not 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 lies
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 =
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',
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
:// . $SERVER_NAME . dirname($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 detect the url my .php lies in.
This is over kill and BS
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
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