SQL - 獲取行的最小值並檢查此最小值是否在行中至少 2 次

Question

我想要實現的是：
1) 獲取表中每個部門的工資最小值。
2) 如果這個最小值在每個部門的表中至少存在兩次，則顯示其部門 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

結果：

department_id
1
2

Answer 1

如果我沒聽錯的話，你需要一個以上員工工資最低的部門。

這是一種使用 window 函數的方法，它通過比較row_number()和rank()來工作：

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

如果我理解正確，您希望最低工資至少出現兩次的部門。 這讓我覺得 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;

請注意，您不需要select distinct ，因為每個部門最多選擇一行。

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