我的SQL Server数据库中有一个表，其中包含以下内容：
Date | Amount
------------|----------
2012-12-17 | 9.00
2012-12-18 | 8.00
2012-12-19 | 0.00
2012-12-20 | 1.50
2012-12-21 | 2.50
2012-12-22 | 0.00
2012-12-23 | 0.00
2012-12-24 | 0.00
2012-12-25 | 0.00
2012-12-26 | 4.00
2012-12-27 | 2.00
2012-12-28 | 7.00
我想要做的是每个选择3行，SUM Amount。如果SUM的总和为0，则应删除这3条记录。否则它应该只留下它们并接下来的3条记录并进行相同的检查。
因此，在这种情况下，只应从表中删除以下三个记录，因为它们是SUM导致0的唯一记录。
2012-12-23 | 0.00
2012-12-24 | 0.00
2012-12-25 | 0.00
如何在SQL Server中执行此操作？
答案 0 :(得分：6)
您可以使用ROW_NUMBER制作3个元素组和calucalte sum。
WITH cte AS
(
SELECT *, rn = (ROW_NUMBER() OVER(ORDER BY [Date]) - 1) / 3
FROM #tab
), cte2 AS
(
SELECT *, [sum] = SUM(Amount) OVER (PARTITION BY rn ORDER BY [Date])
FROM cte
)
SELECT *
FROM cte2
WHERE [sum] = 0;
的 LiveDemo
使用DELETE：
WITH cte AS
(
SELECT *, rn = (ROW_NUMBER() OVER(ORDER BY [Date]) - 1) / 3
FROM #tab
), cte2 AS
(
SELECT *, [sum] = SUM(Amount) OVER (PARTITION BY rn ORDER BY [Date])
FROM cte
)
DELETE t
FROM #tab t
JOIN cte2 c
ON t.[Date] = c.[Date]
WHERE [sum] = 0;
SELECT *
FROM #tab;
修改
如果您的数据可以包含负值，则可以使用：
WITH cte AS
(
SELECT *, rn = (ROW_NUMBER() OVER(ORDER BY [Date]) - 1) / 3
FROM #tab
), cte2 AS
(
SELECT rn, [sum] = SUM(Amount)
FROM cte
GROUP BY rn
)
SELECT c.*
FROM cte c
JOIN cte2 c2
ON c.rn = c2.rn
WHERE [sum] = 0;