Sql 根据日期范围计算天数
我有如下数据Sql 根据日期范围计算天数,sql,sql-server,sql-server-2016,Sql,Sql Server,Sql Server 2016,我有如下数据 create table #Temp(Id int, FromDate date, ToDate date) Insert into #Temp values(1,'9/1/2019','9/1/2019'), (2,'9/2/2019','9/3/2019'), (3,'9/2/2019','9/3/2019'), (4,'9/4/2019','9/6/2019'), (5,'9/7/2019','9/7/2019'
create table #Temp(Id int, FromDate date, ToDate date)
Insert into #Temp
values(1,'9/1/2019','9/1/2019'),
(2,'9/2/2019','9/3/2019'),
(3,'9/2/2019','9/3/2019'),
(4,'9/4/2019','9/6/2019'),
(5,'9/7/2019','9/7/2019')
我试图计算差异并创建天数,即第1天、第2-3天等
预期结果
Id FromDate ToDate Display
1 01/09/2019 01/09/2019 Day 1
2 02/09/2019 03/09/2019 Day 2-3
3 02/09/2019 03/09/2019 Day 2-3
4 04/09/2019 06/09/2019 Day 4-6
5 07/09/2019 07/09/2019 Day 7
我尝试了下面使用datediff的代码,但不确定如何与前一行关联并获取日期范围
select *, DATEDIFF(DAY,FromDate,ToDate)
from #Temp
使用第一个值
select *
, datediff(day, first_value(FromDate) over(order by FromDate), FromDate) + 1
, datediff(day, first_value(FromDate) over(order by FromDate), ToDate) + 1
from #Temp
您不需要前一行的值,而是需要最早的fromdate,然后可以将其与每一行进行比较
select id, min (fromdate) over (order by fromdate asc) as earliest_date,fromDate,todate,
datediff(day,min (fromdate) over (order by fromdate asc),fromdate)+1,
datediff(day,min (fromdate) over (order by fromdate asc),todate)+1
from
#temp
如果您想要完全相同的输出,可以尝试此方法
Select
* ,
case
when (FromDate != ToDate)
then
'Day '+ DATEPART(Day,FromDate) + '-' + DATEPART(Day,ToDate)
else
'Day '+ DATEPART(Day,FromDate)
END AS Display
From #Temp
嘿,塞格。以获得他们想要的输出<代码>选择*,'Day'+转换(datediff(Day,first_value(FromDate)over(order by FromDate),FromDate)+1作为varchar(10))+'-'+转换(datediff(Day,first_value(FromDate)over(order by FromDate),ToDate)+1作为varchar(10))from这是有效的。仍然无法投票支持您的答案,因为我只有不到15次报告
min()
似乎简单多了。