object $DB_DataObject->database ([string $name])
object $DB_DataObject->database
Without any argument, it returns the database that the object deals with. With a string, it will set the database for the instance of the object.
This function can not be called statically.
Example 33-1. Getting the database name
$person = new DataObjects_Person; echo $person->database(); // echo's mydatabase // now use the same object to check on a mirror.. $person->database('mirror1'); $person->id = 12; $person->find(true);