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我有这个表，其中包含一个包含 DateTime 值的 ScheduledTime 列；这些是 3 小时测试的开始时间。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插槽进行 3 小时测试。如果它可以容纳在现有测试或 currentTime+1hr 中，则最好。

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

可选的是，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

查询返回未来一个有效开始时间的列表，前提是它存在。

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。所需的输出是：“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'，非常适合 3 小时测试。我的 CurrentTime 是 '2012-04-20 11:45:23.393'

如果不存在这样的有效时间，我如何确保它返回 CurrentTime+1hr？

它可以在 C# 中完成，但尝试使用 SQL 查询。

请提供任何帮助/提示。

4

2 回答 2

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

        )
于 2012-04-20T11:15:50.027 回答
0

最终起作用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 
            )  
            )

这将返回一个有效时间，或者在最坏的情况下返回最后一次测试结束后的时间。

SELECT COALESCE 的第一部分从现有时隙中返回一个时隙。SELECT COALESCE 的第二部分，在所有测试时隙的末尾返回一个时隙。SELECT COALESCE 的第三部分，返回为 CurrentTime+1hr。

大家对此的讨论对达成这一点非常有帮助。我仍然会寻找更好的建议，截至目前将使用这个。

PS 很抱歉延迟发布此解决方案。

于 2012-04-23T08:43:41.973 回答