[PHP] inspirational

I want to detect the url my .php lies in. This is over kill and BS: $myurlvar = "http://".$SERVER_NAME.parse($SCRIPT_NAME); How do I get "http://foo.com/dir1/dir2/dir3/";. There seems to be nothing in phpinfo() that will give me a full url to play with. Flame me if you will, but I browsed TFM

RE: [PHP] inspirational

be a function that does all this, however, you can try to work around it Thanks, Ray Hunter -Original Message- From: jtjohnston [mailto:[EMAIL PROTECTED]] Sent: Saturday, May 25, 2002 11:39 PM To: [EMAIL PROTECTED] Subject: [PHP] inspirational I want to detect the url my .php li

Re: [PHP] inspirational

try: $url = preg_replace('/^(http:\/\/)[^\/]+(\/.*)$, '\\1$SERVER_NAME\\2', $SCRIPT_URI); Michael "Jtjohnston" <[EMAIL PROTECTED]> schrieb im Newsbeitrag [EMAIL PROTECTED]">news:[EMAIL PROTECTED]... > I want to detect the url my .php lies in. > > This is over kill and BS: > > $myurlvar = "http:

Re: [PHP] inspirational

typo: $url = preg_replace('/^(http:\/\/)[^\/]+(\/.*)$/', '\\1$SERVER_NAME\\2', $SCRIPT_URI); Michael "Michael Virnstein" <[EMAIL PROTECTED]> schrieb im Newsbeitrag [EMAIL PROTECTED]">news:[EMAIL PROTECTED]... > try: > > $url = preg_replace('/^(http:\/\/)[^\/]+(\/.*)$, '\\1$SERVER_NAME\\2', > $S

Re: [PHP] inspirational

but the scriptname itself will be included there. Try this, if you don't want the scriptname to be included.: $url = preg_replace('/^(http:\/\/)[^\/]+((\/[^\/])*\/)([^\/]+)$/', '\\1$SERVER_NAME\\2', $SCRIPT_URI); Haven't tested them, but should work. Michael "Michael Virnstein" <[EMAIL PROTECT

RE: [PHP] inspirational

t;? Try this: $myurlvar = http:// . $SERVER_NAME . dirname($SCRIPT_NAME); ---John Holmes... > -Original Message- > From: jtjohnston [mailto:[EMAIL PROTECTED]] > Sent: Sunday, May 26, 2002 1:39 AM > To: [EMAIL PROTECTED] > Subject: [PHP] inspirational > > I want to det