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