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sketchMedia · 12-17-2007, 08:51 PM

the cause is that he has tried to pass by reference at 'call-time' i.e. when he calls a function he has already defined. To solve it you need to tell PHP that you will be passing by reference in the function declaration and not in the call:

PHP Code:


			
function test(&$arg1, &$arg2)

also the refernece operator (&) has been depreciated for the 'new' keyword, and i believe that PHP now automatically passes objects around by reference.

12-17-2007, 08:51 PM	#8 (permalink)
sketchMedia The Prestige Join Date: Oct 2007 Location: Manchester, UK Posts: 854 Thanks: 32	the cause is that he has tried to pass by reference at 'call-time' i.e. when he calls a function he has already defined. To solve it you need to tell PHP that you will be passing by reference in the function declaration and not in the call: PHP Code: `function test(&$arg1, &$arg2)` also the refernece operator (&) has been depreciated for the 'new' keyword, and i believe that PHP now automatically passes objects around by reference. __________________ mysql> SELECT * FROM `users` WHERE `users`.`clue` > 0; Empty set (0.00 sec)