how to make the json data into jquery code?

Question

the php json_encode($object) result is

[{"price_id":"1","website_id":"0","price_qty":2,"price":"90.0000"},
 {"price_id":"2","website_id":"0","price_qty":5,"price":"80.0000"},
 {"price_id":"3","website_id":"0","price_qty":8,"price":"70.0000"}]

someone tell me, var sorted = arrayWithJsonData.sort(function(a,b){}

how to use the above php json_encode($object) to arrayWithJsonData. i used the following way, but it shows TypeError: object.sort is not a function

first: in php i did: echo '<div style="display:none;" id="object">'.$object.'</div>'; then using in jquery

var object=jQuery("#object").text(); // first sort the array var sorted = object.sort(function(a,b){}

I think the error TypeError: object.sort is not a function is from the javascript object arrayWithJsonData, not php — Diego
– Diego, Commented Nov 23, 2012 at 15:30

aleation · Accepted Answer · 2012-11-23 15:29:30Z

1

You need to parse the object first so javascript(jQuery) can understand it:

JSON.parse(data);

answered Nov 23, 2012 at 15:29

aleation

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Odafe Ojenikoh · Accepted Answer · 2012-11-24 18:38:42Z

1

 var arrayWithJsonData = $.parseJSON($('#object').text());

$.parseJSON is a jQuery utitlity for converting a well-formed json string to a json object (jQuery doc: http://api.jquery.com/jQuery.parseJSON/). In our case, $('#object').text() method of the matched element returns the well-formed json string which is the output of the PHP function json_encode($object). Hope that helps.

edited Nov 24, 2012 at 18:38

answered Nov 23, 2012 at 15:41

Odafe Ojenikoh

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1 Comment

Simon Martin Over a year ago

Perhaps you could explain your answer a little to help the OP understand why it works.

Diego · Accepted Answer · 2012-11-23 15:35:00Z

0

var arrayWithJsonData = <?php json_encode($object) ?>;
var sorted = arrayWithJsonData.sort(function(a, b){
    // Sort criteria between a and b
});

answered Nov 23, 2012 at 15:35

Diego

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