TSQL

Question

我正在开发 SQL Server 2017 (v14.0)。

我有一个 table 这样的：

Key | State    | from       | until      |
----+----------+------------+------------+
100 | open     | 01.01.2021 | 01.01.2021 |
100 | open     | 02.01.2021 | 02.01.2021 |
100 | closed   | 03.01.2021 | 13.01.2021 |
100 | open     | 14.01.2021 | 20.01.2021 |
100 | open     | 20.01.2021 | 30.01.2021 |

我想按 Key 和 State 对其进行分组，但仅限于连续行。

所以我的预期结果是这样的：

Key | State    | from       | until      |
----+----------+------------+------------+
100 | open     | 01.01.2021 | 02.01.2021 |
100 | closed   | 03.01.2021 | 13.01.2021 |
100 | open     | 14.01.2021 | 30.01.2021 |

知道怎么做吗？我有一种强烈的感觉，在 ROW_NUMBER 的帮助下这应该是可能的，但我还没有弄清楚......

（在这个示例数据中，一些奇怪的 group by calendarweek 或类似的东西可能是可能的，但这不是我的意图）

Answer 1

这是一个缺口和孤岛问题。一种解决方案是：

WITH cte1 AS (
    SELECT *, CASE WHEN LAG([state]) OVER (PARTITION BY [key] ORDER BY [from]) = [state] THEN 0 ELSE 1 END AS chg
    FROM t
), cte2 AS (
    SELECT *, SUM(chg) OVER (PARTITION BY [key] ORDER BY [from]) AS grp
    FROM cte1
)
SELECT [key], grp, MIN([state]), MIN([from]), MAX([until])
FROM cte2
GROUP BY [key], grp
ORDER BY [key], grp

Answer 2

一种可能如下补充：

Create table myTable_o1
 (
 [key] int
 ,[state] varchar(100)
 ,[From] date
 ,[Until] date
 )

insert into myTable_o1 values (100, 'open', '2021-01-01','2021-01-01')
insert into myTable_o1 values (100, 'open', '2021-01-02','2021-01-02')
insert into myTable_o1 values (100, 'closed', '2021-01-03','2021-01-13')
insert into myTable_o1 values (100, 'open', '2021-01-4','2021-01-20')
insert into myTable_o1 values (100, 'open', '2021-01-20','2021-01-30')

SELECT 
   [key] 
   ,[state] 
   ,[From] 
   ,[until] 
FROM
(
Select 
  [key] 
 , [state] 
 , [From] 
 , row_number() over (partition by tiles order by [key]) row_num
 , ISNULL(Lead(Until) over (partition by tiles order by [key]) , Until) [until]
 FROM
 (
    SELECT * ,
    Ntile(2) over ( order by [Key]) as [tiles]
    from myTable_o1
   ) AS A
 ) AS B WHERE B.row_num in (1,3)

TSQL - 仅对连续行进行 GROUP BY

TSQL - GROUP BY on continous rows only

sql

sql-server

group-by

sql-server-2017