How do I parse json array using .each with no outer name?

Question

I want to parse this json using jquery and each with a loop. I have only seen json examples where it is wrapped in array with a title such as results. How do I parse json like below...?

$.each(json.results, function() {
    console.log(this['title']);
});

[{"title":"About Us","url_title":"about_us","entry_id":"155","channel_id":"2","author_id":"1","status":"open","entry_date":"1287508538","edit_date":"20101209122240","expiration_date":"0","about_body":"We are the greatest.","about_image":"","about_staff_title":"Joe","about_extended":"More about us.","color":"d12626"},{"title":"Contact Us","url_title":"contact_us","entry_id":"223","channel_id":"2","author_id":"1","status":"open","entry_date":"1291918929","edit_date":"20101209122310","expiration_date":"0","about_body":"Email us.","about_image":"","about_staff_title":"Bob","about_extended":"","color":"080480"}]

What is your expected outcome?

zizozu
– zizozu

2011-12-18 05:18:28 +00:00
Commented Dec 18, 2011 at 5:18 — zizozu
– zizozu, Commented Dec 18, 2011 at 5:18
Like this? jsfiddle.net/TuhHt

Jared Farrish
– Jared Farrish

2011-12-18 05:21:53 +00:00
Commented Dec 18, 2011 at 5:21 — Jared Farrish
– Jared Farrish, Commented Dec 18, 2011 at 5:21

Sudhir Bastakoti · Accepted Answer · 2011-12-18 05:23:14Z

3

$.each(json, function() {     
    console.log(this.title);
});

answered Dec 18, 2011 at 5:23

Sudhir Bastakoti

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3 Comments

Jared Farrish Over a year ago

@digitalbart - This is essentially the same answer as my comment.

Brian Barthold Over a year ago

@JaredFarrish Do you want to post an answer and I will accept it

Jared Farrish Over a year ago

@digitalbart - No, it's ok. We posted it at the same time.

nrabinowitz · Accepted Answer · 2011-12-18 05:21:25Z

1

Shouldn't it just be $.each(json, function() ... ) ? Your JSON is an array, no wrapper, so just pass it directly to $.each().

answered Dec 18, 2011 at 5:21

nrabinowitz

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3 Comments

Brian Barthold Over a year ago

I am not sure what you by passing it directly to $.each()? Can you elaborate

nrabinowitz Over a year ago

What both @Sudhir and @Jared show - if your JSON is an array, and you load it into a variable called json, just pass json as the first argument to $.each, rather than passing json.results.

Brian Barthold Over a year ago

Thanks that makes sense I over thought it.

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