Passing variables to another php script

[Anon]

Hi all,

I am trying use code like this:

include ("http://some.server.com/tuco.php3?var=10");

but there's an error message:

Warning: Failed opening 'http://some.server.com/tuco.php3?var=10' for inclusion in c:\program files\apache group\apache\htdocs\tuc.php3 on line 7

Why?

Of course, script named tuc.php3 exists 🙂 that's the "calling" script...

Thanks, George

[Anon]

"include" opens files, not URL's.

Search the forum, there are a few ways of including remote files in a different way.

[Anon]

The error is shown on line 7. so look at line 6 and 7 for any possible include()'ing.

there might be something out of order there.