这是一个很好的,所以坚持下去。
我有两张桌子跟踪位置的人。我已经使用LEAD和LAG成功合并它们,以在单个表格中创建无缝过渡 我现在的问题是,对于其中一个表格,我需要包含其他活动项目,这些项目位于某些细分市场中 所以为简单起见,我有以下正常情况:
| System | ID | Item | Start | End
| Alpha | 987 | 123 | May, 20 2015 07:00:00 | May, 20 2015 08:00:00
| Alpha | 374 | 123 | May, 20 2015 08:00:00 | May, 20 2015 10:00:00
| Beta | 184 | 123 | May, 20 2015 10:00:00 | May, 20 2015 11:00:00
| Beta | 798 | 123 | May, 20 2015 11:00:00 | May, 20 2015 12:00:00
现在,这些额外的项目位于某些记录中,因此我的数据看起来像这样:
| System | ID | Item | Start | End
| Alpha | 987 | 123 | May, 20 2015 07:00:00 | May, 20 2015 08:00:00
| Alpha | 374 | 123 | May, 20 2015 08:00:00 | May, 20 2015 10:00:00
| Beta | 184 | 123 | May, 20 2015 10:00:00 | May, 20 2015 11:00:00
| Charlie | 874 | 123 | May, 20 2015 10:20:00 | May, 20 2015 10:25:00
| Charlie | 984 | 123 | May, 20 2015 10:37:00 | May, 20 2015 10:54:00
| Beta | 798 | 123 | May, 20 2015 11:00:00 | May, 20 2015 12:00:00
请注意,两个Charlie事件发生在Beta 184中。
我现在已经把桌子分成了每分钟的记录(下一步所需要的,不要问)并且在表格中显示正确的记录,但我可以&#39 ;让我了解如何将它重新组合在一起并在每个段上具有正确的开始和结束时间。 此外,将以某种方式形成新记录以填补空白。
上述情况的最终结果是:
| System | ID | Item | Start | End
| Alpha | 987 | 123 | May, 20 2015 07:00:00 | May, 20 2015 08:00:00 |
| Alpha | 374 | 123 | May, 20 2015 08:00:00 | May, 20 2015 10:00:00 |
| Beta | 184 | 123 | May, 20 2015 10:00:00 | May, 20 2015 10:20:00 |
| Charlie | 874 | 123 | May, 20 2015 10:20:00 | May, 20 2015 10:25:00 |
| Beta | 184 | 123 | May, 20 2015 10:25:00 | May, 20 2015 10:37:00 | new
| Charlie | 984 | 123 | May, 20 2015 10:37:00 | May, 20 2015 10:54:00 |
| Beta | 184 | 123 | May, 20 2015 10:54:00 | May, 20 2015 11:00:00 | new
| Beta | 798 | 123 | May, 20 2015 11:00:00 | May, 20 2015 12:00:00 |
这有意义吗?
希望有人可以帮助我。
答案 0 :(得分:3)
您可以使用以下内容:
DECLARE @Source TABLE (
[System] VARCHAR(50),
ID INT PRIMARY KEY,
Item INT NOT NULL,
Start DATETIME NOT NULL,
[End] DATETIME NOT NULL,
CHECK (Start<[End])
)
INSERT INTO @Source (System, ID, Item, Start, [End]) VALUES ('Alpha', 987, 123, '2015-05-20 07:00', '2015-05-20 08:00')
INSERT INTO @Source (System, ID, Item, Start, [End]) VALUES ('Alpha', 374, 123, '2015-05-20 08:00', '2015-05-20 10:00')
--INSERT INTO @Source (System, ID, Item, Start, [End]) VALUES ('Delta', 555, 123, '2015-05-20 09:30', '2015-05-20 10:00')
INSERT INTO @Source (System, ID, Item, Start, [End]) VALUES ('Beta', 184, 123, '2015-05-20 10:00', '2015-05-20 11:00')
--INSERT INTO @Source (System, ID, Item, Start, [End]) VALUES ('Charlie', 111, 123, '2015-05-20 10:05', '2015-05-20 10:07')
--INSERT INTO @Source (System, ID, Item, Start, [End]) VALUES ('Charlie', 222, 123, '2015-05-20 10:10', '2015-05-20 10:20')
INSERT INTO @Source (System, ID, Item, Start, [End]) VALUES ('Charlie', 874, 123, '2015-05-20 10:20', '2015-05-20 10:25')
INSERT INTO @Source (System, ID, Item, Start, [End]) VALUES ('Charlie', 984, 123, '2015-05-20 10:37', '2015-05-20 10:54')
INSERT INTO @Source (System, ID, Item, Start, [End]) VALUES ('Beta', 798, 123, '2015-05-20 11:00', '2015-05-20 12:00')
;WITH CTE AS (
SELECT *
FROM @Source s1
OUTER APPLY (
SELECT MIN(s2.Start) AS NextStart
FROM @Source s2
WHERE s2.Start>s1.Start AND s2.Start<s1.[End]
) q2
OUTER APPLY (
SELECT MAX(s3.[End]) AS PreviousEnd
FROM @Source s3
WHERE s3.[End]>s1.Start AND s3.[End]<s1.[End]
) q3
)
SELECT System, ID, Item, Start, [End]
FROM CTE WHERE NextStart IS NULL AND PreviousEnd IS NULL
UNION ALL
SELECT System, ID, Item, Start, NextStart
FROM CTE WHERE NextStart IS NOT NULL
UNION ALL
SELECT System, ID, Item, PreviousEnd, [End]
FROM CTE WHERE PreviousEnd IS NOT NULL
UNION ALL
SELECT s4.System, s4.ID, s4.Item, q5.[End], q6.Start
FROM @Source s4
CROSS APPLY (
SELECT *
FROM @Source s5
WHERE s5.Start>s4.Start AND s5.Start<s4.[End]
) q5
CROSS APPLY (
SELECT TOP 1 *
FROM @Source s6
WHERE s6.Start>q5.Start AND s6.Start<s4.[End]
ORDER BY s6.Start
) q6
WHERE q5.[End]<q6.Start
ORDER BY [Start]
UNION的第一部分处理与任何其他间隔不重叠的间隔。
第二部分处理在间隔结束时重叠的行。
第三部分处理在间隔开始时重叠的行。
最后一部分产生与基本区间重叠的两个其他区间之间的间隙(当两个区间不相邻时)。
答案 1 :(得分:1)
似乎@RazvanSocol打败了我,但是因为我做了这个并且它看起来比他简单,我也会在这里发布:
create table #times (
Item int,
EndTime datetime,
primary key (Item, EndTime)
)
insert into #times
select distinct Item, StartTime from timetable
union
select distinct Item, EndTime from timetable
;with CTE as (
select
System, ID, Item, StartTime
from
timetable T1
union all
select
T1.System, T1.ID, T1.Item, T2.EndTime
from
timetable T1
join timetable T2 on T1.Item = T2.Item and
T1.StartTime < T2.StartTime and T1.EndTime > T2.EndTime
where
-- This check added to handle cases with adjacent ranges in the dates
-- as pointed out by Razvan Socol
not exists (select 1 from timetable T3 where T3.StartTime = T2.EndTime)
)
select
System, ID, Item, StartTime, E.EndTime
from
CTE
outer apply (
select top 1 EndTime from #times T
where T.Item = CTE.Item and T.EndTime > CTE.StartTime
order by EndTime asc
) E
order by Item, StartTime
我用了一个临时工。用于收集每个项目的所有不同开始/结束时间的表,然后在CTE中使用第二个选择来创建缺失行,最后通过搜索找到该项目的最早日期来重新计算每行的结束日期。
编辑:添加了对相邻范围的检查
答案 2 :(得分:-2)
试试这个
;with cte as
(select distinct t1.system,t1.ID,t1.item,t2.start,
ROW_NUMBER() over(order by t2.start) rownum
from timetable t1 right join
(select system,start from timetable
union all
select system,[end] from timetable )
as t2 on t1.start = t2.start
)
select
case when c1.system is null
then
'Beta'
else
c1.system
end as system,
case when c1.id is null
then
'184'
else
c1.id
end as id,
case when c1.item is null
then
'123'
else
c1.item
end as item,c1.start,c2.start as [End] from cte c1
join cte c2 on c1.rownum = c2.rownum-1 and c1.start != c2.start
order by c1.rownum;