如果我对您的理解正确,您想按休息号对项目进行分组,然后显示最短日期 + 1 天,以及“下一个”休息号的最短日期。当两个不同地方的余数为 0 时,您期望会发生什么?
with Base as
(
select t.AccNum,
t.Rest,
DATEADD(day, 1, MIN(t.Date)) as [StartDate],
ROW_NUMBER() OVER (ORDER BY MIN(t.Date)) as RowNumber
from Accounts t
where t.Rest <> 0
group by t.AccNum, t.Rest
)
select a.AccNum, a.Rest, a.StartDate, DATEADD(DAY, -1, b.StartDate) as [EndDate]
from Base a
left join Base b
on a.RowNumber = b.RowNumber - 1
order by a.[StartDate]
如果 Rest 数字有可能进一步向下重复,但它需要是一个单独的项目,那么我们需要在我们的初始选择查询中更聪明一点。
with Base as
(
select b.AccNum, b.Rest, MIN(DATEADD(day, 1, b.Date)) as [StartDate], ROW_NUMBER() OVER (ORDER BY MIN(Date)) as [RowNumber]
from (
select *, ROW_NUMBER() OVER (PARTITION BY Rest ORDER BY Date) - ROW_NUMBER() OVER (ORDER BY Date) as [Order]
from Accounts a
-- where a.Rest <> 0
-- If we're still filtering out Rest 0 uncomment the above line
) b
group by [order], AccNum, Rest
)
select a.RowNumber, a.AccNum, a.Rest, a.StartDate, DATEADD(DAY, -1, b.StartDate) as [EndDate]
from Base a
left join Base b
on a.RowNumber = b.RowNumber - 1
order by a.[StartDate]
两个查询的结果:
Account Number REST Start Date End Date
45817840200000057948 2500 2013-01-01 2013-01-14
45817840200000057948 1181 2013-01-15 2013-01-31
45817840200000057948 2431 2013-02-01 2013-02-09
45817840200000057948 1563 2013-02-10 NULL