产品1
3张发票的购买数量为50 + 100 + 50 = 200和
1张发票的销售数量为10
我使用下面的代码将结果作为
Total Purchase - Total Sale = Closing Qty
200 - 10 = 290
但是我得到了错误的结果,如附图所示:
请指导我更正我的代码
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
答案 0 :(得分:0)
如果您想要最终结果,则应使用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
答案 1 :(得分:0)
如果我理解正确,您就会遇到问题,因为您有给定产品的多个购买记录。如果是这种情况,那么只需在 JOIN
之前汇总:
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;
实际上,我不清楚这两个表中哪一个有重复 - 或者两者都有。因此,您可能需要为S
执行类似操作。
答案 2 :(得分:0)
使用来自不同表的聚合时,不要加入表并聚合,然后先聚合并加入聚合:
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;
我已将ifnull
替换为coalesce
,因此该查询完全符合标准SQL。