No response from PHP Script - php

Question

Absolutely new to PHP and so far it isn't pretty.
Anyway, I'm trying to pass a variable over to a PHP script, do a couple things with it, and pass it back to my Javascipt code.
Here's where I pass it off to PHP:
var src=encodeURIComponent("http://www.someonlinesite.com/file.swf");
$.ajax({
url:'test.php?src='+src,
dataType:'json',
success:function(response){
alert(response)
}
});
and here's the script:
<?php
$src=isset($_GET['src'])?$_GET['src']:'';
$size=getimagesize($src);
echo json_encode(array('size'=>$size));
?>
I'm trying to pass the URL of a .SWF video file over to a small PHP script that will use getImagesize() to figure it's dimensions and pass them back.... but I'm not seeing anything in the response and the alert isn't firing.
What's going wrong?
UPDATE:
I've updated the code with the most recent - according to the advice from some nice SO members. When I hardcode the $src variable and navigate directly to the test.php it echoes everything perfectly. So, it looks like the PHP is working. However, it appears like either the callback is never firing or the PHP file isn't returning the data. In the console there still isn't anything in the response.

Answer 1

You need to concatenate your url string parameter in get():
$.get('test.php?src=' + src, function(data){
alert(data);
});
And also, your src variable begins with a double quote and is closed with a single quote. That will cause issues.
var src="http://www.someonelinesite.com/file.swf";
Also, it's probably a bad idea to do this via $_GET since you are passing a URL. $_POST would be better or encode the URL before you pass it. The current url you are passing right now would look like this in a browser:
http://www.mysite.com/test.php?src=http://www.someonelinesite.com/file.swf
That's not pretty. Using encodeURIComponent(), your whole URL will end up looking like this:
http://www.mysite.com/test.php?src=http%3A%2F%2Fwww.someonelinesite.com%2Ffile.swf
Edit to $.ajax
$.get above would work just fine, but going with the implementation of $.ajax works too:
$.ajax({
url:'test.php',
type: 'GET', //Add the type
dataType:'json',
data: {'src': src}, //Add the data, leave it out the url
success:function(data){
alert(data)
}
});

Answer 2

Try this :
In Jquery :
var src="http://www.someonelinesite.com/file.swf";
$.ajax({
url:'test.php?src='+src,
dataType:'json',
success:function(response){
alert(response.size);
}
});
In php
$src=isset($_GET['src'])?$_GET['src']:'';
$size=getimagesize($src);
echo json_encode(array('size'=>$size));
?>