VM2022:2 Uncaught SyntaxError: Unexpected token : on AJAX call
I send a GET request from the back end to get the json response. I got this error.
Uncaught SyntaxError: Unexpected token :
at DOMEval (jquery-3.3.1.js:111)
at Function.globalEval (jquery-3.3.1.js:345)
at text script (jquery-3.3.1.js:9640)
at ajaxConvert (jquery-3.3.1.js:8787)
at done (jquery-3.3.1.js:9255)
at XMLHttpRequest. (jquery-3.3.1.js:9548)
My AJAX request code is:
$.ajax(
type: "GET", //rest Type
dataType: 'jsonp', //mispelled
url: " url_for('live') ",
contentType: "application/json",
success: function (msg)
console.log(msg);
for (var i = 0; i < msg.counters.length; i++)
var counter = msg.counters[i];
console.log(counter);
,
error: ErrorMsg
);
I have no idea where I went wrong. please help.
javascript python jquery json ajax
asked Nov 13 '18 at 10:58
Balakrishnan SathiyakuganBalakrishnan Sathiyakugan
587
|
show 2 more comments
I send a GET request from the back end to get the json response. I got this error.
Uncaught SyntaxError: Unexpected token :
at DOMEval (jquery-3.3.1.js:111)
at Function.globalEval (jquery-3.3.1.js:345)
at text script (jquery-3.3.1.js:9640)
at ajaxConvert (jquery-3.3.1.js:8787)
at done (jquery-3.3.1.js:9255)
at XMLHttpRequest. (jquery-3.3.1.js:9548)
My AJAX request code is:
$.ajax(
type: "GET", //rest Type
dataType: 'jsonp', //mispelled
url: " url_for('live') ",
contentType: "application/json",
success: function (msg)
console.log(msg);
for (var i = 0; i < msg.counters.length; i++)
var counter = msg.counters[i];
console.log(counter);
,
error: ErrorMsg
);
I have no idea where I went wrong. please help.
javascript python jquery json ajax
asked Nov 13 '18 at 10:58
Balakrishnan SathiyakuganBalakrishnan Sathiyakugan
587
msgis already an object so decoding it again will cause the error you see. You just need to remove thevar msg = JSON.decode(msg);line
– Rory McCrossan
Nov 13 '18 at 10:59
@RoryMcCrossan still getting the error
– Balakrishnan Sathiyakugan
Nov 13 '18 at 11:05
Is it the same error?
– Rory McCrossan
Nov 13 '18 at 11:05
@RoryMcCrossan yes same error. VM2335:2 Uncaught SyntaxError: Unexpected token :
– Balakrishnan Sathiyakugan
Nov 13 '18 at 11:10
1
Glad you got it working. I added an answer for you below.
– Rory McCrossan
Nov 13 '18 at 11:17
|
show 2 more comments
I send a GET request from the back end to get the json response. I got this error.
Uncaught SyntaxError: Unexpected token :
at DOMEval (jquery-3.3.1.js:111)
at Function.globalEval (jquery-3.3.1.js:345)
at text script (jquery-3.3.1.js:9640)
at ajaxConvert (jquery-3.3.1.js:8787)
at done (jquery-3.3.1.js:9255)
at XMLHttpRequest. (jquery-3.3.1.js:9548)
My AJAX request code is:
$.ajax(
type: "GET", //rest Type
dataType: 'jsonp', //mispelled
url: " url_for('live') ",
contentType: "application/json",
success: function (msg)
console.log(msg);
for (var i = 0; i < msg.counters.length; i++)
var counter = msg.counters[i];
console.log(counter);
,
error: ErrorMsg
);
I have no idea where I went wrong. please help.
javascript python jquery json ajax
asked Nov 13 '18 at 10:58
Balakrishnan SathiyakuganBalakrishnan Sathiyakugan
587
I send a GET request from the back end to get the json response. I got this error.
Uncaught SyntaxError: Unexpected token :
at DOMEval (jquery-3.3.1.js:111)
at Function.globalEval (jquery-3.3.1.js:345)
at text script (jquery-3.3.1.js:9640)
at ajaxConvert (jquery-3.3.1.js:8787)
at done (jquery-3.3.1.js:9255)
at XMLHttpRequest. (jquery-3.3.1.js:9548)
My AJAX request code is:
$.ajax(
type: "GET", //rest Type
dataType: 'jsonp', //mispelled
url: " url_for('live') ",
contentType: "application/json",
success: function (msg)
console.log(msg);
for (var i = 0; i < msg.counters.length; i++)
var counter = msg.counters[i];
console.log(counter);
,
error: ErrorMsg
);
I have no idea where I went wrong. please help.
javascript python jquery json ajax
javascript python jquery json ajax
asked Nov 13 '18 at 10:58
Balakrishnan SathiyakuganBalakrishnan Sathiyakugan
587
asked Nov 13 '18 at 10:58
Balakrishnan SathiyakuganBalakrishnan Sathiyakugan
587
edited Nov 13 '18 at 11:05
Balakrishnan Sathiyakugan
asked Nov 13 '18 at 10:58
Balakrishnan SathiyakuganBalakrishnan Sathiyakugan
587
asked Nov 13 '18 at 10:58
Balakrishnan SathiyakuganBalakrishnan Sathiyakugan
587
asked Nov 13 '18 at 10:58
Balakrishnan SathiyakuganBalakrishnan Sathiyakugan
587
587
msgis already an object so decoding it again will cause the error you see. You just need to remove thevar msg = JSON.decode(msg);line
– Rory McCrossan
Nov 13 '18 at 10:59
@RoryMcCrossan still getting the error
– Balakrishnan Sathiyakugan
Nov 13 '18 at 11:05
Is it the same error?
– Rory McCrossan
Nov 13 '18 at 11:05
@RoryMcCrossan yes same error. VM2335:2 Uncaught SyntaxError: Unexpected token :
– Balakrishnan Sathiyakugan
Nov 13 '18 at 11:10
1
Glad you got it working. I added an answer for you below.
– Rory McCrossan
Nov 13 '18 at 11:17
|
show 2 more comments
msgis already an object so decoding it again will cause the error you see. You just need to remove thevar msg = JSON.decode(msg);line
– Rory McCrossan
Nov 13 '18 at 10:59
@RoryMcCrossan still getting the error
– Balakrishnan Sathiyakugan
Nov 13 '18 at 11:05
Is it the same error?
– Rory McCrossan
Nov 13 '18 at 11:05
@RoryMcCrossan yes same error. VM2335:2 Uncaught SyntaxError: Unexpected token :
– Balakrishnan Sathiyakugan
Nov 13 '18 at 11:10
1
Glad you got it working. I added an answer for you below.
– Rory McCrossan
Nov 13 '18 at 11:17
msg is already an object so decoding it again will cause the error you see. You just need to remove the var msg = JSON.decode(msg); line– Rory McCrossan
Nov 13 '18 at 10:59
msg is already an object so decoding it again will cause the error you see. You just need to remove the var msg = JSON.decode(msg); line– Rory McCrossan
Nov 13 '18 at 10:59
@RoryMcCrossan still getting the error
– Balakrishnan Sathiyakugan
Nov 13 '18 at 11:05
@RoryMcCrossan still getting the error
– Balakrishnan Sathiyakugan
Nov 13 '18 at 11:05
Is it the same error?
– Rory McCrossan
Nov 13 '18 at 11:05
Is it the same error?
– Rory McCrossan
Nov 13 '18 at 11:05
@RoryMcCrossan yes same error. VM2335:2 Uncaught SyntaxError: Unexpected token :
– Balakrishnan Sathiyakugan
Nov 13 '18 at 11:10
@RoryMcCrossan yes same error. VM2335:2 Uncaught SyntaxError: Unexpected token :
– Balakrishnan Sathiyakugan
Nov 13 '18 at 11:10
1
1
Glad you got it working. I added an answer for you below.
– Rory McCrossan
Nov 13 '18 at 11:17
Glad you got it working. I added an answer for you below.
– Rory McCrossan
Nov 13 '18 at 11:17
|
show 2 more comments
1 Answer
1
active
oldest
votes
You have two issues. Firstly the response will already be deserialised for you buy jQuery. Deserialising it again will cause the error you see. Secondly, the response format appears to be JSON, not JSONP, so the dataType property needs to be amended as well. Try this:
$.ajax(
type: "GET",
dataType: 'json',
url: " url_for('live') ",
contentType: "application/json",
success: function (msg)
for (var i = 0; i < msg.counters.length; i++)
var counter = msg.counters[i];
console.log(counter);
,
error: ErrorMsg
);
answered Nov 13 '18 at 11:17
Rory McCrossanRory McCrossan
242k29207245
add a comment |
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1 Answer
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You have two issues. Firstly the response will already be deserialised for you buy jQuery. Deserialising it again will cause the error you see. Secondly, the response format appears to be JSON, not JSONP, so the dataType property needs to be amended as well. Try this:
$.ajax(
type: "GET",
dataType: 'json',
url: " url_for('live') ",
contentType: "application/json",
success: function (msg)
for (var i = 0; i < msg.counters.length; i++)
var counter = msg.counters[i];
console.log(counter);
,
error: ErrorMsg
);
answered Nov 13 '18 at 11:17
Rory McCrossanRory McCrossan
242k29207245
add a comment |
You have two issues. Firstly the response will already be deserialised for you buy jQuery. Deserialising it again will cause the error you see. Secondly, the response format appears to be JSON, not JSONP, so the dataType property needs to be amended as well. Try this:
$.ajax(
type: "GET",
dataType: 'json',
url: " url_for('live') ",
contentType: "application/json",
success: function (msg)
for (var i = 0; i < msg.counters.length; i++)
var counter = msg.counters[i];
console.log(counter);
,
error: ErrorMsg
);
answered Nov 13 '18 at 11:17
Rory McCrossanRory McCrossan
242k29207245
add a comment |
You have two issues. Firstly the response will already be deserialised for you buy jQuery. Deserialising it again will cause the error you see. Secondly, the response format appears to be JSON, not JSONP, so the dataType property needs to be amended as well. Try this:
$.ajax(
type: "GET",
dataType: 'json',
url: " url_for('live') ",
contentType: "application/json",
success: function (msg)
for (var i = 0; i < msg.counters.length; i++)
var counter = msg.counters[i];
console.log(counter);
,
error: ErrorMsg
);
answered Nov 13 '18 at 11:17
Rory McCrossanRory McCrossan
242k29207245
You have two issues. Firstly the response will already be deserialised for you buy jQuery. Deserialising it again will cause the error you see. Secondly, the response format appears to be JSON, not JSONP, so the dataType property needs to be amended as well. Try this:
$.ajax(
type: "GET",
dataType: 'json',
url: " url_for('live') ",
contentType: "application/json",
success: function (msg)
for (var i = 0; i < msg.counters.length; i++)
var counter = msg.counters[i];
console.log(counter);
,
error: ErrorMsg
);
answered Nov 13 '18 at 11:17
Rory McCrossanRory McCrossan
242k29207245
answered Nov 13 '18 at 11:17
Rory McCrossanRory McCrossan
242k29207245
answered Nov 13 '18 at 11:17
Rory McCrossanRory McCrossan
242k29207245
answered Nov 13 '18 at 11:17
Rory McCrossanRory McCrossan
242k29207245
242k29207245
add a comment |
add a comment |
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msgis already an object so decoding it again will cause the error you see. You just need to remove thevar msg = JSON.decode(msg);line– Rory McCrossan
Nov 13 '18 at 10:59
@RoryMcCrossan still getting the error
– Balakrishnan Sathiyakugan
Nov 13 '18 at 11:05
Is it the same error?
– Rory McCrossan
Nov 13 '18 at 11:05
@RoryMcCrossan yes same error. VM2335:2 Uncaught SyntaxError: Unexpected token :
– Balakrishnan Sathiyakugan
Nov 13 '18 at 11:10
1
Glad you got it working. I added an answer for you below.
– Rory McCrossan
Nov 13 '18 at 11:17