我有下表的数据
id departmentid ischanged date
3 22 0 2014-01-04
3 101 0 2014-01-05
3 125 1 2014-01-06
3 169 1 2014-01-07
3 175 0 2014-01-08
3 176 0 2014-01-09
3 177 0 2014-01-10
5 22 0 2014-01-04
5 101 0 2014-01-05
5 125 0 2014-01-06
5 169 0 2014-01-07
5 175 0 2014-01-08
5 176 0 2014-01-09
5 177 0 2014-01-10
我目前的查询是
insert into #temp1(id, startdate, enddate)
SELECT t1.id, '2014-1-4' as startdate, min(isnull(enddate,'2014-01-10')) as endDate
FROM (
SELECT id, departmentid, ischanged
FROM dbo.[table] where date = '2014-1-4'
) AS t1
left join
(
SELECT id, departmentid, ischange , date
FROM dbo.[table] where date >= '2014-1-4'
) as t2
on t1.id = t2.id and (t1.ischange <> t2.ischange)
group by t1.id
并将产生以下输出,如果ischange
没有更改,则它将从查询中获取硬编码的结束日期,否则它将获取最小的ischange
值更改日期
id startdate enddate
3 2014-01-04 2014-01-05
5 2014-01-04 2014-01-10
但是我在找一个像这样的结果集
id startdate enddate
3 2014-01-04 2014-01-05
3 2014-01-08 2014-01-10
5 2014-01-04 2014-01-10
您需要ischanged
为0的句点。您可以使用row_number()
:的差异来完成此操作
select id, min(date) as startdate, max(date) as enddate
from (select t.*,
(row_number() over (partition by id order by date) -
row_number() over (partition by id, ischanged order by date)
) as grp
from table t
) t
where ischanged = 0
group by id, grp;