[英]Calculate days based on date range
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')
我正在尝试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
我使用 datediff 尝试了下面的代码,但不确定如何关联上一行并获取日期范围
select *, DATEDIFF(DAY,FromDate,ToDate)
from #Temp
使用 first_value
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
如果你想要完全相同的 output,你可以试试这个
Select
* ,
case
when (FromDate != ToDate)
then
'Day '+ DATEPART(Day,FromDate) + '-' + DATEPART(Day,ToDate)
else
'Day '+ DATEPART(Day,FromDate)
END AS Display
From #Temp
您不想要前一行的值,您想要最早的起始日期,然后您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
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