Friendly URL like grooveshark/twitter: example.com/#!/whatever

Question

I am trying to get the requested URL from PHP.

The URL is like this: 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/

How Grooveshark and Twitter handle this?

Somewhat related: http://stackoverflow.com/questions/3009380/whats-the-shebang-hashbang-in-facebook-and-new-twitter-urls-for — Pierre-Olivier, Mar 29 '12 at 20:56
As far as I'm aware, anything after the hash is not sent to PHP as it's completely client side. — Corbin, Mar 29 '12 at 20:57
@Pierre-OlivierBourgeois thank you! i have a lot of info to read about it. — AngelBroz, Mar 29 '12 at 21:54

score 2 · Accepted Answer · answered Mar 29 '12 at 20:55

2

Twitter, Grooveshark and Facebook (before) uses Javascript to get this Friendly URL.

Maybe you can use $_SERVER['REQUEST_URI'] to get it with php, i am not sure.

answered Mar 29 '12 at 20:55

Estefano Salazar

You can't get it with $_SERVER['REQUEST_URI'] or any other Server variable because is not sent to the server. But, yes! i will use Javascript to handle the Friendly URl, i will load the header and the footer, and the javascript, then i will get that friendly url to load it with ajax. – AngelBroz Mar 29 '12 at 21:59

1 Answers1