SQL

Question

我想要实现的是：
1) 获取每个部门的 table 薪水的最小值。
2) 如果对于每个个部门，这个最小值在table中至少存在两次次，然后显示它的部门id。

示例：

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，你想要工资最低的部门。

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

如果我没理解错的话，你希望最低工资至少出现两次的部门。这让我觉得 window 函数：

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，因为这样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

SQL - 获取行的最小值并检查此最小值是否在行中至少 2 次

SQL - get MIN value of row and check this MIN value to be in row at least 2 times

oracle

group-by

aggregate-functions

greatest-n-per-group