在一个结果行中选择具有给定时间间隔的 id history 的行

Question

A table StatusHistory 包含：

ID      Status  DtSince DtUntil    
2415891  0     20200816 20200917
2415891  4     20200917 NULL
2415892  0     20200904 20200905
2415892  4     20200905 NULL

我需要 select 每个 id 的每个月初和月底的状态。用户将输入所需的日期。例如：

用户输入 --> @Month = '202009'

select 语句必须 return dates 20200901 and 20201001

的状态

对于新的且 dtSince>20200901 的记录，它应该 return 2415892 的活动 row.For 示例它应该 return 4。

所以@Month = 202009 的结果集应该是：

Id       BeginningOfMonth_status EndOfMonth_status 
2415891       0                         4
2415892       4                         4

这是我的解决方案

select * from
    (
    select * from 
        (
         select *, ROW_NUMBER() over (partition by id  order by dtSince desc) r from  
         dbo.StatusHistory
         where DtUntil  is null or (DtUntil <20201001and dtSince>=20200901) --2432290
        )x
     where x.r=1
)s1
inner join 
   (
   select * from 
       (
        select *, ROW_NUMBER() over (partition by id  order by dtSince desc) r from  
        dbo.StatusHistory
        where DtUntil  is null or (DtUntil <20201001and dtSince>=20200901) --2432290
       )x
   where x.r=2
   )s2
on s1.id = s2.id

问题：

1)有比我的代码更简单的吗？

2) 以上，我没有涵盖 For records that are new and have dtSince>20200901 要求

Answer 1

嗯。 . .您可以使用 where 子句获取涵盖特定月份的所有行。然后使用row_number()枚举行和条件聚合：

select id,
       max(case when seqnum_asc = 1 then status end) as beginning_of_month,
       max(case when seqnum_desc = 1 then status end) as end_of_month
from (select t.*,
             row_nunber() over (partition by id order by dtsince) as seqnum_asc,
             row_nunber() over (partition by id order by dtsince desc) as seqnum_desc
      from t
      where dtsince < dateadd(month, 1, convert(date, @Month + '01'))) and
            (dtuntil >= convert(date, @Month + '01') or dtuntil is null)
     ) t
group by id

在一个结果行中选择具有给定时间间隔的 id history 的行

selecting rows with id history for a given time interval in one result row

sql

sql-server

sql-server-2008-r2