评论区里的答案详解

评论区里的答案详解:

-- 方法一
select s.emp_no, s.salary, e.last_name, e.first_name
from salaries s join employees e
on s.emp_no = e.emp_no
where s.salary =              -- 第三步: 将第二高工资作为查询条件
    (
    select max(salary)        -- 第二步: 查出除了原表最高工资以外的最高工资(第二高工资)
    from salaries
    where salary <    
        (
        select max(salary)    -- 第一步: 查出原表最高工资
        from salaries
        where to_date = '9999-01-01'   
        )
    and to_date = '9999-01-01'
    )
and s.to_date = '9999-01-01'



-- 方法二
select s.emp_no, s.salary, e.last_name, e.first_name
from salaries s join employees e
on s.emp_no = e.emp_no
where s.salary =
    (
    select s1.salary
    from salaries s1 join salaries s2      -- 自连接查询
    on s1.salary <= s2.salary
    group by s1.salary                     -- 当s1<=s2链接并以s1.salary分组时一个s1会对应多个s2
    having count(distinct s2.salary) = 2   -- (去重之后的数量就是对应的名次)
    and s1.to_date = '9999-01-01'
    and s2.to_date = '9999-01-01'
    )
and s.to_date = '9999-01-01'

表自连接以后:

当s1<=s2链接并以s1.salary分组时一个s1会对应多个s2

对s2进行去重统计数量, 就是s1对应的排名