regex to fetch path out of $SCRIPT_FILE

Anon

Hello

I need a small regural expression to get the path out of the $SCRIPT_FILENAME. $SCRIPT_FILENAME returns the scriptname and the path. something like:
/home/www/myapps/webarrchives/search.php But I want to
get the path. Not the script name. I could use something like:
$checker = (substr(\"$SCRIPT_NAME\", -10)); but only
when I know the name of the script beforehand. I don\'t
know the name of the script beforehand. There are so
many scripts. So could I get the path. Any help I will
appreciate

thanks

Anon

THE simplest solution to this particular problem is:

$path = dirname(FILE) . '/';

(dirname returns the path without a trailing slash, but I prefer to handle paths with their trailing slash).

André

Anon

You mean simple directory name or a path to the directory? Fo example, if my directory is hassan then I would write
$path = dirname(hassan) . '/';

???

Anon

Even the directory name is not known beforehand. I am talking about when the script is hosted by some other system and they define their own directory structure and the script has to be able to build that structure. That is what I thought I need to use $SCRIPT_FILE and remove the file name by way of regular expression and there I will have the directory. But I don't know HOW?

thanks again

Anon

I tried to explain how. FILE is a predefined constant in PHP, always pointing at the current file being processed. In some cases it might be to extreme, because if what currently is being parsed is an include, FILE will point to this include file rather than the script doing the include, if this is undesired behaviour, use $SCRIPT_FILENAME.

Anyway, if you actually spend 10 seconds looking at the manual entry for dirname() you will understand how it works; here's an example taken directly from the manual:

$path = "/etc/passwd";
$file = dirname ($path); // $file is set to "/etc"

As you can see, no trailing slash, and in most cases you will probably want the slash, thus i would write the second line as:
$file = dirname($path).'/';

Hope this helps