查找每个部门中最年轻和最年长的员工的薪水

Question

有两张桌子

1) Employee
id | Name |  Department | Dob 

2) Salary
id | salary

我想找到每个部门中最年轻和最年长的员工的薪水。 但使用以下查询我无法获得正确的 id、salary。

SELECT salary.id,employee.Dept,salary.salary,MIN(employee.DoB) 
from employee 
INNER JOIN salary ON salary.id = employee.id  GROUP by Dept

上面的查询返回正确的 Dob，但 ids 和薪水与出生日期不匹配。

Answer 1

如果您正在运行 MySQL 8.0，只需使用 window 函数：

select *
from (
    select 
        e.*, 
        s.salary,
        row_number() over(partition by department order by dob asc) rn_asc,
        row_number() over(partition by department order by dob desc) rn_desc
    from employee e
    inner join salary s on s.id = employee.id
) t
where 1 in (rn_asc, rn_desc)

在早期版本中，一种选择是加入聚合查询：

select e.*, s.salary
from employee e
inner join salary s on s.id = employee.id
inner join (
    select department, min(dob) min_dob, max(dob) max_dob 
    from employee
    group by department
) d on d.department = e.department and e.dob in (d.min_dob, d.max_dob)

Answer 2

我想我会在相关子查询中使用=两次：

select s.*, e.department
from salary join
     employee e
     on s.id = e.id
where e.dob = (select min(e2.dob) from employee e where e2.department = e.department) or
      e.dob = (select max(e2.dob) from employee e where e2.department = e.department) ;

有了employee(department, dob)的索引，我希望这会有很好的表现。