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

2002-12-11 Thread Philip Olson
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, Phil

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

2002-12-11 Thread Chris Wesley
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 "m

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

2002-12-11 Thread Brad Bulger
if you have URL wrappers enabled, do $fp = fopen("http://path/to/your/file/$calendar_file";, 'r'); 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&

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

2002-12-11 Thread John W. Holmes
> 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 .= $b