How to convert a json dump array dump to javascript array object
Multi tool use
-1
the json string is:
x = "['test', 'test1', 'test2']"
I used:
y = JSON.parse(x);
When i get the first element:
alert(y[0]);
The result is:
[
How can I change the above json string to a Javascript array object?
javascript json
asked Nov 16 '18 at 4:04
almost a beginneralmost a beginner
9621724
add a comment |
-1
the json string is:
x = "['test', 'test1', 'test2']"
I used:
y = JSON.parse(x);
When i get the first element:
alert(y[0]);
The result is:
[
How can I change the above json string to a Javascript array object?
javascript json
asked Nov 16 '18 at 4:04
almost a beginneralmost a beginner
9621724
3
That's not JSON. Strings in JSON are delimited by double quotes not single quotes.
– Paulpro
Nov 16 '18 at 4:06
2
Fix whatever is trying to serve that broken JSON and then you can useJSON.parse.
– CertainPerformance
Nov 16 '18 at 4:07
1
wrong json with ' character. try use " as var x = "["test","test1", "test2"]"; or var x = '["test", "test1", "test2"]';
– Haidangdevhaui
Nov 16 '18 at 4:21
add a comment |
-1
-1
-1
the json string is:
x = "['test', 'test1', 'test2']"
I used:
y = JSON.parse(x);
When i get the first element:
alert(y[0]);
The result is:
[
How can I change the above json string to a Javascript array object?
javascript json
asked Nov 16 '18 at 4:04
almost a beginneralmost a beginner
9621724
the json string is:
x = "['test', 'test1', 'test2']"
I used:
y = JSON.parse(x);
When i get the first element:
alert(y[0]);
The result is:
[
How can I change the above json string to a Javascript array object?
javascript json
javascript json
asked Nov 16 '18 at 4:04
almost a beginneralmost a beginner
9621724
asked Nov 16 '18 at 4:04
almost a beginneralmost a beginner
9621724
asked Nov 16 '18 at 4:04
almost a beginneralmost a beginner
9621724
asked Nov 16 '18 at 4:04
almost a beginneralmost a beginner
9621724
asked Nov 16 '18 at 4:04
almost a beginneralmost a beginner
9621724
9621724
3
That's not JSON. Strings in JSON are delimited by double quotes not single quotes.
– Paulpro
Nov 16 '18 at 4:06
2
Fix whatever is trying to serve that broken JSON and then you can useJSON.parse.
– CertainPerformance
Nov 16 '18 at 4:07
1
wrong json with ' character. try use " as var x = "["test","test1", "test2"]"; or var x = '["test", "test1", "test2"]';
– Haidangdevhaui
Nov 16 '18 at 4:21
add a comment |
3
That's not JSON. Strings in JSON are delimited by double quotes not single quotes.
– Paulpro
Nov 16 '18 at 4:06
2
Fix whatever is trying to serve that broken JSON and then you can useJSON.parse.
– CertainPerformance
Nov 16 '18 at 4:07
1
wrong json with ' character. try use " as var x = "["test","test1", "test2"]"; or var x = '["test", "test1", "test2"]';
– Haidangdevhaui
Nov 16 '18 at 4:21
3
3
That's not JSON. Strings in JSON are delimited by double quotes not single quotes.
– Paulpro
Nov 16 '18 at 4:06
That's not JSON. Strings in JSON are delimited by double quotes not single quotes.
– Paulpro
Nov 16 '18 at 4:06
2
2
Fix whatever is trying to serve that broken JSON and then you can use
JSON.parse.– CertainPerformance
Nov 16 '18 at 4:07
Fix whatever is trying to serve that broken JSON and then you can use
JSON.parse.– CertainPerformance
Nov 16 '18 at 4:07
1
1
wrong json with ' character. try use " as var x = "["test","test1", "test2"]"; or var x = '["test", "test1", "test2"]';
– Haidangdevhaui
Nov 16 '18 at 4:21
wrong json with ' character. try use " as var x = "["test","test1", "test2"]"; or var x = '["test", "test1", "test2"]';
– Haidangdevhaui
Nov 16 '18 at 4:21
add a comment |
2 Answers
2
active
oldest
votes
1
Try the other way around with quotes.
i.e x should be
x = '["test", "test1", "test2"]'
answered Nov 16 '18 at 5:10
GayanthaGayantha
9726
add a comment |
-1
You can solve your problem by using the following code
const alert_message = x.split(',').map(d=> d.replace(/[|]|'|'/g, ''));
edited Nov 16 '18 at 6:42
Quick learner
2,63511026
answered Nov 16 '18 at 5:16
Biplab MalakarBiplab Malakar
42738
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
);
);
draft saved
draft discarded
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%2f53331272%2fhow-to-convert-a-json-dump-array-dump-to-javascript-array-object%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
1
Try the other way around with quotes.
i.e x should be
x = '["test", "test1", "test2"]'
answered Nov 16 '18 at 5:10
GayanthaGayantha
9726
add a comment |
1
Try the other way around with quotes.
i.e x should be
x = '["test", "test1", "test2"]'
answered Nov 16 '18 at 5:10
GayanthaGayantha
9726
add a comment |
1
1
1
Try the other way around with quotes.
i.e x should be
x = '["test", "test1", "test2"]'
answered Nov 16 '18 at 5:10
GayanthaGayantha
9726
Try the other way around with quotes.
i.e x should be
x = '["test", "test1", "test2"]'
answered Nov 16 '18 at 5:10
GayanthaGayantha
9726
answered Nov 16 '18 at 5:10
GayanthaGayantha
9726
answered Nov 16 '18 at 5:10
GayanthaGayantha
9726
answered Nov 16 '18 at 5:10
GayanthaGayantha
9726
9726
add a comment |
add a comment |
-1
You can solve your problem by using the following code
const alert_message = x.split(',').map(d=> d.replace(/[|]|'|'/g, ''));
edited Nov 16 '18 at 6:42
Quick learner
2,63511026
answered Nov 16 '18 at 5:16
Biplab MalakarBiplab Malakar
42738
add a comment |
-1
You can solve your problem by using the following code
const alert_message = x.split(',').map(d=> d.replace(/[|]|'|'/g, ''));
edited Nov 16 '18 at 6:42
Quick learner
2,63511026
answered Nov 16 '18 at 5:16
Biplab MalakarBiplab Malakar
42738
add a comment |
-1
-1
-1
You can solve your problem by using the following code
const alert_message = x.split(',').map(d=> d.replace(/[|]|'|'/g, ''));
edited Nov 16 '18 at 6:42
Quick learner
2,63511026
answered Nov 16 '18 at 5:16
Biplab MalakarBiplab Malakar
42738
You can solve your problem by using the following code
const alert_message = x.split(',').map(d=> d.replace(/[|]|'|'/g, ''));
edited Nov 16 '18 at 6:42
Quick learner
2,63511026
answered Nov 16 '18 at 5:16
Biplab MalakarBiplab Malakar
42738
edited Nov 16 '18 at 6:42
Quick learner
2,63511026
edited Nov 16 '18 at 6:42
Quick learner
2,63511026
edited Nov 16 '18 at 6:42
Quick learner
2,63511026
2,63511026
answered Nov 16 '18 at 5:16
Biplab MalakarBiplab Malakar
42738
answered Nov 16 '18 at 5:16
Biplab MalakarBiplab Malakar
42738
answered Nov 16 '18 at 5:16
Biplab MalakarBiplab Malakar
42738
42738
add a comment |
add a comment |
draft saved
draft discarded
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.
draft saved
draft discarded
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%2f53331272%2fhow-to-convert-a-json-dump-array-dump-to-javascript-array-object%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
vObpf3Ch0 7AVc,71eFPFl
3
That's not JSON. Strings in JSON are delimited by double quotes not single quotes.
– Paulpro
Nov 16 '18 at 4:06
2
Fix whatever is trying to serve that broken JSON and then you can use
JSON.parse.– CertainPerformance
Nov 16 '18 at 4:07
1
wrong json with ' character. try use " as var x = "["test","test1", "test2"]"; or var x = '["test", "test1", "test2"]';
– Haidangdevhaui
Nov 16 '18 at 4:21