sql仅按顺序分组

2021-12-13 00:00:00 sql group-by sql-server-2008 sql-server gaps-and-islands

假设我有下表:

MyTable
---------
| 1 | A |
| 2 | A |
| 3 | A |
| 4 | B |
| 5 | B |
| 6 | B |
| 7 | A |
| 8 | A |
---------

我需要 sql 查询输出以下内容:

I need the sql query to output the following:

---------
| 3 | A |
| 3 | B |
| 2 | A |
---------

基本上我正在做一个 group by 但只针对在序列中在一起的行.有什么想法吗?

Basically I'm doing a group by but only for rows which are together in the sequence. Any ideas?

请注意，数据库位于 sql server 2008 上.有一篇关于此主题的帖子，但它使用了 oracle 的 lag() 函数.

Note that the database is on sql server 2008. There is a post on this topic however it uses oracle's lag() function.

推荐答案

这被称为孤岛"问题.使用 Itzik Ben Gan 的方法:

This is known as the "islands" problem. Using Itzik Ben Gan's approach:

;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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