help with inheritance problem

Hi. I'm working with PHP3 and am trying to create some classes. Something isn't working as expected while attempting to access variables in base classes. class my_db { var $host_name, $user_name, $user_pwd, $db_name, $connection; function my_db() { $this->db_connect(); } function db_connect() # this works ok { $this->ts = time(); $this->host_name = ... $this->user_name = ... $this->user_pwd = ... $this->db_name = ... $this->connection = @mysql_connect( $this->host_name, $this->user_name, $this->user_pwd ) or die( "Couldn't connect to host" ); $db = @mysql_select_db( $this->db_name, $this->connection ) or die( "Couldn't select database" ); return $this->connection; } } # my_db class base_table { var $db; // constructor // pass this an instance of a mydb object function db_table( $db_obj ) { $db = $db_obj; } function so_something_with_base() { // here, I want to access $this->db->connection in base class, but I cannot seem to get it. $bdb2 = $this->db; $conn = $bdb2->$connection; # <<--------------- line 116 . . . } . . . } # base_tbl class derived_tbl extends base_table { // constructor // pass this an instance of an mydb object function derived_tbl( $db_obj ) { $this->base_table( $db_obj ); # is this proper syntax? It seems to work... } } //===================== called in PHP3 script like so: . . . . $mydb = new my_db(); # this works ok $d_tbl = new derived_tbl( $mydb ); # this _seems_ to work, but... $db2 = $d_tbl->db; # scalar? Should be db object in base_tbl emit("db2->connection = [$db2->connection]<br>"); # <<------- line 55 //===================== Here's what I get trying to run the above script: Warning: Using scalar variable $db2 as an array or object in ..script file.. on line 55 Warning: Object bdb2 not found in classes.php on line 116 All the books I have describe PHP4, with a few caveats for PHP3. Can anybody give me an idea of what's wrong here? Thanks -David