Question

我像这样有两个表table1和table2。您会看到时代空白

table1
date      item time amount
----------------------------
1/1/2000  a    1    100
1/1/2000  a    2    100
1/1/2000  a    3    200
1/1/2000  a    6    300
1/1/2000  b    1    100
1/1/2000  b    2    100
1/1/2000  b    5    200
2/1/2000  a    1    500
2/1/2000  a    3    500
2/1/2000  a    4    550

我也有table2可以填补空白

table2
date      item time amount new
-------------------------------------------
1/1/2000  a    1    100    N
1/1/2000  a    2    100    N
1/1/2000  a    3    200    N
1/1/2000  a    4           Y  <-- added amount should be 200
1/1/2000  a    5           Y  <-- added amount should be 200
1/1/2000  a    6    300    N
1/1/2000  b    1    100    N
1/1/2000  b    2    100    N
1/1/2000  b    3           Y  <-- added amount should be 100 
1/1/2000  b    4           Y  <-- added amount should be 100
1/1/2000  b    5    200    N
2/1/2000  a    1    500    N
2/1/2000  a    2    500    N
2/1/2000  a    3           Y  <-- added amount should be 500
2/1/2000  a    4    550    N

金额差距行应采用上次/上次时间的值。我能够识别丢失的行并添加缺口行，但是我尝试将金额复制到缺口行，但未成功。我查看了我认为是stackoverflow中类似问题的内容，并尝试了解决方案，但它没有起作用，例如：

update t2
set t2.amount = t1.amount
from table2 t2
inner join table1 t1 on t2.date = t1.date and t1.item = t2.item
where t2.new = 'Y'
and t2.time > (select t2.time 
              from table1 t3
              where max(t3.time) < t2.time)


update t2
set t2.amount = t1.amount
from table1 t1
inner join table2 t2 on t1.date = t2.date and t1.item = t2.item
where t2.new = 'Y' and max(t1.time) < t2.time

有人知道如何访问上一行的金额吗？游标有效，但这是不得已的解决方案。感谢您抽出宝贵的时间来帮助您。

添加我的创建表代码

create table #table1  (or #table2)
(
   date smalldatetime,
   item char(1),
   [time] int,
   amount int
   ,new char(1) -- for new row flag 
)

Answer 1

您需要找到先前的amount的非空值：

update t
set amount = (
  select amount from table2 
  where
  date = t.date and item = t.item and time = (
    select max(time) from table2
    where 
    date = t.date and item = t.item 
    and time < t.time and amount is not null and new = 'N'
  ) 
)
from table2 t
where t.amount is null and t.new = 'Y'

请参见demo。

Answer 2

您应该使用窗口查询：

select 
*,sum(amount) over (partition by time order by time) as previous_amount_for_null_values
from table2

Answer 3

update tab2
set tab2.amount=abc.PREV_AMOUNT
from
table2 tab2
join
(SELECT *,T2.AMOUNT PREV_AMOUNT
FROM TABLE1 T1
JOIN
(
SELECT
MAX(TIME) TIME,DATE,ITEM,MAX(AMOUNT) AMOUNT
FROM TABLE1
WHERE TIME!=(SELECT MAX(TIME) FROM TABLE1
GROUP BY DATE,ITEM)
GROUP BY DATE,ITEM) T2
ON T1.DATE=T2.DATE
AND T1.ITEM=T2.ITEM
AND T1.TIME=T2.TIME) abc
on tab2.date=abc.date
and tab2.item=abc.item
where tab2.new='Y' and tab2.amount is null

打破它：找到每个日期和项目组的第二个最大时间。将第二个最大时间与日期和项目上的原始table1结合起来，将得到的金额为上一个金额使用此上一个金额根据日期和项目更新表2，只要new ='Y'

PS：我尚未在sql Developer上运行此查询。但是我使用的想法应该可行。

Answer 4

您可以尝试...

    UPDATE T SET T.Amount = ( SELECT MAX(T2.Amount) FROM table2  T2 
    WHERE T2.Amount IS NOT NULL AND T2.[time] <= T.[time] and T2.item = T.item and t2.[date]=T.[date]
                      ) from table2  as T
      WHERE T.Amount IS NULL

标记它接受，如果可以的话。

SQL：使用上一行的值填充当前行

4 个答案: