在Oracle中，你如何执行取决于先前计算的结果计算

Question

我想执行一个简单的计算如下

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 

结果是依赖于前一个SR> = 2，而始终以0.99开始。

执行计算时循环。 数据库Oracle。

Answer 1

如果你想要计算运行总和，它会很容易使用分析函数来执行：

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“产品”聚合功能，我们可以在这里使用，所以我们要绕道使用EXP和LN（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; 

这应该给你所期望的结果[未得到充分系列多