http://softtime.ru/forumhttp://softtime.ru/forum/read.php?id_forum=3&id_theme=73746#post434460сделал так: $sql = "(SELECT photo.id_factory AS id, photo.id AS id_photo, photo.url AS url_photo, factory.id AS id_factory, factory.name AS...оператор UNIONFri, 7 May 2010 08:54:35 +0300Bvz