Question

我想按如下方式执行一个简单的计算

数据集

SR      value_one      result 
1         null          0.99 
2          1            0.99*1 = 0.99 
3         0.75          0.99*0.75 = 0.7425 
4         0.75          0.7425*0.75 = 0.556875 
5          1            0.556875*1 = 0.556875 
6          1            0.556875*1 = 0.556875 
7          1            0.556875*1 = 0.556875 
8          1            0.556875*1 = 0.556875 
9          1            0.556875*1 = 0.556875 
10         1            0.556875*1 = 0.556875

结果取决于前一个SR＆gt; = 2，而始终以0.99开始。

执行计算时循环。数据库Oracle。

Answer 1

如果您想计算一个运行总和，那么使用分析函数很容易：

with d as (
  select 1 as sr, cast(null as number) as value_one from dual union all 
  select 2 as sr, 1 as value_one from dual union all 
  select 3 as sr, 0.75 as value_one from dual union all 
  select 4 as sr, 0.75 as value_one from dual union all 
  select 5 as sr, 1 as value_one from dual union all 
  select 6 as sr, 1 as value_one from dual union all 
  select 7 as sr, 1 as value_one from dual union all 
  select 8 as sr, 1 as value_one from dual union all 
  select 9 as sr, 1 as value_one from dual union all
  select 10 as sr, 1 as value_one from dual 
)
select d.*, sum(nvl(value_one, 0.99)) over (order by sr)
from d;

不幸的是，没有＆＃34;产品＆＃34;我们可以在这里使用的聚合函数，所以我们必须使用EXP和LN绕道而行（见https://stackoverflow.com/a/3912248/1230592）：

with d as (
  select 1 as sr, cast(null as number) as value_one from dual union all 
  select 2 as sr, 1 as value_one from dual union all 
  ..
)
select d.*, nvl(exp (sum (ln (value_one)) over (order by sr)), 1) * 0.99
from d;

这应该会给你想要的结果

在Oracle中，如何根据先前的计算结果执行计算

1 个答案: