Я пытаюсь подавить повторяющиеся значения в столбце TotalCarton. Попытка заменить значение пустым или нулевым, но потерпела неудачу. Любая помощь?
Вот сценарий SQL:
SELECT ORDERS.StorerKey,
ORDERS.OrderKey,
PackKey = (SELECT MAX(PackKey) FROM BAX_PACK_DTL WITH (NOLOCK) WHERE ORderKey = ORDERS.OrderKey),
PackHU = BAX_PACK_DTL.OuterPackID,
SalesOrderNum = ( SELECT Upper(Max(ORDERDETAIL.CustShipInst01)) FROM ORDERDETAIL WITH (NOLOCK) WHERE OrderKey = ORDERS.OrderKey),
DeliveryNum = Upper(ORDERS.ExternOrderKey),
TotalCarton = ( CASE BAX_PACK_DTL.PackType WHEN 'C' THEN Count(DISTINCT(BAX_PACK_DTL.OuterPackID))
ELSE 0 END ),
TotalPallet = ( CASE BAX_PACK_DTL.PackType WHEN 'P' THEN Count(DISTINCT(BAX_PACK_DTL.OuterPackID))
ELSE 0 END ),
SumCarton = (SELECT COUNT(DISTINCT(OuterPackSeq)) FROM BAX_PACK_DTL WITH (NOLOCK) WHERE PackType = 'C' AND PackKey = '0000000211'),
SumPallet = (SELECT COUNT(DISTINCT(OuterPackSeq)) FROM BAX_PACK_DTL WITH (NOLOCK) WHERE PackType = 'P' AND PackKey = '0000000211'),
AddWho = Upper(ORDERS.EditWho),
ORDERS.AddDate
FROM ORDERS WITH (NOLOCK) INNER JOIN ORDERDETAIL WITH (NOLOCK) ON ORDERS.StorerKey = ORDERDETAIL.StorerKey
AND ORDERS.OrderKey = ORDERDETAIL.OrderKey
INNER JOIN PICKDETAIL WITH (NOLOCK) ON ORDERDETAIL.StorerKey = PICKDETAIL.StorerKey
AND ORDERDETAIL.OrderKey = PICKDETAIL.OrderKey
AND ORDERDETAIL.OrderLineNumber = PICKDETAIL.OrderLineNumber
INNER JOIN BAX_PACK_DTL WITH (NOLOCK) ON PICKDETAIL.OrderKey = BAX_PACK_DTL.OrderKey
AND PICKDETAIL.PickDetailKey = BAX_PACK_DTL.PickDetailKey
WHERE (SELECT COUNT(DISTINCT(ORDERKEY)) FROM PICKDETAIL WITH (NOLOCK) WHERE OrderKey = ORDERS.OrderKey ) > 0
AND BAX_PACK_DTL.PackKey = '0000000211'
AND BAX_PACK_DTL.OuterPackID IN
('P111111111',
'P22222222',
'P33333333')
GROUP BY ORDERS.StorerKey,
ORDERS.OrderKey,
ORDERS.ExternOrderKey,
ORDERS.HAWB,
ORDERS.SO,
ORDERS.EditWho,
ORDERS.AddDate,
PICKDETAIL.WaveKey,
BAX_PACK_DTL.OuterPackID,
BAX_PACK_DTL.PackKey,
BAX_PACK_DTL.PackType
ORDER BY BAX_PACK_DTL.OuterPackID ASC
Ниже приведен текущий набор результатов на основе запроса выше.