Question

如何获取以前的记录取决于unique_id和日期？我想制作我的桌子（dbo：Daily_Summary）：

ID    Partner     part_no    Date      Periode_Sum     Previous_Sum
1      aa            12    2011-12-21       40             
2      aa            12    2011-12-22       30              40
3      bb2           13    2011-12-22       20             
4      bb2           13    2011-12-23       30              20
5                          2011-12-24
6                          2011-12-25
7      aa            12    2011-12-26       30              70
8      bb2           13    2011-12-27       40              50
and so on

其中'Previous_Sum'是函数更新，该函数仅更新Periode_Sum的上一条记录。按伙伴分组，part_no和日期

我的显示查询有错误。当我按伙伴，part_no和日期显示上一条记录时。 previous_sum的记录仅显示第一个伙伴。这是我的疑问：

 SELECT ds.partner_id, ds.part_no, ds. periode_in
        , ds.periode_out, ds.periode_date, ds.periode_sum, 
(   
  SELECT F1.periode_sum
  FROM Daily_Summary as F1 where 
  F1.periode_sum =
    ( 
      SELECT  Max(F2.periode_sum) 
      FROM Daily_Summary as F2 where F2.periode_date < ds.periode_date
      and F2.partner_id=ds.partner_id and F2.part_no = ds.part_no
    ) 
) AS Prev_Value
FROM Daily_Summary ds
where stsrc='A'
order by ds.partner_id, ds.periode_date

我尝试使用declare new param但不工作：

DECLARE @prev_sum INT = 0
DECLARE @dudut INT = 2

SELECT @prev_sum = 
(select sum(periode_sum) 
    from Daily_Summary t1 
    where t1.partner_id = ds.partner_id and t1.part_no = ds.part_no
        and t1.periode_date < ds.periode_date --and t1.id < t.id
) 
FROM Daily_Summary ds
where stsrc='A'
order by ds.partner_id, ds.periode_date 

select partner_id,part_no,model,periode_sum,
case when @prev_sum is null then @dudut else @prev_sum end 
FROM daily_summary
where stsrc='A'

Answer 1

select * into #tab from (
    select 1 as id, 'aa' as partner, 12 as part_no, 
        cast('2011-12-21' as datetime) as date, 40 as periode_sum union all
    select 2, 'aa', 12, '2011-12-22', 30 union all
    select 3, 'bb2', 13, '2011-12-22', 20 union all
    select 4, 'bb2', 13, '2011-12-23', 30 union all
    select 5, null, null, '2011-12-24', null union all
    select 6, null, null, '2011-12-25', null union all
    select 7, 'aa', 12, '2011-12-26', 30 union all
    select 8, 'bb2', 13, '2011-12-27', 40
) t


select *, (select sum(periode_sum) 
    from #tab t1 
    where t1.partner = t.partner and t1.part_no = t.part_no
        and t1.date < t.date and t1.id < t.id
) as previous_sum
from #tab t

如果允许特定对（partner，part_no）每天超过一行，则应使用and t1.date < t.date and t1.id < t.id代替and t1.date <= t.date and t1.id < t.id而不是and t1.id < t.id。如果id确保所有行的正确顺序（及时），则只能使用id partner part_no date periode_sum previous_sum 1 aa 12 2011-12-21 00:00:00.000 40 NULL 2 aa 12 2011-12-22 00:00:00.000 30 40 3 bb2 13 2011-12-22 00:00:00.000 20 NULL 4 bb2 13 2011-12-23 00:00:00.000 30 20 5 NULL NULL 2011-12-24 00:00:00.000 NULL NULL 6 NULL NULL 2011-12-25 00:00:00.000 NULL NULL 7 aa 12 2011-12-26 00:00:00.000 30 70 8 bb2 13 2011-12-27 00:00:00.000 40 50。

结果：

{{1}}

如何使用以前的数据更新列

1 个答案: