Описание

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.

Заметка

Эта функция не должна вызываться статически.

Пример

$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);