找到最重叠的日子？

Question

我试着写一个会发出的查询:

最多重叠的日期/秒.

格式是 d/m/yyyy

所以我在这里有日期范围:

dateStart ----- dateEnd

  1/1---7/1                          

         8/1--15/1                   

               16/1------20/1               

         8/1--------------21/1       

                17/1---19/1 

                 18/1--19/1  

这是理想的结果分析:

左边的2个常见日是8/1和9/1 (出现在2个范围内)

在右边的4个常见日是18/1和19/1(出现在4个范围...... 4> 2所以它应该赢.)

期望的结果:

18/1

19/1

它们看起来最重叠.

编辑

这是日期时间范围的脚本.

DECLARE @t table( dt1 DATETIME , dt2 DATETIME)

INSERT INTO @t
SELECT '20110101','20110107'
UNION ALL
SELECT '20110108','20110115'
UNION ALL
SELECT '20110116','20110120'
UNION ALL
SELECT '20110108','20110121'
UNION ALL
SELECT '20110117','20110119'
UNION ALL
SELECT '20110118','20110119'

Answer 1

恐怕这并不完全是您想要的，但也许它对您有帮助（我的时间不多了）：

DECLARE @tbl table( startdate DATETIME , enddate DATETIME)

INSERT INTO @tbl
SELECT '20110101','20110107'
UNION ALL
SELECT '20110108','20110115'
UNION ALL
SELECT '20110116','20110120'
UNION ALL
SELECT '20110108','20110121'
UNION ALL
SELECT '20110117','20110119'
UNION ALL
SELECT '20110118','20110119'

;with overlapping_events as(
    select startdate, enddate
       , (select sum(inTimeSpan) 
    from (
       select case when startdate<=events.startdate then 1 else 0 end
         + case when enddate <= events.startdate then -1 else 0 end as inTimeSpan
       from @tbl 
       where startdate <= events.startdate
         or enddate <= events.startdate) as previous
    ) as overlapping
    from @tbl events
)
select oe.* 
from overlapping_events oe 
order by overlapping desc, startdate asc, enddate asc

startdate                     enddate                     overlapping
2011-01-18 00:00:00.000   2011-01-19 00:00:00.000         4
2011-01-17 00:00:00.000   2011-01-19 00:00:00.000         3
2011-01-08 00:00:00.000   2011-01-15 00:00:00.000         2
2011-01-08 00:00:00.000   2011-01-21 00:00:00.000         2
2011-01-16 00:00:00.000   2011-01-20 00:00:00.000         2
2011-01-01 00:00:00.000   2011-01-07 00:00:00.000         1