Question

我有一个包含ScheduledTime列的表，它包含DateTime值;这些是开始的时间 3小时测试。该列看起来像这样：

SELECT ScheduledTime from Schedule;

2012-03-16 16:34:36.000
2012-03-16 16:34:36.000
2012-03-16 16:34:36.000
2012-03-16 16:34:36.000
2012-03-16 16:34:36.000
2012-03-16 16:34:36.000
2012-03-16 16:34:36.000
2012-03-16 16:34:36.000
2012-03-20 09:14:48.000
2012-03-20 13:14:48.000
2012-03-20 17:14:48.000
2012-03-20 21:14:48.000
2012-04-14 17:13:19.000
2012-04-14 21:13:19.000
2012-04-15 01:13:19.000
2012-04-15 05:13:19.000
2012-04-15 09:13:19.000
2012-04-15 13:13:19.000
2012-04-15 17:13:19.000
2012-04-15 21:13:19.000
2012-04-16 01:13:19.000
2012-04-16 05:13:19.000
2012-04-16 13:13:19.000
2012-04-16 17:13:19.000
2012-04-17 01:13:19.000
2012-04-17 05:13:19.000
2012-04-17 13:13:19.000
2012-04-18 18:29:16.000
2012-04-19 15:48:12.000
2012-04-19 19:10:00.000
2012-04-19 23:47:00.000
2012-04-20 10:27:28.000
2012-04-20 17:27:00.000

上面列出的日期时间值是3小时测试的开始时间。所以我需要获得下一个可用的插槽进行3小时测试。如果它可以在现有测试或currentTime + 1hr内容纳，那将是最好的。

任务是生成一个列（比如SlotsColumn），它保存从ScheduledTime列获得的3小时槽的开始时间，并且SlotsColumn值> {fn Now（）}即CurrentTime。

可选是如果它没有任何这样的值，它应该给CurrentTime + 1hr

使用SQL Server 2008。

I have tried this:  
WITH rows AS 
        (
        SELECT  *, ROW_NUMBER() OVER (ORDER BY [ScheduledTime] ) AS rn 
        FROM    [dbo].[Schedule]  
        WHERE [ScheduledTime] IS NOT NULL               
        ) 
SELECT  TOP(1)
DATEADD(second,10800, mc.ScheduledTime) AS NextTime 
FROM    rows mc 
JOIN    rows mp   
ON      mc.rn = mp.rn - 1   
WHERE (DATEDIFF(SECOND, mc.ScheduledTime, mp.ScheduledTime) - 10800) >= 10800 
order by mc.ScheduledTime DESC

查询返回将来一个有效开始时间的列表，仅当它存在时才会返回。

对于上面的示例数据，它返回'2012-04-20 03：47：00.000'这是错误的，因为它不认为在'2012-04-20 17：27：00.000'有另一个测试。它们不应重叠。期望的输出是：'2012-04-20 12：27：00.000'，因为它是在'2012-04-20 10：27：28.000'测试之后和'2012-04-20 17测试之前： 27：00.000'，如此完美的插槽进行3小时测试。对我来说，CurrentTime是'2012-04-20 11：45：23.393'

如果没有这样的有效时间，我如何确保它返回CurrentTime + 1hr？

可以在C＃中完成，但尝试SQL查询。

请提供任何帮助/提示。

Answer 1

DECLARE @now DATETIME2 = '2012-04-20 11:45:23.393'
DECLARE @schedule TABLE
        (
        ScheduledTime DATETIME2 NOT NULL
        )

INSERT
INTO    @schedule
VALUES 
('2012-03-16 16:34:36.000'),
('2012-03-16 16:34:36.000'),
('2012-03-16 16:34:36.000'),
('2012-03-16 16:34:36.000'),
('2012-03-16 16:34:36.000'),
('2012-03-16 16:34:36.000'),
('2012-03-16 16:34:36.000'),
('2012-03-16 16:34:36.000'),
('2012-03-20 09:14:48.000'),
('2012-03-20 13:14:48.000'),
('2012-03-20 17:14:48.000'),
('2012-03-20 21:14:48.000'),
('2012-04-14 17:13:19.000'),
('2012-04-14 21:13:19.000'),
('2012-04-15 01:13:19.000'),
('2012-04-15 05:13:19.000'),
('2012-04-15 09:13:19.000'),
('2012-04-15 13:13:19.000'),
('2012-04-15 17:13:19.000'),
('2012-04-15 21:13:19.000'),
('2012-04-16 01:13:19.000'),
('2012-04-16 05:13:19.000'),
('2012-04-16 13:13:19.000'),
('2012-04-16 17:13:19.000'),
('2012-04-17 01:13:19.000'),
('2012-04-17 05:13:19.000'),
('2012-04-17 13:13:19.000'),
('2012-04-18 18:29:16.000'),
('2012-04-19 15:48:12.000'),
('2012-04-19 19:10:00.000'),
('2012-04-19 23:47:00.000'),
('2012-04-20 10:27:28.000'),
('2012-04-20 17:27:00.000')

SELECT  COALESCE
        (
        (
        SELECT  TOP 1
                DATEADD(hour, 3, scheduledTime)
        FROM    @schedule s
        WHERE   s.scheduledTime >= DATEADD(hour, -3, @now)
                AND NOT EXISTS
                (
                SELECT  NULL
                FROM    @schedule sn
                WHERE   sn.scheduledTime > s.scheduledTime
                        AND sn.scheduledTime < DATEADD(hour, 6, s.scheduledTime)
                )

        ORDER BY
                scheduledTime
        ),
        DATEADD(hour, 1, @now)

-- This is redundant

        )

Answer 2

终于工作并达到预期结果的查询是：

    WITH rows AS 
                (
                SELECT  *, ROW_NUMBER() OVER (ORDER BY [ScheduledTime] ) AS rn 
                FROM    Schedule
                WHERE [ScheduledTime] IS NOT NULL                     
                ) 
    SELECT  COALESCE 
            ( 
            (
            SELECT  Top 1
            DateAdd(SECOND,-10800,mp.ScheduledTime) AS NewNextTime
            FROM    rows mc 
            JOIN    rows mp   
            ON      mc.rn = mp.rn - 1   
            WHERE (DATEDIFF(SECOND, mc.ScheduledTime, mp.ScheduledTime) - 10800) >= 10800 
            And mp.ScheduledTime >= GETDATE()
            And DateAdd(SECOND,-10800,mp.ScheduledTime)>= GETDATE()
            order by mc.ScheduledTime
            ),
            ( 
            SELECT  TOP 1 
                    DATEADD(hour, 3, scheduledTime) AS NewNextTime 
            FROM    Schedule s 
            WHERE   s.scheduledTime >= DATEADD(hour, -3, GetDate()) 
                    AND NOT EXISTS 
                    ( 
                    SELECT  NULL 
                    FROM    Schedule sn 
                    WHERE   sn.scheduledTime > s.scheduledTime 
                            AND sn.scheduledTime < DATEADD(hour, 6, s.scheduledTime) 
                    ) 

            ORDER BY 
                    scheduledTime 
            ),
           (
                SELECT  TOP 1 
                DATEADD(SECOND, (3600), GetDate()) AS NewNextTime 
            )  
            )

这将返回有效时间，或者在最坏情况下返回上一次测试结束后的时间。

SELECT COALESCE的第一部分，从现有的插槽中返回一个时隙。 SELECT COALESCE的第二部分在所有测试槽的末尾返回一个时隙。 SELECT COALESCE的第三部分，返回CurrentTime + 1hr。

你们所有人的讨论对于达成这个目标非常有帮助。我仍然会寻找更好的建议，现在将使用这个。

P.S。很抱歉发布此解决方案的延迟。

获取时隙列（SQL查询）

2 个答案: