Jquery JSON response handling

Question

I have an ajax query written in jQuery that is returning valid JSON in this format

$.ajax({
   type     : 'POST',
   url      : 'ajax/job/getActiveJobs.php',
   success  : function(data){
       if(data[''] === true){
           alert('json decoded');
       }

       $('#waiting').hide(500);
       $('#tableData').html(data['content']);
       $('#message').removeClass().addClass((data.error === true)
       ?'error':'success').text(data.msg);
       if(data.error === true)
           $('#message')

   },
   error    : function(XMLHttpRequest, textStatus, errorThrown){
       $('#waiting').hide(500);
       $('#message').removeClass().addClass('error').html(data.msg);
   } })

I take it this is not correct as it is not displaying the data, if I use

$('#mydiv').html(data);

I get all of the data back and displayed.

any help is really appreciated

It would help a lot if you'd post the complete code of your ajax "success" handler. What you've posted here does not look wrong, so it must be something else. — Pointy, Jun 17 '11 at 08:03
Try an `alert(data)` and check what type of object you get. May be it's not interpreted as JSON. — DanielB, Jun 17 '11 at 08:05

score 4 · Answer 1 · answered Jun 17 '11 at 08:04

4

Set the dataType to be json so jQuery will convert the JSON to a JavaScript Object.

Alternatively, use getJSON() or send the application/json mime type.

answered Jun 17 '11 at 08:04

alex

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score 1 · Accepted Answer · answered Jun 17 '11 at 08:07

1

Either set dataType to json or use var json = JSON.parse(data) to do it manually.

EDIT:

I'm adding this because somebody else suggested eval, don't do this because it gets passed straight into a JSON object without any sanitation first, allowing scripts to get passed leading straight into an XSS vulnerability.

answered Jun 17 '11 at 08:07

Connor Smith

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Thanks Connor Smith I will be accepting your answer in precisely 6 minutes lol how weird I can't accept an answer when yours has solved the problem. I will not be using eval() dont worry :) Thanks again – Deviland Jun 17 '11 at 08:12
No, it uses: `new Function("return " + data))();` if JSON.parse is not available. Source, jQuery source code: https://gist.github.com/277432 – Connor Smith Jun 17 '11 at 08:30
@Connor I deleted it after I realised you didn't use `$.parseJSON()` (I was imagining things). BTW, passing a string to `new Function()` is basically an `eval()`. – alex Jun 17 '11 at 08:35
@alex It seems that it is indeed as bad security-wise but that new Function() is better, smarter, at parsing the data and is faster: http://stackoverflow.com/questions/2449220/jquery-uses-new-functionreturn-data-instead-of-evaldata-to-parse-j – Connor Smith Jun 17 '11 at 08:39

score 0 · Answer 3 · answered Jun 17 '11 at 08:05

0

The data is the Json so you will want to do this:

success: function (data) {
  var newobject = eval(data);
  alert(newobject.msg);

}

or do this:

$ajax({
    url: url,
    data: {},
   dataType: "json",
   success: function (newObject) {
   alert(newobject.msg);
}
});

answered Jun 17 '11 at 08:05

Richard

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1

Do not do this. Eval is very bad! http://stackoverflow.com/questions/1843343/json-parse-vs-eval – Connor Smith Jun 17 '11 at 08:07
don't worry had no intention of using eval :) although I do appreciate the attempt to help :) – Deviland Jun 17 '11 at 08:09

Jquery JSON response handling

3 Answers3