SQL - 获取行的最小值并检查此最小值是否在行中至少 2 次

Question

我想要实现的是：
1) 获取表中每个部门的工资最小值。
2) 如果这个最小值在每个部门的表中至少存在两次，则显示其部门 id。

例子：

column1 name  salary department_id
id1     John1 10000  1
id2     John2 10000  1
id3     John3 30000  2
id4     John4 30000  2
id5     John5 50000  3
id6     John6 20000  4

结果：

department_id
1
2

Answer 1

如果我没听错的话，你需要一个以上员工工资最低的部门。

这是一种使用 window 函数的方法，它通过比较row_number()和rank()来工作：

select distinct department_id
from (
    select 
        t.*, 
        row_number() over(partition by department_id order by salary) rn,
        rank()       over(partition by department_id order by salary) rnk
    from mytable t
) t
where rnk = 1 and rn > 1

Answer 2

如果我理解正确，您希望最低工资至少出现两次的部门。 这让我觉得 window 功能：

select t.department_id
from (select t.*,
             count(*) over (partition by department_id, salary) as cnt,
             row_number() over (partition by department_id order by salary) as seqnum
      from t
     ) t
where seqnum = 1 and cnt > 1;

请注意，您不需要select distinct ，因为每个部门最多选择一行。

Answer 3

    SELECT department_id
      FROM Employee
     WHERE Employee.salary = (select min(emp.salary) from Employee emp where emp.department_id = Employee.department_id) 
  GROUP BY department_id
    HAVING COUNT(1) >=2