列出在超过 2 名员工的团队中从平均收入中获得平均值（准确率高达 30%）的员工（姓名、base_salary）

Posted 2023-03-31

【中文标题】列出在超过 2 名员工的团队中从平均收入中获得平均值（准确率高达 30%）的员工（姓名、base_salary）

【英文标题】：List the employees (name, base_salary) that earn an average value (with accuracy up to 30%) from the average earnings in teams of >2 employees

【发布时间】：2020-06-18 16:44:41

【问题描述】：

这是我目前得到的：

SELECT surname, base_salary from emp p LEFT JOIN (select id_team, avg(base_salary) as s, count(*) as c from emp group by id_team) as o ON(p.id_team = o.id_team)
where p.base_salary between o.s*0.7 and o.s*1.3 and o.c >=2

在 Oracle LIVE SQL 上，我收到 ORA-00905：缺少关键字错误。

这是桌子的样子。

【参考方案1】：

您的问题是使用as 关键字为子查询设置别名。甲骨文不支持。您的查询应该可以删除它：

select ...
from emp p 
left join (
    select id_team, avg(base_salary) as s, count(*) as c from emp group by id_team
) as o on p.id_team = o.id_team)
--^-- here  
where ...

另一方面，我认为使用窗口函数可以更有效地表达您的查询：

select surname, base_salary
from (
    select 
        surname, 
        base_salary, 
        avg(base_salary) over(partition by id_team) avg_base_salary,
        count(*) over(partition by id_team) no_emps
    from emp 
) e
where no_emps > 2 and base_salary between avg_base_salary * 0.7 and avg_base_salary * 1.3 

【讨论】：

非常感谢！