Question

我正在尝试在我的表格中创建两个字段，以显示视频的时间是否与任何其他视频时间冲突，以及视频的冲突。

示例数据：

SELECT vid, timein, timeout
FROM mytable

vid timein                      timeout
1   2015-04-15 06:00:00.000     2015-04-16 17:00:00.000
2   2015-04-17 03:00:00.000     2015-04-17 18:00:00.000
3   2015-04-16 16:00:00.000     2015-04-17 06:00:00.000
4   2015-04-12 12:00:00.000     2015-04-12 22:00:00.000
5   2015-03-25 01:00:00.000     null

期望的输出：

vid timein                      timeout                     Clash   Clashwith
1   2015-04-15 06:00:00.000     2015-04-16 17:00:00.000     CLASH   3
2   2015-04-17 03:00:00.000     2015-04-17 18:00:00.000     CLASH   3
3   2015-04-16 16:00:00.000     2015-04-17 06:00:00.000     CLASH   1, 2
4   2015-04-12 12:00:00.000     2015-04-12 22:00:00.000     OK
5   2015-03-25 01:00:00.000     null                        OK

我做过什么：

SELECT vid, timein, timeout,
CASE WHEN (SELECT tin.timein 
           FROM mytable tin 
           WHERE tin.vid = mytable.vid 
           AND mytable.timeout IS NOT NULL) 
           BETWEEN mytable.timein AND mytable.timeout
     THEN 'CLASH' 
     ELSE 'OK' 
     END AS inclash,
CASE WHEN (SELECT tout.timeout 
           FROM mytable tout 
           WHERE tout.vid = mytable.vid 
           AND mytable.timeout IS NOT NULL) 
           BETWEEN mytable.timein AND mytable.timeout
     THEN 'CLASH' 
     ELSE 'OK' 
     END AS outclash
FROM mytable

这不起作用，因为它会对所有结果发生冲突，我也不知道如何与列编写冲突。

Answer 1

你可以用一个简单的apply()来做，并连接所有冲突的id你可以使用xml技巧：

select
    t1.vid,
    t1.[timein],
    t1.[timeout],
    case when c.Clashwith is not null then 'CLASH' else 'OK' end as Clash,
    c.Clashwith
from mytable as t1
    outer apply (
        select
            stuff(
                (
                    select ',' + cast(t2.vid as nvarchar(max))
                    from mytable as t2
                    where
                        t2.vid <> t1.vid and
                        t2.[timein] <= t1.[timeout] and
                        t2.[timeout] >= t1.[timein]
                     for xml path(''), type
                ).value('.', 'nvarchar(max)')
            ,1,1,'') as Clashwith
    ) as c

<强> sql fiddle demo

SQL时间＆amp;超时表中的任何ID之间的冲突

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