Ambiguous field name
Добавлено: 12 май 2006, 19:47
Код: Выделить всё
SELECT ID_SERVICE, NAME, COST FROM SERVICE s
join PRICE p on s.id_service = p.id_service
Ambiguous field name between table SERVICE and table PRICE .
ID_SERVICE.
Код: Выделить всё
SELECT ID_SERVICE, NAME, COST FROM SERVICE s
join PRICE p on s.id_service = p.id_service