"Kevin Bridges" <[EMAIL PROTECTED]> wrote in message news:[EMAIL PROTECTED] > Greetings php-general, > > I'm just starting out with php ... judging from the posts I've been > reading on this list I'm thinking this question is so basic it might > almost be pathetic! I'm assuming the answers are in the php manual, > but I have not found the correct pages ... any pointers to manual > pages, tutorials, or an explanation would be greatly appreciated. > > <?php > require_once 'library/functions/dbConnect.php'; > class Content { > var $db = null; // PEAR::DB pointer > function Content(&$db) { > $this->db = $db; > } > } > ?> > > I'm writing my first class following an example I found on the web > and I have 2 questions. The constructor for Content accepts an > object ... why is &$db used instead of $db? > > I'm used to a this.varName syntax from other languages ... I grasp > what is happening with $this->db, but I don't understand the literal > translation of -> and would like help understanding.
Hi Kevin, with PHP4 &$db means the value is passed by reference. Otherwise a copy of $db would be passed to the method. See here: http://de2.php.net/references With PHP5 variable assignments or passing variables to functions by reference is standard. $this->db is the PHP equal to this.varName. You access property db of the current object. Methods are called this way: $this->methodName() Hope this clears things up a bit. Best regards, Torsten Roehr -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
