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May 9th, 2004, 01:02 PM
#1
Thread Starter
Ex-Super Mod'rater
Files filename
Is there a way to find out what file the user is viewing?
I'm hoping there is a fuction for this, its not a simple as as you might think. I have a set of files that the user can open, Say: index.php, request.php, staff.php, ect.... But when those files are opened they will all include the template.php file and inside this I want to beable to tell what the original file was entered. Also the template file sometimes includes files and I will need to still access the file name from these too.
I know I could do this with a variable in every file but I was wondering if there was a function that could do it?
Thanx
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May 9th, 2004, 01:32 PM
#2
Frenzied Member
PHP Code:
<?
$currentFile = $_SERVER["SCRIPT_NAME"];
$parts = Explode('/', $currentFile);
print $parts[count($parts) - 1];
?>
That should do the trick.
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May 9th, 2004, 01:44 PM
#3
Thread Starter
Ex-Super Mod'rater
Thanx, just what I needed.
Last edited by Electroman; May 9th, 2004 at 01:48 PM.
When your thread has been resolved please edit the original post in the thread (  )
and amend "-[RESOLVED]-" to the end of the title and change the icon to  , Thank you.
When posting Code use the [VBCode]Code Here[/VBCode] tags to be able to use the code highlighting.

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