Question

Here is the part of my script I need help with, the PHP Variable works, just not for the links.

$source_file = 'http://domain.tld/subfolder/<?php echo $_GET["m"]; ?>".png';

It doesn’t work. When the script runs, it runs it as , instead of what was typed in for the m value, however when I have this not in a link, it works just fine. What can I do to fix this?

Answer 1

Because you are using single quotes, PHP takes the literal value. You can use this:

$source_file = 'http://domain.tld/subfolder/' . $_GET["m"] . '.png';