Question

我正在使用Oracle。我有这样的表：

Company   Employee   salary
   A1       Jim     122000
...

我想要返回薪水超过2个标准差（〜> 95％）的员工人数最多的公司。这是我的代码

With Com_2Std as (
Select company-name, AVG(salary)+2*STDDEV(salary) as AboveS
From works
Group By company-name)

Select company-name, count(employee-name) as ENumber
From works
Where ENumber=MAX(
Select  count(a.employee-name)
From works a, Com_2Std b
Where  a.company-name=b.company-name
And a.salary>b.AboveS;
Group by a.company-name)
Group by company-name;

我有两个问题：

（1）我今天无法访问oracle而无法测试它。请问我的代码是否正确？

（2）它看起来很复杂，有什么方法可以简化它吗？

Answer 1

with Com_2Std as (
    select company-name, AVG(salary)+2*STDDEV(salary) as AboveS
    from works
    group by company-name
),
CompanyCount as (
    select a.company-name, count(*) as CountAboveS
    from
        works a
        inner join Com_2Std b on a.company-name=b.company-name
    where
        a.salary > b.AboveS
    group by a.company-name
)
select company-name
from CompanyCount
where CountAboveS = (select max(CountAboveS) from CompanyCount)

这应该是接近的。它也会产生联系。

STDDEV函数和with子句

1 个答案: