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 doesn't pull the file from the web server - it should refer to a file on the filesystem instead.

Calling include or require just tells PHP to paste the contents of the given file in your code at this place, nothing more than that.