Re: Using fopen() to open a php file with variables.
| From: | Brad Bulger | Date: | Thu, 12 Dec 2002 02:01:22 +0000 |
| Subject: | Re: Using fopen() to open a php file with variables. | ||
| References: | 1 | Groups: | php.general |
| Request: | Send a blank email to php-general+get-127946@lists.php.net to get a copy of this message | ||
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
>
>