对于此表：
+----+--------+-------+
| ID | Status | Value |
+----+--------+-------+
| 1 | 1 | 4 |
| 2 | 1 | 7 |
| 3 | 1 | 9 |
| 4 | 2 | 1 |
| 5 | 2 | 7 |
| 6 | 1 | 8 |
| 7 | 1 | 9 |
| 8 | 2 | 1 |
| 9 | 0 | 4 |
| 10 | 0 | 3 |
| 11 | 0 | 8 |
| 12 | 1 | 9 |
| 13 | 3 | 1 |
+----+--------+-------+
我需要将相同的顺序组相加Status才能产生此结果。
+--------+------------+
| Status | Sum(Value) |
+--------+------------+
| 1 | 20 |
| 2 | 8 |
| 1 | 17 |
| 2 | 1 |
| 0 | 15 |
| 1 | 9 |
| 3 | 1 |
+--------+------------+
如何在SQL Server中做到这一点？
注意：该ID列中的值是连续的。
根据我在您的问题中添加的标签，这是一个空白问题。
最好的解决方案可能是
WITH T
AS (SELECT *,
ID - ROW_NUMBER() OVER (PARTITION BY [STATUS] ORDER BY [ID]) AS Grp
FROM YourTable)
SELECT [STATUS],
SUM([VALUE]) AS [SUM(VALUE)]
FROM T
GROUP BY [STATUS],
Grp
ORDER BY MIN(ID)
如果ID值不能保证如所述连续，那么您将需要使用
ROW_NUMBER() OVER (ORDER BY [ID]) -
ROW_NUMBER() OVER (PARTITION BY [STATUS] ORDER BY [ID]) AS Grp
而是在CTE定义中。
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