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

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"); Oh my! That's not going to work, because "make_calendar.php3?month=$month&year=$year" is not a file in your filesystem. I'd start making suggestions on what you can do to achieve the same effect, but I think it'd be easier if you told us what it is you want to achieve by doing that. (It would for me, at least. Then I'll start making suggestions.) THX, ~Chris