像groovehark / twitter这样友好的网址：example.com/#!/whatever
[英]Friendly URL like grooveshark/twitter: example.com/#!/whatever
问题描述
I am trying to get the requested URL from PHP. 
我试图从PHP获取请求的URL。
The URL is like this: http://example.com/#!/controller/method/var1 URL如下所示： http://example.com/#!/controller/method/var1 ： http://example.com/#!/controller/method/var1
I want to get the /controller/method/var1 part from PHP but all I get is http://example.com/ 我想从PHP获取/controller/method/var1部分，但我得到的只是http://example.com/
How Grooveshark and Twitter handle this? Grooveshark和Twitter如何处理这个问题？
1 个解决方案
解决方案1
2 已采纳 2012-03-29 20:55:26
Twitter, Grooveshark and Facebook (before) uses Javascript to get this Friendly URL. Twitter，Grooveshark和Facebook（之前）使用Javascript来获取此友好URL。
Maybe you can use $_SERVER['REQUEST_URI'] to get it with php, i am not sure. 也许你可以使用$ _SERVER ['REQUEST_URI']来获取它，我不确定。
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