Re: Loading a File into Variable - How??

Andre, here's the function that worked for me... function file_get_contents($filename) { $fd = fopen($filename, "r", 1); $contents = fread($fd, 12000); fclose($fd); return $contents; } $page_string = file_get_contents("my_file.php"); The third parameter in fopen() [1] can be removed if you don't want to look for files in your include_path. The second parameter in fread() [12000] is where I hardcoded the filesize. Increase that number if you'll be opening larger files. Originally I had "12000" replaced with "filesize($filename)" but if the $filename was opened from the include_path, this seems to always return zero, which is why I hardcoded the byte size into fread(). Monty > From: aajdubuc@webhart.net (Andre Dubuc) > Reply-To: aajdubuc@webhart.net > Newsgroups: php.general > Date: Thu, 18 Jul 2002 17:45:14 -0400 > To: Monty <monty3@hotmail.com> > Cc: php-general@lists.php.net > Subject: Re: [PHP] Loading a File into Variable - How?? > > Hi Monty, > > I've been trying to do the same thing with no success. Would you be so kind > as to show me how you finally did it? I'm not too clear what you meant by: > > "So, if I replace the filesize($filename) command with a hard-coded number, > it works." > > Tia, > Andre > > On Thursday 18 July 2002 04:28 pm, you wrote: >> I just want to load an entire file into a >> single string variable. >> >> However, I figured out the problem shortly after posting that first message >> (of course). Because the file being opened is in the include_path, it seems >> filesize() doesn't see those files. So, if I replace the >> filesize($filename) command with a hard-coded number, it works. >> >> Monty