Question

我有下表：

╔═════════════════════════╦═══╗
║       DATETIME          ║ M ║
╠═════════════════════════╬═══╣
║ 2015-06-17 15:03:44.840 ║ 1 ║
║ 2015-06-17 15:03:33.940 ║ 1 ║
║ 2015-06-17 15:03:32.950 ║ 1 ║
║ 2015-06-17 15:03:22.060 ║ 1 ║
║ 2015-06-17 15:03:11.210 ║ 0 ║
║ 2015-06-17 14:50:12.950 ║ 0 ║
║ 2015-06-17 13:50:12.850 ║ 0 ║
║ 2015-06-17 12:50:12.790 ║ 0 ║
║ 2015-06-17 12:49:10.410 ║ 1 ║
║ 2015-06-17 12:48:59.510 ║ 1 ║
║ 2015-06-17 12:48:48.630 ║ 1 ║
║ 2015-06-17 12:48:37.730 ║ 1 ║
║ 2015-06-17 12:48:26.830 ║ 1 ║
║ 2015-06-17 12:48:15.940 ║ 1 ║
║ 2015-06-17 12:48:14.950 ║ 1 ║
║ 2015-06-17 12:48:04.060 ║ 1 ║
║ 2015-06-17 12:47:53.170 ║ 1 ║
║ 2015-06-17 12:47:42.280 ║ 0 ║
║ 2015-06-17 12:44:14.350 ║ 0 ║
║ 2015-06-17 12:43:08.970 ║ 1 ║
╚═════════════════════════╩═══╝

我的目标是选择零和一组，并计算日期和时间的差异。

我怎样才能做到这一点？

我的最终结果是能够找到连续出现的零并计算相应的持续时间。

＃| T |持续时间
1 | 0 | 2小时13分钟
2 | 0 | 3分钟

Answer 1

这是一个解决方案：

DECLARE @t TABLE ( d DATETIME, m INT )

INSERT  INTO @t
VALUES  ( '2015-06-17 15:03:44.840', 1 ),
        ( '2015-06-17 15:03:33.940', 1 ),
        ( '2015-06-17 15:03:32.950', 1 ),
        ( '2015-06-17 15:03:22.060', 1 ),
        ( '2015-06-17 15:03:11.210', 0 ),
        ( '2015-06-17 14:50:12.950', 0 ),
        ( '2015-06-17 13:50:12.850', 0 ),
        ( '2015-06-17 12:50:12.790', 0 ),
        ( '2015-06-17 12:49:10.410', 1 ),
        ( '2015-06-17 12:48:59.510', 1 ),
        ( '2015-06-17 12:48:48.630', 1 ),
        ( '2015-06-17 12:48:37.730', 1 ),
        ( '2015-06-17 12:48:26.830', 1 ),
        ( '2015-06-17 12:48:15.940', 1 ),
        ( '2015-06-17 12:48:14.950', 1 ),
        ( '2015-06-17 12:48:04.060', 1 ),
        ( '2015-06-17 12:47:53.170', 1 ),
        ( '2015-06-17 12:47:42.280', 0 ),
        ( '2015-06-17 12:44:14.350', 0 ),
        ( '2015-06-17 12:43:08.970', 1 );

WITH    cte
          AS ( SELECT   * ,
                        ROW_NUMBER() OVER ( ORDER BY d DESC )
                        - ROW_NUMBER() OVER ( PARTITION BY m ORDER BY d DESC ) rn
               FROM     @t
             )
    SELECT  ROW_NUMBER() OVER(ORDER BY MIN(d) desc) AS [#],
            DATEDIFF(mi, MIN(d), MAX(d)) / 60 Hours,
            DATEDIFF(mi, MIN(d), MAX(d)) - ( DATEDIFF(mi, MIN(d), MAX(d)) / 60 ) * 60 Minutes
    FROM    cte
    WHERE   m = 0
    GROUP BY rn

输出：

#   Hours   Minutes
1   2       13
2   0       3

在CTE内部有一些间隙和岛屿解决方案，然后按等级分组并计算结果。

SQL Server交替组行

1 个答案: