Re: DB_Object linking question

DB_DataObject, not DB_Object On Wed, 22 Dec 2004 11:39:34 -0800 (PST), l Burnerheimerton <lburnerheimerton@yahoo.com> wrote: > I have two tables, building and rooms. There can be more than one room per building. The > structure is shown below in database.ini, as well as the database.links.ini and code. > > The problem is that the rooms are displaying but not the linked building names. > > Any ideas? > > code ======================================= > <?php > require_once 'dao/Room.php'; You should be requiring DB/DataObject.php here require_once 'DB/DataObject.php'; > $dbconfig = parse_ini_file('dao/db_dataobject.ini', true); > // Load Database Settings (note main PEAR class is loaded by Room.php) > foreach ($dbconfig as $class => $values) { > $options = &PEAR::getStaticProperty($class, 'options'); > $options = $values; > } > $room = new DataObject_Room(); and using the factory here $room =& DB_DataObject::factory('room'); > $room->building_ID = '2'; > $room->find(); > echo 'Room - Buildings:<br />'; > while ($room->fetch()) { > $room->getLink('building_ID','building','ID'); getLink *returns* the linked dataobject. You also don't need the second and third params as long as your db.links.ini is set up. $building =& $room->getLink('building_ID'); > echo $room->number . ' - '. $building->name . '<br/>'; > > } > ?> > -- Justin Patrin