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

require() does not make an HTTP call. All it does is open the file from disk and include the code in the position of the call. So simple local variables are enough.