Re: quick php for perl coder question

Thomas Bolioli <tpblists@terranovum.com> wrote on 09/11/2008 01:10:18 PM: > I want to return an array from a function. > I have this: > return array($found, $username, $email, $nickname); > as my code. I have a bug in the code and I am not sure where yet. Should > the above statement with this: > > array($vars); > $vars = function($type, $abc, $xyz); > $found = $vars[0]; > $username = $vars[1]; > $email = $vars[2]; > $nickname = $vars[3]; > > on the other end work? If so, then the bug is somewhere else. You probably want: $vars = array(); instead of: array($vars); which will throw an undefined variable error (if error reporting is turned up), but otherwise, yes, that should work. Kirk