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Posted by Alan Jones on 05/13/07 17:37
On Sun, 13 May 2007 09:44:20 +0200, Schraalhans Keukenmeester
<invalid@invalid.spam> wrote:
>> echo ($_SERVER['REQUEST_URI']);
>>
>> and...
>>
>> echo basename($_SERVER['REQUEST_URI']);
>>
>> ...returned the name of the file it is in, the include 'child' file,
>> and not the actual page URL accessed; /index.php. I need the
>> include script to know if it is being run within index.php.
>>
>
>It _should_ tell you the top level calling script. Show us what you
>are doing exactly please.
Thanks for the help. Please see my response to Jerry's post. Just
want to consolidate the thread. :)
>Btw, I am no fan of include files behaving
>differently depending on where they're called from. What if your script
>name changes ?
I hope to get the request_uri result working correctly for now, and
then compose a script simple enough to not cause future problems.
>Open_basedir is a php setting telling the server the top_level directory
>from which files can be opened in your scripts. It's commonly set to point
>to your httpdocs dir. (In php.ini)
>You got this warning because you tried linking files from your server's
>root directory. Link takes a filepath, not a url. In other words, remove
>the / from the arguments. (Or provide a full path to the file):
>
>link ("index.php","index_body.php");
Thanks, I'll put that aside for now. Hopefully, I won't need to use
link().
>Using the phpinfo() function you can see if open_basedir is set and if so,
>to what directory.
Apparently it is: /home/httpd/vhosts/jalanjones.com/httpdocs:/tmp
I assume that's the correct setting.
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