Eval() = Unexpected token : error

Question

I tried this simple JavaScript code:

eval('{"Topics":["toto","tata","titi"]}')

In the Chrome console, for example, this returns

SyntaxError: Unexpected token :

I tried the JSON on JSONLint and it's valid.

Do you see the bug?

score 52 · Answer 1 · edited Jan 15 '21 at 22:36

52

You have to write like this

eval('('+stringJson+')' );

to convert an string to Object

Hope I help!

edited Jan 15 '21 at 22:36

Jonas Sourlier

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answered Jan 17 '13 at 11:15

Martin Varta

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4

This was the only solution that worked in my case. Thank you! – Kevin Beal Aug 28 '14 at 19:24
1

thx, this is much better than accepted answer. It is good to point out that eval is evil :), but still, this answers the question. – apocalypz Sep 19 '14 at 09:10
I had the same issue evaluating a normal javascript function and this solved my problem. Why / how does wrapping an expression in parenthesis fix the problem? – Stephen Paul Mar 30 '17 at 06:07
This works for me – Mayur Gupta Nov 14 '21 at 21:55

score 30 · Answer 2 · 2012-07-09T14:44:48.230

30

Because eval does not force an expression context and the string provided is an invalid JavaScript program, thus the first three tokens (and how they are looked at) are:

{            // <-- beginning of a block, and NOT an Object literal
"Topics"     // <-- string value, okay (note this is NOT a label)
:            // <-- huh? expecting ";" or "}" or an operator, etc.

Happy coding.

edited Jul 09 '12 at 14:44

answered Nov 02 '11 at 18:34

may I ask you why eval('function(){}') throws an exception too ? – BiAiB Jan 09 '13 at 17:38
4

@BiAiB For the same reason as above :) The contents of `eval` run in a *statement* context and thus it is taken as a FunctionDeclaration grammar construct. The error generated by that is "SyntaxError: function statement requires a name". Either give it a name (`eval('function f(){}'); f()`) or force it into a FunctionExpression construct (`f = eval('(function(){alert("hi")})'); f()`). See http://es5.github.com/x13.html – Jan 09 '13 at 20:53
thanks! the tricky part to me was because strings like '3' are correctly evaluated, and not 'function(){}'. The second cannot be evaluated as an ExpressionStatement: `an ExpressionStatement cannot start with the function keyword because that might make it ambiguous with a FunctionDeclaration` (http://es5.github.com/x12.html#x12.4) – BiAiB Jan 10 '13 at 10:54

score 14 · Accepted Answer · edited Jun 19 '15 at 01:39

14

FWIW, use JSON.parse instead. Safer than eval.

edited Jun 19 '15 at 01:39

Andrew Marshall

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answered Nov 02 '11 at 18:30

Jonathan M

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score 9 · Answer 4 · answered Nov 02 '11 at 18:30

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Number one: Do not use eval.

Number two. Only use eval to make something, well be evaluated. Like for example:

eval('var topics = {"Topics":["toto","tata","titi"]}');

answered Nov 02 '11 at 18:30

Naftali

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score 4 · Answer 5 · edited Dec 28 '12 at 02:40

4

USE:

function evalJson(jsArray){ eval("function x(){ return "+ jsArray +"; }"); return x(); }

var yourJson =evalJson('{"Topics":["toto","tata","titi"]}');

console.log(yourJson.Topics[1]); // print 'tata''

edited Dec 28 '12 at 02:40

competent_tech

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answered Dec 28 '12 at 02:21

Pica Mio

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works for me..don't know if it is the best practice, but it got me up and running – Patrick Jackson Jan 16 '13 at 19:50

score 4 · Answer 6 · answered Nov 02 '11 at 18:30

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Because that's evaluating an object. eval() requires you to pass in syntactically valid javascript, and all you're doing is passing in a bare object. The call should be more like:

eval('var x = {"Topics":etc...}');

answered Nov 02 '11 at 18:30

Marc B

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score 0 · Answer 7 · answered Jun 07 '13 at 15:22

0

if you are using JQuery use the function $.parseJSON(), worked for me, had the same problem

answered Jun 07 '13 at 15:22

Mark E

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score 0 · Answer 8 · answered Nov 24 '20 at 20:21

0

if you want to make an array use the below code

var jsonObject = eval('(' +"["+ response + "]"+')');

answered Nov 24 '20 at 20:21

Abdur Rahman

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Eval() = Unexpected token : error

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