我有下表,它显示了 SalesQty
和 StockQty
按 Article
分组, Supplier
, Branch
和 Month
.
╔════════╦════════╦══════════╦═════════╦══════════╦══════════╗
║ Month ║ Branch ║ Supplier ║ Article ║ SalesQty ║ StockQty ║
╠════════╬════════╬══════════╬═════════╬══════════╬══════════╣
║ 201811 ║ 333 ║ 2 ║ 3122 ║ 4 ║ 11 ║
║ 201811 ║ 345 ║ 1 ║ 1234 ║ 2 ║ 10 ║
║ 201811 ║ 345 ║ 1 ║ 4321 ║ 3 ║ 11 ║
║ 201812 ║ 333 ║ 2 ║ 3122 ║ 2 ║ 4 ║
║ 201812 ║ 345 ║ 1 ║ 1234 ║ 3 ║ 12 ║
║ 201812 ║ 345 ║ 1 ║ 4321 ║ 4 ║ 5 ║
║ 201901 ║ 333 ║ 2 ║ 3122 ║ 1 ║ 8 ║
║ 201901 ║ 345 ║ 1 ║ 1234 ║ 6 ║ 9 ║
║ 201901 ║ 345 ║ 1 ║ 4321 ║ 2 ║ 8 ║
║ 201902 ║ 333 ║ 2 ║ 3122 ║ 7 ║ NULL ║
║ 201902 ║ 345 ║ 1 ║ 1234 ║ 4 ║ 13 ║
║ 201902 ║ 345 ║ 1 ║ 4321 ║ 1 ║ 10 ║
╚════════╩════════╩══════════╩═════════╩══════════╩══════════╝
现在我想对 SalesQty
求和并获取最新 StockQty
并按 Article, Supplier, Branch
分组.
最终结果应该是这样的:
╔════════╦══════════╦═════════╦═════════════╦════════════════╗
║ Branch ║ Supplier ║ Article ║ SumSalesQty ║ LatestStockQty ║
╠════════╬══════════╬═════════╬═════════════╬════════════════╣
║ 333 ║ 2 ║ 3122 ║ 14 ║ NULL ║
║ 345 ║ 1 ║ 1234 ║ 15 ║ 13 ║
║ 345 ║ 1 ║ 4321 ║ 10 ║ 10 ║
╚════════╩══════════╩═════════╩═════════════╩════════════════╝
我已经试过了,但它给了我一个错误,我不知道在这种情况下我必须做什么。
我做了这个例子,所以你可以自己试试。 db<>fiddle
SELECT
Branch,
Supplier,
Article,
SumSalesQty = SUM(SalesQty),
-- my attempt
LatestStockQty = (SELECT StockQty FROM TestTable i
WHERE MAX(Month) = Month
AND TT.Branch = i. Branch
AND TT.Supplier = i.Branch
AND TT.Article = i.Branch)
FROM
TestTable TT
GROUP BY
Branch, Supplier, Article
感谢您的帮助!
最佳答案
我们可以在这里尝试使用 ROW_NUMBER
来隔离每个组的最新记录:
WITH cte AS (
SELECT t.*, ROW_NUMBER() OVER (PARTITION BY Branch, Supplier, Article
ORDER BY Month DESC) rn,
SUM(SalesQty) OVER (PARTITION BY Branch, Supplier, Article) SumSalesQty
FROM TestTable t
)
SELECT
Month,
Branch,
Supplier,
Article,
SumSalesQty,
StockQty
FROM cte
WHERE rn = 1;
在我们计算的 CTE 中,为每个 Branch
/Supplier
/Article
分组一个行号值,最近的从 1 开始月。我们还计算同一分区的销售量总和。然后,我们只需要从该 CTE 中选择行号等于 1 的所有行。
Demo
关于sql-server - T-SQL : I'm trying to get the latest row of a column but also the sum of another column,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/54923664/