CODEWITHSUNDEEP
ajaxapachejsonjquery
Andrew Bashtannik
Andrew Bashtannik

Reputation: 972

jQuery, JSON and Apache problem

I have an jQuery JSON request, that loads some JSON from another server (ex. foo.com):

$.getJSON("http://foo.com/json.php",function(data) { alert(data); });

But I receive data as null. This is not cross-domain issue, I tried following:

$.getJSON("http://twitter.com/users/usejquery.json?callback=?",
    function(data) { alert(data); });

and received nice JSON object. So, I think there is problem with backend, Apache 2.2.14. Here are HTTP headers, sent from server:

Date: Sun, 07 Mar 2010 16:08:38 GMT
Server: Apache/2.2.14 (CentOS)
X-Powered-By: PHP/5.3.1
Content-Length: 2
Keep-Alive: timeout=15, max=99
Connection: Keep-Alive
Content-Type: application/json; charset=UTF-8

The headers are the same in each case: regular HTTP-request or AJAX. But I receive empty content with AJAX, and normal JSON with browser request. I'm using Firebug for tests, PHP5 for forming JSON.

Somebody have any ideas? Thank you!

Upvotes: 3

Views: 3804

Answers (2)

Andrew Bashtannik
Andrew Bashtannik

Reputation: 972

More about cross-domain JSON throw JSONP and jQuery.

Upvotes: 0

John Boker
John Boker

Reputation: 83699

I'm pretty sure that in order to do cross domain calls like this you have to have a callback, it's what's needed to do JSONP.

here is some more info on jsonp http://www.insideria.com/2009/03/what-in-the-heck-is-jsonp-and.html

For jsonp to work you have to have a callback for the server to wrap the json string in. for example:

$.getJSON("http://foo.com/json.php?callback=?", function(data){});

here a callback function is generated by jquery and passed into the request, so it would be something like:

http://foo.com/json.php?callback=generatedFunction

then what's returned by the server should be:

generatedFunction("{key:value, key2:value2}");

where the parameter in that function is the actual json string.

in the php to return this it would be something like:

$callback = $_GET['callback'];
print($callback."(".json_encode($theobject).");");

Upvotes: 4

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