Question

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

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

非常感谢任何帮助。

Answer 1

我简化了您的方案以隔离主要问题。让我们在这张表中填补空白：

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，您期望得到的结果是：

ini fi 
--- -- 
1   3  
8   9

为了得到这个结果，我按照这种方式按摩数据：首先我用开始和结束时段创建两个视图，然后我加入两个视图以获得最终结果。请注意，我自己连接表以查找开始和结束时段：

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
;

您可以将此技术传输到数据和数据类型。如果您有任何问题翻译查询可以自由询问。

Check query and results.

Answer 2

这是适合我的解决方案。

SELECT EID, AbsenceType, MIN(AbsenceStartDate) AS AbsenceStartDate, MAX(AbsenceEndDate) AS AbsenceEndDate
FROM (SELECT EID, AbsenceType, AbsenceStartDate, AbsenceEndDate,
      DATEADD(dd, - ROW_NUMBER() OVER (PARTITION BY EID, AbsenceType ORDER BY EID,AbsenceStartDate), AbsenceStartDate)
      FROM AbsenceDetails
      GROUP BY EID,AbsenceType,AbsenceStartDate,AbsenceEndDate
      ) a(EID, AbsenceType, AbsenceStartDate, AbsenceEndDate, Grp)
GROUP BY EID, AbsenceType, Grp;

Answer 3

我会这样做：

确定缺席日期的顺序列表。

SELECT
    ad1.EID, ad1.StartDate, ad2.EndDate
FROM 
    AbsenceDetails ad1
    JOIN AbsenceDetails ad2
    ON ad1.EID = ad2.EID
WHERE
    DATEDIFF(ss, ad1.EndDate, ad2.StartDate) = 1

结果如下：

769 2012-06-25 00:00:00.000 2012-06-26 23:59:59.000
769 2012-09-03 00:00:00.000 2012-09-04 23:59:59.000
769 2012-09-04 00:00:00.000 2012-09-05 23:59:59.000
769 2012-09-05 00:00:00.000 2012-09-06 23:59:59.000
769 2012-09-06 00:00:00.000 2012-09-07 23:59:59.000

遍历列表并确定每个延伸的开始和结束持续时间。最好在应用层完成。

Answer 4

如果我正确理解了您的问题，您希望在记录中找到连续的时间间隔主要问题是确定实际构成连续时间间隔的内容：
如果你正在寻找工作缺席而不是

的任何序列

date1.09:00 to date1.18:00  
date2.09:00 to date2.18:00

其中date2是下一个工作日，date1可以被认为是连续的。

在您的情况下，它相对容易，但您无法在单个查询中执行此操作。至少我现在想不出办法。

P.S。 “podiluska”建议的“Islands and Gaps”算法将帮助您在单个查询/存储过程中编写它。

SQL Server查询将顺序日期分组

4 个答案: