Ajax return call for error is not returning [duplicate]
This question already has an answer here:
How do I return the response from an asynchronous call?
33 answers
I have an Ajax call inside a function like so:
function send_data(url, data) {
$.ajax({
url: url,
type: 'POST',
data: data,
success: function (data) {
//Empty for now
},
error: function (xhr) {
return xhr.responseJSON;
},
});
}
and then I trigger this function like so:
console.log(send_data(url, data));
I should receive the response output, but I don't. I instead, get undefined. I know JavaScript is asynchronous, but shouldn't this work? I added a console.log output inside the error function, before the return statement, and it outputs the data correctly. So why doesn't it return the data?
javascript jquery ajax
asked Jan 1 at 7:57
user2896120user2896120
9861025
marked as duplicate by adiga, CertainPerformance javascript
Users with the javascript badge can single-handedly close javascript questions as duplicates and reopen them as needed.
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Jan 1 at 8:13
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
This question already has an answer here:
How do I return the response from an asynchronous call?
33 answers
I have an Ajax call inside a function like so:
function send_data(url, data) {
$.ajax({
url: url,
type: 'POST',
data: data,
success: function (data) {
//Empty for now
},
error: function (xhr) {
return xhr.responseJSON;
},
});
}
and then I trigger this function like so:
console.log(send_data(url, data));
I should receive the response output, but I don't. I instead, get undefined. I know JavaScript is asynchronous, but shouldn't this work? I added a console.log output inside the error function, before the return statement, and it outputs the data correctly. So why doesn't it return the data?
javascript jquery ajax
asked Jan 1 at 7:57
user2896120user2896120
9861025
marked as duplicate by adiga, CertainPerformance javascript
Users with the javascript badge can single-handedly close javascript questions as duplicates and reopen them as needed.
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$hover.hover(
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$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
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Jan 1 at 8:13
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
This question already has an answer here:
How do I return the response from an asynchronous call?
33 answers
I have an Ajax call inside a function like so:
function send_data(url, data) {
$.ajax({
url: url,
type: 'POST',
data: data,
success: function (data) {
//Empty for now
},
error: function (xhr) {
return xhr.responseJSON;
},
});
}
and then I trigger this function like so:
console.log(send_data(url, data));
I should receive the response output, but I don't. I instead, get undefined. I know JavaScript is asynchronous, but shouldn't this work? I added a console.log output inside the error function, before the return statement, and it outputs the data correctly. So why doesn't it return the data?
javascript jquery ajax
asked Jan 1 at 7:57
user2896120user2896120
9861025
This question already has an answer here:
How do I return the response from an asynchronous call?
33 answers
I have an Ajax call inside a function like so:
function send_data(url, data) {
$.ajax({
url: url,
type: 'POST',
data: data,
success: function (data) {
//Empty for now
},
error: function (xhr) {
return xhr.responseJSON;
},
});
}
and then I trigger this function like so:
console.log(send_data(url, data));
I should receive the response output, but I don't. I instead, get undefined. I know JavaScript is asynchronous, but shouldn't this work? I added a console.log output inside the error function, before the return statement, and it outputs the data correctly. So why doesn't it return the data?
This question already has an answer here:
How do I return the response from an asynchronous call?
33 answers
javascript jquery ajax
javascript jquery ajax
asked Jan 1 at 7:57
user2896120user2896120
9861025
asked Jan 1 at 7:57
user2896120user2896120
9861025
asked Jan 1 at 7:57
user2896120user2896120
9861025
asked Jan 1 at 7:57
user2896120user2896120
9861025
asked Jan 1 at 7:57
user2896120user2896120
9861025
9861025
marked as duplicate by adiga, CertainPerformance javascript
Users with the javascript badge can single-handedly close javascript questions as duplicates and reopen them as needed.
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$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
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Jan 1 at 8:13
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by adiga, CertainPerformance javascript
Users with the javascript badge can single-handedly close javascript questions as duplicates and reopen them as needed.
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$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
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Jan 1 at 8:13
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
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1 Answer
1
active
oldest
votes
what you do is basically declare function inside function
function foo(){
function bar(){console.log('in bar')}
}
console.log(foo())as you can see, the bar is not called (which in your case would be called by jQuery)
answered Jan 1 at 8:01
apple appleapple apple
2,6721823
I see, what is the best way of fixing this?
– user2896120
Jan 1 at 8:11
the basic would bereturn $.ajax(...), but you should learn aboutPromiseto make use of it.
– apple apple
Jan 1 at 8:13
Hmm, I see. Is there a way to handle 400 bad request errors? I'm getting a bad request because let's say not all the form elements are used. so a bad request will be thrown. Is there a way to handle them?
– user2896120
Jan 1 at 8:20
@user2896120 I don't use jQuery, but isn't theerrorcallback just for this?
– apple apple
Jan 1 at 8:22
So is it normal for the console to output 400 bad request each time? I was wondering if I could hide that
– user2896120
Jan 1 at 8:24
|
show 6 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
what you do is basically declare function inside function
function foo(){
function bar(){console.log('in bar')}
}
console.log(foo())as you can see, the bar is not called (which in your case would be called by jQuery)
answered Jan 1 at 8:01
apple appleapple apple
2,6721823
I see, what is the best way of fixing this?
– user2896120
Jan 1 at 8:11
the basic would bereturn $.ajax(...), but you should learn aboutPromiseto make use of it.
– apple apple
Jan 1 at 8:13
Hmm, I see. Is there a way to handle 400 bad request errors? I'm getting a bad request because let's say not all the form elements are used. so a bad request will be thrown. Is there a way to handle them?
– user2896120
Jan 1 at 8:20
@user2896120 I don't use jQuery, but isn't theerrorcallback just for this?
– apple apple
Jan 1 at 8:22
So is it normal for the console to output 400 bad request each time? I was wondering if I could hide that
– user2896120
Jan 1 at 8:24
|
show 6 more comments
what you do is basically declare function inside function
function foo(){
function bar(){console.log('in bar')}
}
console.log(foo())as you can see, the bar is not called (which in your case would be called by jQuery)
answered Jan 1 at 8:01
apple appleapple apple
2,6721823
I see, what is the best way of fixing this?
– user2896120
Jan 1 at 8:11
the basic would bereturn $.ajax(...), but you should learn aboutPromiseto make use of it.
– apple apple
Jan 1 at 8:13
Hmm, I see. Is there a way to handle 400 bad request errors? I'm getting a bad request because let's say not all the form elements are used. so a bad request will be thrown. Is there a way to handle them?
– user2896120
Jan 1 at 8:20
@user2896120 I don't use jQuery, but isn't theerrorcallback just for this?
– apple apple
Jan 1 at 8:22
So is it normal for the console to output 400 bad request each time? I was wondering if I could hide that
– user2896120
Jan 1 at 8:24
|
show 6 more comments
what you do is basically declare function inside function
function foo(){
function bar(){console.log('in bar')}
}
console.log(foo())as you can see, the bar is not called (which in your case would be called by jQuery)
answered Jan 1 at 8:01
apple appleapple apple
2,6721823
what you do is basically declare function inside function
function foo(){
function bar(){console.log('in bar')}
}
console.log(foo())as you can see, the bar is not called (which in your case would be called by jQuery)
function foo(){
function bar(){console.log('in bar')}
}
console.log(foo())function foo(){
function bar(){console.log('in bar')}
}
console.log(foo())answered Jan 1 at 8:01
apple appleapple apple
2,6721823
answered Jan 1 at 8:01
apple appleapple apple
2,6721823
answered Jan 1 at 8:01
apple appleapple apple
2,6721823
answered Jan 1 at 8:01
apple appleapple apple
2,6721823
2,6721823
I see, what is the best way of fixing this?
– user2896120
Jan 1 at 8:11
the basic would bereturn $.ajax(...), but you should learn aboutPromiseto make use of it.
– apple apple
Jan 1 at 8:13
Hmm, I see. Is there a way to handle 400 bad request errors? I'm getting a bad request because let's say not all the form elements are used. so a bad request will be thrown. Is there a way to handle them?
– user2896120
Jan 1 at 8:20
@user2896120 I don't use jQuery, but isn't theerrorcallback just for this?
– apple apple
Jan 1 at 8:22
So is it normal for the console to output 400 bad request each time? I was wondering if I could hide that
– user2896120
Jan 1 at 8:24
|
show 6 more comments
I see, what is the best way of fixing this?
– user2896120
Jan 1 at 8:11
the basic would bereturn $.ajax(...), but you should learn aboutPromiseto make use of it.
– apple apple
Jan 1 at 8:13
Hmm, I see. Is there a way to handle 400 bad request errors? I'm getting a bad request because let's say not all the form elements are used. so a bad request will be thrown. Is there a way to handle them?
– user2896120
Jan 1 at 8:20
@user2896120 I don't use jQuery, but isn't theerrorcallback just for this?
– apple apple
Jan 1 at 8:22
So is it normal for the console to output 400 bad request each time? I was wondering if I could hide that
– user2896120
Jan 1 at 8:24
I see, what is the best way of fixing this?
– user2896120
Jan 1 at 8:11
I see, what is the best way of fixing this?
– user2896120
Jan 1 at 8:11
the basic would be
return $.ajax(...), but you should learn about Promise to make use of it.– apple apple
Jan 1 at 8:13
the basic would be
return $.ajax(...), but you should learn about Promise to make use of it.– apple apple
Jan 1 at 8:13
Hmm, I see. Is there a way to handle 400 bad request errors? I'm getting a bad request because let's say not all the form elements are used. so a bad request will be thrown. Is there a way to handle them?
– user2896120
Jan 1 at 8:20
Hmm, I see. Is there a way to handle 400 bad request errors? I'm getting a bad request because let's say not all the form elements are used. so a bad request will be thrown. Is there a way to handle them?
– user2896120
Jan 1 at 8:20
@user2896120 I don't use jQuery, but isn't the
error callback just for this?– apple apple
Jan 1 at 8:22
@user2896120 I don't use jQuery, but isn't the
error callback just for this?– apple apple
Jan 1 at 8:22
So is it normal for the console to output 400 bad request each time? I was wondering if I could hide that
– user2896120
Jan 1 at 8:24
So is it normal for the console to output 400 bad request each time? I was wondering if I could hide that
– user2896120
Jan 1 at 8:24
|
show 6 more comments
