I am loading a file from a 'same level' directory, relative to the script.
I am more familiar with the syntax of Perl which is simply
This doesn't work in PHP and the solutions I have seen ie. realpath or the rather odd looking substr(dirname(__FILE__), strlen($_SERVER['DOCUMENT_ROOT']))
appear far too complicated for a simple problem.
What is the correct syntax?
Keith
I am more familiar with the syntax of Perl which is simply
Code:
$Menu = "../chargrill.kjw";
This doesn't work in PHP and the solutions I have seen ie. realpath or the rather odd looking substr(dirname(__FILE__), strlen($_SERVER['DOCUMENT_ROOT']))
appear far too complicated for a simple problem.
What is the correct syntax?
Keith