PHP Include with VARIABLES

[Anon]

I would like to do something like

<?PHP
include("mysinclude.php?var=001");
?>
but I get an error ( I dont get an error when I leave off the ?var=001 )

The error I get is
Warning: Failed opening 'pd_001.php?itemid=20' for inclusion (include_path='') in /disk2/httpd/newworld/html/shop2/product.php on line 3 ...

any ideas how I can get around this, or is it only possible by using fopen ?

ANY help would be greatly appreciated...

[smarlowe]

Do it this way:

$var="001";
include "somefilethatusesvar.php";

And the include file should see the $var that you set.

[[deleted]]

Just to elaborate:
include makes the included script part of the script that does the include, at the point where the include occurs.

So any vars declared in the parent script are immediately available in the included script, because the two scripts are really one script through the include().

[Anon]

ahhhh.... of course... it simply shares the variables become it "becomes part of" (well actually is part of) the main file 🙂 DOH!

thnx guys...