Posted by PI on 02/02/07 09:43

Hi guys, I need some urgent help here please:
I am using an include_once() command to include a file that sits a
level above the file view.php. when the code is run, I have an error
as follows:

Warning: include_once(../common.php) [function.include-once]: failed
to open stream: No such file or directory in C:\Program Files\..
\view.php on line 2

Warning: include_once() [function.include]: Failed opening '../
common.php' for inclusion (include_path='.;C:\Program Files\xampp\php
\pear\') in C:\Program Files\...\view.php on line 2

The include command is simply: include_once(../common.php) - the '../'
before the 'common.php' to indicate a level above the current file
level. Is this right? I could not find a resource to confirm this.

The strange thing is that while the error seems to imply that the
calling file, view.php cannot 'see' the common.php file, other
functions called from the 'common.php' file executes correctly and on
the same calling file/page (view.php).

Appears to the a double take - quite confusing - a PHP newbie and I'd
appreciate your prompt response, thanks

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