Sql表左外部聯接結果分組依據

Question

產品1

3張發票的購買數量為50 + 100 + 50 = 200和

1張發票的銷售數量為10

我正在使用下面的代碼來獲取結果

Total Purchase - Total Sale = Closing Qty 
     200       -      10    = 290

但是我得到了錯誤的結果，如下圖所示：

請指導我更正我的代碼

SELECT 
    P.PRODUCT as PRODUCTNAME,
    P.QUANTITY AS PURCHASE,
    ISNULL(S.QUANTITY, 0) AS SALE,
    ISNULL(P.QUANTITY, 0) - ISNULL(s.QUANTITY, 0) AS CLOSINGQTY
FROM 
    [PurchaseData] P 
LEFT OUTER JOIN 
    [DeliveryData1] S ON P.Product = s.PRODUCT

Answer 1

如果您想要最終結果，則應使用sum和group by

      SELECT 
        P.PRODUCT as PRODUCTNAME,
        sum(P.QUANTITY AS PURCHASE),
        sum(isnull(S.QUANTITY,0)) AS SALE,
        sum(isnull(P.QUANTITY,0))-sum(isnull(s.QUANTITY,0)) AS CLOSINGQTY

    FROM [PurchaseData] P LEFT OUTER JOIN [DeliveryData1] S ON P.Product = s.PRODUCT
    GROUP BY P.PRODUCT

Answer 2

如果我理解正確，則您有問題，因為給定產品具有多個購買記錄。 如果是這種情況，則只需在 JOIN 之前進行聚合：

SELECT p.*, COALESCE(S.QUANTITY, 0) AS SALE,
       COALESCE(P.QUANTITY, 0)-COALESCE(s.QUANTITY, 0) AS CLOSINGQTY
FROM (SELECT p.product, p.quantity
      FROM PurchaseData P
      GROUP BY p.product
     ) p LEFT OUTER JOIN 
     DeliveryData1 S
     ON P.product = s.producct;

實際上，我不清楚兩個表中的哪個表都有重復項，或者即使兩個都重復。 因此，您可能需要對S執行類似的操作。

Answer 3

當使用來自不同表的聚合時，不要先合並表再聚合，而先聚合並加入聚合：

select
  p.product as productname,
  p.sum_quantity as purchase,
  coalesce(s.sum_quantity, 0) as sale,
  p.sum_quantity - coalesce(s.sum_quantity, 0) as closingqty
from
(
  select product, sum(quantity) as sum_quantity
  from purchasedata
  group by product
) p
left join
(
  select product, sum(quantity) as sum_quantity
  from deliverydata1
  group by product  
) s on s.product = p.product;

我已經將ifnull替換為coalesce ，因此查詢完全符合標准SQL。