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

2002-12-11 Thread Jay \(PHP List\)
Okay, I need to retrieve the output of a php file with variables being passed to it. Example: $calendar_file = make_calendar.php3?month=$monthyear=$year; $fp = $fp = fopen($calendar_file, r); while (!feof ($fp)) { $buffer = fgets($fp, 4096); $calendar_html .= $buffer; }

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: $calendar_file = make_calendar.php3?month=$monthyear=$year; $fp = $fp = fopen($calendar_file, r); while (!feof ($fp)) { $buffer = fgets($fp, 4096); $calendar_html .= $buffer; }

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 =

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: $calendar_file = make_calendar.php3?month=$monthyear=$year; $fp = $fp = fopen($calendar_file, r); Oh my! That's not going to work, because

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,