Then if i make this $foo = new foo(); $foo->ID = 12; what is happening?? if __Set and __Get only works if a member isn't exits, How come I do this? $foo->ID = 12; if the member $ID is private??? I thought that this was possible because i have declared the two method set and get :P "Jochem Maas" <jochem@xxxxxxxxxxxxx> escribió en el mensaje news:47417E21.6000203@xxxxxxxxxxxxxxxx > Kiketom wrote: >> Hi all. >> Yesterday i have looking for the overloading members >> >> Member overloading >> void __set ( string name, mixed value ) >> mixed __get ( string name ) >> >> As an example i put this code: >> >> class foo >> { >> private $ID; >> private $Name; >> private $LastName; > > when you declare these three as 'real' members, __get() and __set() > will no longer be called - they are only called for non-existent members. > > so instead dump your data in an array or something > > private $data = array( > 'ID' => null, > 'Name' => null, > 'LastName' => null, > ); > >> >> private function __get($var) >> { >> return $var; > > there is no such thing as 'implicit class scope' $var will not refer to > $this->var as you seem to expect. > > return isset($this->data[$var]) ? $this->data[$var] : null; > >> } >> >> private function __set($var,$value) >> { >> $var = $value; > > same thing here. > > if (array_key_exists($var, $this->data) > $this->data[$var] = $value; > >> } >> } >> -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
