RE: [PHP] Using fopen() to open a php file with variables.

> Okay, I need to retrieve the output of a php file with variables being > passed to it. Example: > > $calendar_file = "make_calendar.php3?month=$month&year=$year"; > $fp = $fp = fopen("$calendar_file", "r"); > while (!feof ($fp)) { > $buffer = fgets($fp, 4096); > $calendar_html .= $buffer; > } > fclose($fp); > > It gives me an error that the file doesn't exist, but it does. Any > suggestions? No, it doesn't exist. You have a file make_calendar.php3 on your system, not what you've listed above. If you want to do it like that, fopen() it with a full address: fopen("http://yourdomain.com/make_calendar.php3?month=$month&year=$year" ); Hopefully the email doesn't mangle that link... ---John W. Holmes... PHP Architect - A monthly magazine for PHP Professionals. Get your copy today. http://www.phparch.com/