require code question

[Parker Clack]

If I wanted to require a .cgi file in one of the .php files would I just use:

require("/cgi-bin/myfile.cgi");

Thanks,
Parker

[Mike Sullivan]

Nope - that will just display the script I believe.

You can use passthru - http://www.php.net/manual/function.passthru.php3 but when I tried that, it returned the "Content-type: text/html" line, and obviously I didn't want that, so I used this little code:

exec("/absolute/path/to/cgi/file", $test);
while(list($key,$val)=each($test)) {
if ($val!="Content-type: text/html") {
$test2 .= "$val";
}
}
echo "$test2";

Probably grossly inefficient, but it works...

[Michael]

First of all I am not sure why you would want to include a .cgi in a PHP script - but requiring it shouldn't post it on the screen should it? If you "included" the file it would place text in the script, but require is different I think. I think that simply saying

require("/cgi-bin/myfile.cgi");

Does work. I would use the full path to the file though.