* Thus wrote Ben Edwards ([EMAIL PROTECTED]): > Wan I insatiate an object I want to pass a database resource that can be > used throughout the object. The below code should indicate what I want but > I get the error "mysql_query(): supplied argument is not a valid MySQL-Link > resource". I guess this is because "$this->$db = $db" is wrong.
When referencing 'this' vars you dont use the $ on the var, instead use: $this->db = $db; Curt -- "I used to think I was indecisive, but now I'm not so sure." -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
