Quote:
$foo = preg_replace('/foo/', 'bar', $foo);preg_replace('/foo/', 'bar', $foo);
I think you can leave "$foo = " out. Can't be bothered testing it, but I'm fairly sure that would work.
I don't see what the big deal about making the PHP Frontend to Links SQL. The backend is the hard part -> The front end is the simple bit which is easier in PHP.
Cheers,
Michael Bray
