JSONP request to PHP page not working (cross domain)

Question

This is my PHP page (on a different URL)...

<?php
header('Content-Type: application/json');
?>
stat({"online":1});

And this is my jQuery:

$(document).ready(function(){

    var url = 'http://blah.com/jsontest.php?callback=stat';

    $.getJSON(url, function(data) {
        if (data.online !== undefined) {
            console.log('yay');
        }
    }).error(function() {
        console.log('no');
    });

});

For some reason it is always logging 'no' i.e. its running the error function.

Using jQuery 1.5.2 any ideas?

@Nicola — nothing, because the error function is being called not the success function. — Quentin
– Quentin, Commented May 5, 2011 at 11:41
@David - i also don't understand the "stat({"online":1});" part in the server side script. It might be obvious, but usually i echo('{"online":1}');die(); on server side to have things work P.S. ok, my first comment was really stupiud, sorry! :) — Nicola Peluchetti
– Nicola Peluchetti, Commented May 5, 2011 at 11:43
I believe that if you want to name your own callback function you need to use $.ajax() and specify the jsonpCallback in the call setup. — epascarello
– epascarello, Commented May 5, 2011 at 11:45

Quentin · Accepted Answer · 2011-05-05 11:38:07Z

4

First, JSON-P is not JSON. The content type should be application/javascript. Some browsers may reject JSON-P served as JSON for being unsafe.

Second, getJSON expects that the URL you request to have a ? for the callback method name (and you'll need to get your PHP to pay attention to $_GET['callback']).

Third, if fixing that doesn't work, look at the Net tab in Firebug / Chrome debugger / Dragonfly / etc and see what data is actually going across the wire.

answered May 5, 2011 at 11:38

Quentin

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3 Comments

fire Over a year ago

Can I use JSON for cross domain or do I have to use JSON-P?

Marcel Korpel Over a year ago

You'll have to use JSONP, that's why it was ‘invented’ in the first place.

epascarello Over a year ago

+1 for content type You really should use the built in callback=? and read the value on the server. If you use your own hard coded callback, you will most likely run into problems if there are multiple requests open. So the second point here is valid.

Chris Dutrow · Accepted Answer · 2011-07-27 23:44:17Z

There's some shenanigans with having with include a "callback" function. Apparently you're not returning an object, but a function that was submitted in the original client request. I only vaguely understand what all that means, however, I do have some code to do this that actually works:

Server Side:

<?php
$headers = get_headers($toGetUrl,1);
$return["pop_singer"] = "Britney Spears";
// Right here is where the json object gets wrapped in a function that was submitted under the name "callback"
echo $_GET['callback']."(".json_encode($return).")";
?>

Client side ( this is the same thing as $.getJSON() ):

$.ajax({
type: "GET",
url: serverUrl,
dataType: 'jsonp',
error: function(request, status) {
    // Do some error stuff
},
success: function(data, textStatus, jqXHR) {
    var property = data.pop_singer;   // equals "Britney Spears"
    // Do some successful stuff
}

});

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