Question

假设我有A列，并且当前获取的A值为null。我需要返回到上一行，并找到列A的非null值。然后我需要从看到非值的点到当前点，找到另一列B的总和。之后，我需要将B与A相加，这将是A的新值。

要查找列A非空值，我已将查询写为

nvl(last_value(nullif(A,0)) ignore nulls over (order by A),0)

但是我需要如上所述进行B的计算。

nvl(last_value(nullif(A,0)) ignore nulls over (order by A),0)

有人可以帮我吗？

样本数据

A        B      date

null     20     14/06/2019

null     40     13/06/2019

10       50     12/06/2019

此处应将A的14/06/2019的值替换为B + A的12/06/2019的值（这是A的第一个非空值）= 20 + 40 + 50 + 10 = 120

Answer 1

我认为您要使用

处理

sum(<column>) over (...)和last_value over (...)的功能如下：

with t( A,B, "date" ) as
(
 select null, 20, date'2019-06-14' from dual union all
 select null, 40, date'2019-06-13' from dual union all
 select 10   ,50, date'2019-06-12' from dual
)
select nvl(a,sum(b) over (order by 1)+
             last_value(a) ignore nulls
                    over (order by 1 desc)
          ) as a,
       b, "date" 
  from t; 

A   B   date
--- --  ----------
120 20  14.06.2019
120 40  13.06.2019
 10 50  12.06.2019

Demo

Answer 2

如果您具有12c或更高版本：

with t( A,B, dte ) as
(
 select null, 20, date'2019-06-14' from dual union all
 select null, 40, date'2019-06-13' from dual union all
 select 10   ,50, date'2019-06-12' from dual
)
select * from t
match_recognize(
  order by dte desc
  measures
    nvl(
      first(a),
      y.a + sum(b)
    ) as a,
    first(b) as b,
    first(dte) as dte
  after match skip to next row
  pattern(x* y{0,1})
  define x as a is null,
    y as a is not null
);

     A          B DTE       
------ ---------- ----------
   120         20 2019-14-06
   100         40 2019-13-06
    10         50 2019-12-06

Answer 3

使用条件计数将数据分为不同的组，然后使用该组进行分析计算：

select a, b, dt, grp, sum(nvl(a, 0) + nvl(b, 0)) over (partition by grp order by dt) val
  from (
    select a, b, dt, count(case when a is not null then 1 end) over (order by dt) grp
      from t order by dt desc)
  order by dt desc

抽样结果：

     A          B DT                 GRP        VAL
------ ---------- ----------- ---------- ----------
               20 2019-06-14           4        120
               40 2019-06-13           4        100
    10         50 2019-06-12           4         60
     5          2 2019-06-11           3          7
     6          1 2019-06-10           2          7
                3 2019-06-09           1         14
     7          4 2019-06-08           1         11

^demo

如何找到一列的最后一个非null值以及如何递归地找到另一列的总和值

3 个答案: