumar
512210
answered Nov 22 at 18:47
Banujan Balendrakumar
512210
512210
add a comment |
add a comment |
What you can do is, in the controller action of saving the data, use a try catch block.
PHP File
try
{
save_data()
}
catch(Exception $e)
{
return response->json(['message' => $e->getMessage() ], 500);
}
500 means server error status. You can define it yourself or get it from the Exception object. Using a try catch block ensures the ajax call to know that a success/ error has occured.
JS File
In the error function of ajax, you can add data argument, and console.log(data), it will catch the error from server side.
Goodluck!
edited Nov 22 at 19:44
Marius
867
answered Nov 22 at 19:06
adis
157114
add a comment |
What you can do is, in the controller action of saving the data, use a try catch block.
PHP File
try
{
save_data()
}
catch(Exception $e)
{
return response->json(['message' => $e->getMessage() ], 500);
}
500 means server error status. You can define it yourself or get it from the Exception object. Using a try catch block ensures the ajax call to know that a success/ error has occured.
JS File
In the error function of ajax, you can add data argument, and console.log(data), it will catch the error from server side.
Goodluck!
edited Nov 22 at 19:44
Marius
867
answered Nov 22 at 19:06
adis
157114
add a comment |
What you can do is, in the controller action of saving the data, use a try catch block.
PHP File
try
{
save_data()
}
catch(Exception $e)
{
return response->json(['message' => $e->getMessage() ], 500);
}
500 means server error status. You can define it yourself or get it from the Exception object. Using a try catch block ensures the ajax call to know that a success/ error has occured.
JS File
In the error function of ajax, you can add data argument, and console.log(data), it will catch the error from server side.
Goodluck!
edited Nov 22 at 19:44
Marius
867
answered Nov 22 at 19:06
adis
157114
What you can do is, in the controller action of saving the data, use a try catch block.
PHP File
try
{
save_data()
}
catch(Exception $e)
{
return response->json(['message' => $e->getMessage() ], 500);
}
500 means server error status. You can define it yourself or get it from the Exception object. Using a try catch block ensures the ajax call to know that a success/ error has occured.
JS File
In the error function of ajax, you can add data argument, and console.log(data), it will catch the error from server side.
Goodluck!
edited Nov 22 at 19:44
Marius
867
answered Nov 22 at 19:06
adis
157114
edited Nov 22 at 19:44
Marius
867
edited Nov 22 at 19:44
Marius
867
edited Nov 22 at 19:44
Marius
867
867
answered Nov 22 at 19:06
adis
157114
answered Nov 22 at 19:06
adis
157114
answered Nov 22 at 19:06
adis
157114
157114
add a comment |
add a comment |
Thanks for contributing an answer to Stack Overflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53436458%2fhow-to-handle-two-separate-responses-from-the-server-to-a-jquery-ajax-request%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Most db writes are very fast. Do you have some long process running before you can write to db?
– charlietfl
Nov 22 at 18:33
After you send a response you usually can't send another response, however you may be able to broadcast something to a client through a socket. However they can be tricky to set up properly
– apokryfos
Nov 22 at 18:46
Could you not wait until after the write to send the response?
– Ghostrydr
Nov 22 at 19:41
You can send short messages in either direction through sockets, laravel is set up nicely for this, I highly recommend checkout out the related videos on laracasts for this.
– wheelmaker
Nov 22 at 21:02
How did u go???
– adis
Nov 24 at 8:31