为什么PHP脚本无响应？

Question

我试图将变量传递给PHP脚本，用它做一些事情，然后将其传递回我的Javascipt代码。 在这里我将其传递给PHP：

   var src=encodeURIComponent("http://www.someonlinesite.com/file.swf");
   $.ajax({
       url:'test.php?src='+src,
       dataType:'json',
       success:function(response){
           alert(response)
       }
   });

这是脚本：

<?php
   $src=isset($_GET['src'])?$_GET['src']:'';
   $size=getimagesize($src);
   echo json_encode(array('size'=>$size));
?>

我正在尝试将.SWF视频文件的URL传递到一个小的PHP脚本，该脚本将使用 getImagesize（）来确定尺寸并将其传回。但是我没有在响应中看到任何内容，并且 alert 也没有触发——出了什么问题？

更新:

根据一些成员的建议，我已经用最新的代码进行了更新。当我对$ src变量进行硬编码并直接导航到test.php时，它会完美地呼应所有内容。因此，看起来PHP正在运行。但是，似乎是回调从未触发或PHP文件未返回数据，在控制台中，响应中仍然没有任何内容。

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)
   }
});