Passing arguments in an include

Is it possible to pass an argument when using an include statement?
ie

This is causing errors for me. Is there another way of doing this?
I'm very new to PHP, so please don't get too technical 😉

Thanks!

With an include, any vars declared are also visible from the include file (except in the functions).

For the example you have given, just declaring arg=test is enough for it to be visible in the include file.

When including files the variables from the main file will pass on to the included file. Try this

<?php
$arg="test";
include "table.inc"
?>

table.inc will be able to use/modify $arg.

If you need $arg in a function try this!

<?php
$arg="test";
include "table.inc"
?>

<snip>
this is table.inc
<?php
function whatever() {
globel $arg;

--> rest of your code...
}
?>

Thanks guys! Got it working.