sql group仅按行顺序排列

Pav*_*vel 14 sql sql-server group-by sql-server-2008 gaps-and-islands

说我有下表:

MyTable
---------
| 1 | A |
| 2 | A |
| 3 | A |
| 4 | B |
| 5 | B |
| 6 | B |
| 7 | A |
| 8 | A |
---------
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我需要sql查询输出以下内容:

---------
| 3 | A |
| 3 | B |
| 2 | A |
---------
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基本上我正在做一个group by但只针对序列中的行.有任何想法吗?

请注意,数据库位于sql server 2008上.有关此主题的帖子,但它使用oracle的lag()函数.

Mar*_*ith 24

这被称为"岛屿"问题.使用Itzik Ben Gan的方法:

;WITH YourTable AS
(
SELECT 1 AS N, 'A' AS C UNION ALL
SELECT 2 AS N, 'A' AS C UNION ALL
SELECT 3 AS N, 'A' AS C UNION ALL
SELECT 4 AS N, 'B' AS C UNION ALL
SELECT 5 AS N, 'B' AS C UNION ALL
SELECT 6 AS N, 'B' AS C UNION ALL
SELECT 7 AS N, 'A' AS C UNION ALL
SELECT 8 AS N, 'A' AS C
),
     T
     AS (SELECT N,
                C,
                DENSE_RANK() OVER (ORDER BY N) - 
                DENSE_RANK() OVER (PARTITION BY C ORDER BY N) AS Grp
         FROM   YourTable)
SELECT COUNT(*),
       C
FROM   T
GROUP  BY C,
          Grp 
ORDER BY MIN(N)
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