Re: Using fopen() to open a php file with variables.

Why don't you just use include? $month = 12; $year = 2002; // This include has access to the above variables include 'make_calendar.php3'; Include it wherever you intended to print $caledar_html. It doesn't exist in your code because the url is seen as part of the filename. Regards, Philip Olson On Wed, 11 Dec 2002, Jay (PHP List) wrote: > 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? > > Jay Douglas > Systems Consultant > Fort Collins, CO > >