how to pass a variable through the require() or include() function of php?

Question

when I use this:

require(\"diggstyle_code.php?page=$page_no\");

the warning is :failed to open stream: No error in C:\\xampp\\htdocs\\4ajax\\ga

Answer 1

this should work, but it's quite a dirty hack:

$_GET['page'] = $page_no;
require('diggstyle_code.php');

you probably want to refactor your code to use functions and/or objects and call them from your files instead of including them (spaghetti code alert)