Parsing json data failing - jquery

Question

I don't know where I ma missing something but I have this

var myvar = [{"id":1,"name":"name1"},{"id":2,"name":"name2"}];

and I tried this

$(jQuery.parseJSON(JSON.stringify(myvar))).each(function() {  
        console.log(this.name);
});

But I have an error in my console : Syntax error, unrecognized expression [{"id":1,"name":"name1"},{"id":2,"name":"name2"]

I am missing something, but I don't know what ?

Edit : in fact when I copy paste the myvar in my console and run the parsing then it works ?? But, when I refresh my page and when I retrieve myvar as so : console.log(myvar), I get [{"id":1,"name":"name1"},{"id":2,"name":"name2"}], without normally told by chrome's console that it is an object

var myvar = [{"id":1,"name":"name1"},{"id":2,"name":"name2"}]; — Antony
– Antony, Commented Jan 25, 2013 at 13:56
@Newben there's still the one in the error message? Did you not copy this directly from the console? — Rory McCrossan
– Rory McCrossan, Commented Jan 25, 2013 at 14:00
@RoryMcCrossan, yes error while copy pasting. I have the edited error message — epsilones
– epsilones, Commented Jan 25, 2013 at 14:01
@Newben I have name1 and name2 in the console now. Is the problem solved? — Antony
– Antony, Commented Jan 25, 2013 at 14:02

Aram Kocharyan · Accepted Answer · 2013-01-25 13:56:05Z

3

You're not closing the object.

var myvar = [{"id":1,"name":"name1"},{"id":2,"name":"name2"}];

answered Jan 25, 2013 at 13:56

Aram Kocharyan

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4 Comments

epsilones Over a year ago

sorry I have edited my post (error while copy-pasting), so I always have the problem

Aram Kocharyan Over a year ago

I'm not quite sure why you're serializing and then deserializing though (perhaps for testing or sanitising), but it should work now.

epsilones Over a year ago

Quite strange, in fact if I copy paste the myvar in my console and runt he parsing then it works ?? But, when I refresh my page and when I retrive the myvar variable as so console.log(myvar), I get [{"id":1,"name":"name1"},{"id":2,"name":"name2"}], without normally told by chrome that it is an object

Aram Kocharyan Over a year ago

That means it's still a string

Anton · Accepted Answer · 2013-01-25 13:56:27Z

2

you are missing a } at the end

var myvar = [{"id":1,"name":"name1"},{"id":2,"name":"name2"}];

answered Jan 25, 2013 at 13:56

Anton

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1 Comment

epsilones Over a year ago

sorry I have edited my post (error while copy-pasting), so I always have the problem

Robin van Baalen · Accepted Answer · 2013-01-25 14:13:24Z

0

That's weird; I get an error as well trying your code. It seems perfectly fine though.

Why don't you just try this:

jQuery(myvar).each(function () {
    console.log(this.name);
});

this outputs

name1
name2

in my console. This seems to be a solution for you since you're already converting the object to a string and back to an object (array).

answered Jan 25, 2013 at 14:13

Robin van Baalen

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Comments

Robin van Baalen · Accepted Answer · 2013-01-25 14:18:31Z

0

This is the correct syntax :

$(jQuery.parseJSON(myvar)).each(function() {  
    console.log(this.name);
});

(In fact, when I copy paste the myvar in my console and run the parsing then it works. But, when I refresh my page and when I retrieve myvar as so : console.log(myvar), I get [{"id":1,"name":"name1"},{"id":2,"name":"name2"}], without normally told by chrome's console that it is an object )

edited Jan 25, 2013 at 14:18

Robin van Baalen

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answered Jan 25, 2013 at 14:12

epsilones

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