> How do you pass by reference? I tried > $var = myFunction(&$var1, &$var2); > but it gave me a warning saying that pass by reference is depreciated
function myFunction( &$var1, &$var2 ) { } You now have to do it in the function definition, not the function call. Chris -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php