从php获取脚本URL

I have a PHP file (mymainfile.php)

inside this PHP file, i have include some other PHP files (include.php).

The URL is:

http://www.domain.com/my/mymainfile.php

Now i want to get the URL into a variable inside the include.php file.

All my tries with the $_SERVER variables will give me "domain.com/my/include.php" back and that's not what I'm looking for.

Is it possible, and if it's, how?

Have you tried getcwd()? http://php.net/manual/en/function.getcwd.php Or dirname(__FILE__) (or just __FILE__ ) Aand 1 more try

function get_file_dir() {
    global $argv;
    return realpath($argv[0]);
}