SQL：使用前一行中的值填充当前行

我有两张桌子，一张桌子和二张桌子一样。你可以看到时间有差距

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

我还有表2，我在其中填补了空白

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

---

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 

---

也许这个Oracle解决方案会有所帮助，我希望sql server有类似rownum或类似的解决方案。首先，对数据进行排序，并使用唯一的numbers rownum伪列。第二，对于每个数字，计算非空值的最大前一个数字。连接数据

也许有一个解析函数的解，但不，我脑子里没有

-- create test data
drop table tab1;
create table tab1 (
    ym varchar2(7),
    val number
);
insert into tab1 values('2016/01',3);
insert into tab1 values('2016/04',6);
insert into tab1 values('2016/08',4);
insert into tab1 values('2016/09',2);
insert into tab1 values('2016/01',5);
insert into tab1 values('2016/09',8);
insert into tab1 values('2016/05',7);
insert into tab1 values('2016/12',3);
insert into tab1(ym) values('2016/03');
insert into tab1(ym) values('2016/11');
insert into tab1(ym) values('2016/12');
insert into tab1(ym) values('2016/12');

-- solution
with q0 as (-- get rownum for each row
    select a.*, rownum as rnm -- get rownums
    from (
        select *
        from tab1
        order by ym -- order by, to further get proper order numbers from rownum
    ) a
),

q1 as (-- for each rnm get previous (maximal) rnm with non missing value
    select a.rnm, max(b.rnm) as rnm_prev
    from q0 a
    left join q0 b on a.rnm > b.rnm and b.val is not null -- get only smaller rnms with values
    group by a.rnm
)


select q0.ym, q0.rnm, q1.rnm_prev,
q0.val, pv.val as val_prev, nvl(q0.val, pv.val) as val_cor
from q0
left join q1 on q0.rnm = q1.rnm
left join q0 pv on q1.rnm_prev = pv.rnm
order by q0.rnm

细分： 查找每组日期和项目的第二个最长时间。 将第二个最大时间与日期和项目的原始表1连接，以获得上一个金额 使用此上一个金额根据日期和项目更新表2，只要是新的='Y'

---

PS：我还没有在SQLDeveloper上运行这个查询。但是我使用的想法应该有效。

您需要找到前面的非空值amount:

请参阅。

您可以试试这个

    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

---

如果工作正常，请将其标记为已接受。

这不起作用，我没有使用运行总数，只是直接复制。有人有其他建议吗？谢谢。此表2是根据您的查询创建的。是的，我有原始的表1，然后我创建了一个新的表2，填补了空白。我希望看到用于错误检查的表的不同版本。请共享您的查询以创建表2…既然您已经用tab1和gaps中的值填充了tab2，为什么您要加入tab1以尝试获取值以填补空白？所需的值已经存在于tab2中，或者我缺少simething？这就是我所做的。我在问题中发布了我的代码，但sql server不喜欢我的查询。非常感谢，这是有效的，但是我必须在第二个select语句中添加和time-- create test data drop table tab1; create table tab1 ( ym varchar2(7), val number ); insert into tab1 values('2016/01',3); insert into tab1 values('2016/04',6); insert into tab1 values('2016/08',4); insert into tab1 values('2016/09',2); insert into tab1 values('2016/01',5); insert into tab1 values('2016/09',8); insert into tab1 values('2016/05',7); insert into tab1 values('2016/12',3); insert into tab1(ym) values('2016/03'); insert into tab1(ym) values('2016/11'); insert into tab1(ym) values('2016/12'); insert into tab1(ym) values('2016/12'); -- solution with q0 as (-- get rownum for each row select a.*, rownum as rnm -- get rownums from ( select * from tab1 order by ym -- order by, to further get proper order numbers from rownum ) a ), q1 as (-- for each rnm get previous (maximal) rnm with non missing value select a.rnm, max(b.rnm) as rnm_prev from q0 a left join q0 b on a.rnm > b.rnm and b.val is not null -- get only smaller rnms with values group by a.rnm ) select q0.ym, q0.rnm, q1.rnm_prev, q0.val, pv.val as val_prev, nvl(q0.val, pv.val) as val_cor from q0 left join q1 on q0.rnm = q1.rnm left join q0 pv on q1.rnm_prev = pv.rnm order by q0.rnm

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

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'

    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