sof_user sof_user - 3 months ago 9
PHP Question

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

when I use this:


the warning is :failed to open stream: No error in C:\xampp\htdocs\4ajax\gallery_core.php on line 198

and the error is:

Failed opening required 'diggstyle_code.php?page=1' (include_path='.;C:\xampp\php\PEAR') in C:\xampp\htdocs\4ajax\gallery_core.php on line 198

value of the variable
is collected beforehand.

But if I omit the
'?page=$page_no part'
from the argument of the require function, then no error or warning is shown.

I need to pass the variable when I use the require() function.


require (and include) will open the file corresponding to the path/name they receive.

Which means that, with your code, you would have to have, on your disk, a file called diggstyle_code.php?page=1 -- it's obviously not the case, so it fails.

Quoting the Variable scope page of the PHP manual :

The scope of a variable is the context within which it is defined.
For the most part all PHP variables only have a single scope.
This single scope spans included and required files as well.

So, in your case, you do not need to pass the variable : if you have a variable in your current script, it will also exist in the script you include.

(Outside of functions, of course, which have their own scope)

So, in your main script, you should have :

$page_no = 10;
require 'diggstyle_code.php';

And, in diggstyle_code.php :

echo $page_no;
// Or work with $page_no the way you have to

Remember that including/requiring a file is exactly the same as copy-pasting its content at the line it's required.