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SQL:使用前一行中的值填充当前行

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ithoughtso · 技术社区 · 7 年前

我有两张桌子,一张和二张一样。你可以看到时代有差距

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

我也有表2,我填补了空白

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

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

有人知道如何获取前一行的金额吗?光标可以工作,但这是最后的解决方案。感谢您在忙碌的一天中抽出时间来帮忙。

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 
)

1 回复 | 直到 7 年前

forpas 7 年前

amount :

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'

看到了吗 demo

saman tr 7 年前

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

Tomasz 7 年前

也许这个Oracle解决方案会有所帮助(我希望sqlserver有类似rownum或类似的解决方案)。首先,对数据进行排序并获得唯一的数字(使用rownum pseudocolumn)。第二,计算每个数的最大前一个数的非空值。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

BlackSwan 7 年前

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

找到每组日期和项目的第二个最长时间。使用此上一个金额根据日期和项目更新表2,其中new='Y'

PS:我还没有在SQLDeveloper上运行这个查询。但我的想法应该行得通。

DarkRob 7 年前

你可以试试这个。。。

    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,则标记为已接受。