[SOLVED] Getting the current domain

MikeHopley · November 22nd, 2007, 12:54 PM

I want to assign a variable to the current domain of the web page. When I'm developing the site, the domain is "http://localhost/", but when I upload it the domain is "http://www.badmintonbible.com/".

So I'm using this variable to make links work both locally and on the site itself.

I've tried using $_SERVER['DOCUMENT_ROOT'], but this goes too deep: it returns all the server's file structure.

There must be a pre-assigned variable for this, but I haven't found it. Any ideas?

c010depunkk · November 22nd, 2007, 01:00 PM

I had the same problem and here's my solution. This function returns the current domain plus the folder structure to where the file is located plus a trailing slash:

Code:


			
function getLinkPrefix() {
    preg_match('#^/([^/]+)/[.]*#',$_SERVER['PHP_SELF'],$regs);
    return 'http://'.$_SERVER['HTTP_HOST'].'/'.($regs[1]==''?'':$regs[1].'/');
}

Hope this helps, you should be able to modify it to work. I think the $_SERVER['HTTP_HOST'] is interesting 4 u.

karinne · November 22nd, 2007, 01:02 PM

Hmmmm ... I can't test any of this as I don't have access but... have you tried

'HTTP_HOST'

@lun · November 22nd, 2007, 01:22 PM

what karinne said:

this has always worked for me

Code:


			
$url = "http://".$_SERVER['HTTP_HOST']."/";

MikeHopley · November 22nd, 2007, 01:40 PM

Quote:

Originally Posted by @lun

what karinne said:

this has always worked for me

Code:


			
$url = "http://".$_SERVER['HTTP_HOST']."/";

Fantastic, that's solved it -- and quickly, too.

Thanks everyone!