从CLI生成可执行的PHP脚本并包含？

Consider this:

#!/usr/bin/php
<?php
class Foo {
        static function bar () {
                echo "Foo->bar
";
        }
}
if (PHP_SAPI === 'cli') {
        Foo::bar();
}
?>

I can execute this from CLI, but when I include it in, say, a CGI-run PHP script, the shebang ends up in the output.

I like simple scripts compact: I guess I could put the class part in a separate "lib"-file and have a simple wrapper for CLI use. BUT I'd like to keep it all in one place without having to worry about include paths etc.

Is this possible without ob_*-wrapping the include to capture the shebang (if this is even possible), or is it dumb to cram all of this into one file anyway? Alternatives/Thoughts/Best Practices welcome!

Edit: I'd like to put the script in my PATH, so calling I'd rather not call it by php file.php. See my comment to @misplacedme's answer

It's actually easy.
Remove the shebang and when you run the script, run it as
php scriptname.php OR /path/to/php scriptname.php
instead of
./scriptname.php

Running php script.php will look in only the current directory, or any directory within PATH. If you absolutely have to run it that way, add it. export PATH=$PATH:/path/to/php/script/folder(in bash)
That will mess up includes unless you're using full paths within the script.
No matter what you do, you'll have to use full paths somewhere.