SQL Server 查询以对连续日期进行分组

Question

我有一个名为 Absence Details 的表，我想对连续日期进行分组。这里是数据

EID        AbsenceType  AbsenceStartDate               AbsenceEndDate
769     Holiday     2012-06-25  00:00:00.000            2012-06-25 23:59:59.000
769     Holiday     2012-06-26  00:00:00.000            2012-06-26 23:59:59.000
769     Holiday     2012-09-03  00:00:00.000            2012-09-03 23:59:59.000
769     Holiday     2012-09-04  00:00:00.000            2012-09-04 23:59:59.000
769     Holiday     2012-09-05  00:00:00.000            2012-09-05 23:59:59.000
769     Holiday     2012-09-06  00:00:00.000            2012-09-06 23:59:59.000
769     Holiday     2012-09-07  00:00:00.000            2012-09-07 23:59:59.000

我想要得到的结果是

EID     AbsenceType AbsenceStartDate          AbsenceEndDate
769     Holiday     2012-06-25  00:00:00.000         2012-06-26 23:59:59.000
769     Holiday     2012-09-03  00:00:00.000         2012-09-07 23:59:59.000

任何帮助深表感谢。

Answer 1

我已经简化了您的场景以隔离主要问题。让我们假设这个表有间隙:

with ns as (
select 1 as n union
select 2 as n union
select 3 as n union
select 8 as n union    --gap
select 9 as n )
select * 
into #ns
from ns;

现在，您期望的结果是:

ini fi 
--- -- 
1   3  
8   9  

为了得到这个结果，我以这种方式处理数据:首先，我创建了两个具有开始和结束时间段的 View ，然后，我加入了两个 View 以获得最终结果。请注意，我将 table 与其自身连接以定位开始和结束时间段:

with 
inis as                                     -- identifying start periods
(
   select n1.n
   from #ns n1
   left outer join #ns n2 
       on n1.n = n2.n + 1
   where n2.n is null
   ),
fis as                                      -- identifying ends periods
( 
   select n1.n 
   from #ns n1
   left outer join #ns n2 
       on n1.n = n2.n - 1
   where n2.n is null
   )  
select inis.n as ini, min( fis.n ) as fi    -- joining starts and ends
from inis 
inner join fis 
  on inis.n <= fis.n
group by inis.n
;
   

您可以将此技术转移到您的数据和数据类型。如果您在翻译查询时有任何问题，请随时提出。
Check query and results.