sql - sql 仅按顺序的行分组

Question

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 查询来输出以下内容：

---------
| 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。有任何想法吗？

请注意，该数据库位于 sql server 2008 上。关于此主题有一篇文章，但它使用了 oracle 的 lag() 函数。

Answer 1

这就是所谓的“孤岛”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)

Answer 2

这对你有用......

SELECT 
  Total=COUNT(*), C 
FROM 
(
 SELECT 
 NGroup = ROW_NUMBER() OVER (ORDER BY N) - ROW_NUMBER() OVER (PARTITION BY C ORDER BY N),
 N,
 C
 FROM MyTable 
)RegroupedTable
GROUP BY C,NGroup

Answer 3

只是为了好玩，没有任何特定于 SQL 的函数，并且不假设 ID 列是单调递增的：

WITH starters(name, minid, maxid) AS (
    SELECT
        a.name, MIN(a.id), MAX(a.id)
    FROM
        mytable a RIGHT JOIN
        mytable b ON
            (a.name <> b.name AND a.id < b.id) 
    WHERE 
        a.id IS NOT NULL
    GROUP BY 
        a.name
),
both(name, minid, maxid) AS (
    SELECT
        name, minid, maxid
    FROM
        starters
    UNION ALL
    SELECT
        name, MIN(id), MAX(id)
    FROM
        mytable
    WHERE
        id > (SELECT MAX(maxid) from starters)
    GROUP BY
        name
)
SELECT
    COUNT(*), m.name, minid
FROM 
    both INNER JOIN 
    mytable m ON
        id BETWEEN minid AND maxid
GROUP BY
    m.name, minid

结果（忽略 midid 列）：

(No column name)    name    minid
3   A   1
3   B   4
2   A   7