Katharina Wolkwitz wrote:
Hi again!
I think I've found the solution to my problem:
Changing the following line in accesscontrol.php from:
function doControlUserAccess( $input, $argv, &$parser )
prototype means the third parameter is a reference.
Since it was given a value, per the error message, changing the
prototype to
function doControlUserAccess( $input, $argv, $parser )
fixes it.