php获取通过url的json数据，返回的是空值,该如何处理

php获取通过url的json数据，返回的是空值
http://php.net/api/apply.php?&apply={"apply":{"name":"4747474","site":"http:\/\/4444444444444444","icon":"47","information":"474","about":"747","weibo":"74444444444","email":"74444444444444444","moblie":"89742589633","isfree":"y","applytime":1349939108}}

这个是一个链接，我在这个页面上要怎么把那段json格式给接下出来呢？
------解决方案--------------------
$x = json_decode($_GET['apply']);
------解决方案--------------------
var_dump(json_decode($_GET['apply'],true));
看輸出什麽？
------解决方案--------------------
var_dump(json_decode(urldecode($_GET['apply']),true));
------解决方案--------------------

print_r(json_decode('{"apply":{"name":"4747474","site":"http:\/\/4444444444444444","icon":"47","information":"474","about":"747","weibo":"74444444444","email":"74444444444444444","moblie":"89742589633","isfree":"y","applytime":1349939108}}'));

stdClass Object<br />(<br />    [apply] => stdClass Object<br />        (<br />            [name] => 4747474<br />            [site] => http://4444444444444444<br />            [icon] => 47<br />            [information] => 474<br />            [about] => 747<br />            [weibo] => 74444444444<br />            [email] => 74444444444444444<br />            [moblie] => 89742589633<br />            [isfree] => y<br />            [applytime] => 1349939108<br />        )<br /><br />)<br /><br />

------解决方案--------------------
LZ地址中参数apply前面多了个&符号，你看会不会是这个的问题