Connecting to PHP on different domain to HTML with Jquery (AJAX)

Question

Trying to receive a response from a PHP which is on a different domain to my HTML:

 $.get('getstate.php', {
                email: $('#game-email').val(),
                country: 'DE',
                lang: lang,
                source: 'promotion'
            }, function (data) {
                console.log(data);
             });

In chrome I get this error:

XMLHttpRequest cannot load ..... Origin null is not allowed by Access-Control-Allow-Origin.

What do I need to do to achieve this?

http://stackoverflow.com/questions/1653308/access-control-allow-origin-multiple-origin-domains — Eli, Mar 22 '13 at 11:00
surely if you're trying to get `getstate.php` it should be the full path of the url if its located on another domain — chriz, Mar 22 '13 at 11:01
the fastest way is to proxy the request through the local php server and a curl request — David Fregoli, Mar 22 '13 at 11:06

score 1 · Accepted Answer · edited May 23 '17 at 12:28

1

A quick google shows me that this has been answered here many times before... It's not possible with ajax for security reasons, so you will have to either

call another domain from within your PHP script, or
look at JSONP

edited May 23 '17 at 12:28

Community

1
1

answered Mar 22 '13 at 11:03

msturdy

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score 0 · Answer 2 · answered Mar 22 '13 at 11:05

0

Try dataType: 'jsonp'
$.get('page.php, {}, callback, "jsonp");

answered Mar 22 '13 at 11:05

the server needs to support jsonp obviously – David Fregoli Mar 22 '13 at 11:07
is there any other way without the server having to support jsonp? – user1937021 Mar 22 '13 at 12:33
@1937021 Do you has access to that server? – Mar 22 '13 at 13:01

score 0 · Answer 3 · answered Mar 22 '13 at 11:10

a simple PHP proxy from http://jquery-howto.blogspot.co.uk/2009/04/cross-domain-ajax-querying-with-jquery.html

<?php
// Set your return content type
header('Content-type: application/xml');

// Website url to open
$daurl = 'http://feeds.feedburner.com/jQueryHowto';

// Get that website's content
$handle = fopen($daurl, "r");

// If there is something, read and return
if ($handle) {
    while (!feof($handle)) {
        $buffer = fgets($handle, 4096);
        echo $buffer;
    }
    fclose($handle);
}
?>

Connecting to PHP on different domain to HTML with Jquery (AJAX)

3 Answers3