用于查找次高的SQL

Question

说我有一份员工表，我在一栏中保留员工的姓名，另一栏列出员工的薪水。应该写什么SQL来找到绘制第二高工资的员工的姓名???

Answer 1

我会使用row_number()：

select e.*
from (select e.*, row_number() over (order by salary desc) as seqnum
      from employees e
     ) e
where seqnum = 1

其他人提到了limit / top / rownum方法：

select e.*
from (select e.*
      from employees e
      order by salary desc
      limit 2
     ) t
order by salary
limit 1

你可以用偏移做同样的事情：

select e.*
from employees e
order by salary desc
limit 2, 1

然而，如果你有最高工资的平局，所有都有问题。通过将row_number()替换为dense_rank()：

，可以轻松修复第一种方法

select e.*
from (select e.*, dense_rank() over (order by salary desc) as seqnum
      from employees e
     ) e
where seqnum = 1

而另一种方法是：

select e.*
from employees e
where e.salary < (select max(salary) from employees)
order by salary desc
limit 1  -- or top 1 or rownum = 1 etc.

Answer 2

您可以尝试从“前2名”中选择，按升序排序，然后选择“前1名”，这样可以为您提供正确的行。

Answer 3

您可以执行以下操作：

select * from (select * from Employees order by salary desc limit 2) as A order by salary asc limit 1;