CODEWITHSUNDEEP
jqueryarraysjsonparsing
Brian Barthold
Brian Barthold

Reputation: 1623

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

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"}]

Upvotes: 1

Views: 982

Answers (2)

Sudhir Bastakoti
Sudhir Bastakoti

Reputation: 100175

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

Upvotes: 3

nrabinowitz
nrabinowitz

Reputation: 55688

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

Upvotes: 1

Related Questions