Question

我正在尝试为特定商店的特定员工获取MAX计数。我可以很容易地得到每个员工的数量以及他们在哪个商店工作，但我只想在每个商店中排名前1位。

SELECT Emp_ID || ': ' || First_Name || ' ' || Last_Name AS "Emp. Name",
        Store_Num || ': ' || Store_Name AS "Store",
        COUNT(Emp_ID) AS "Num Rentals"
FROM Employee JOIN rental USING (Emp_ID)
        JOIN store USING (Store_Num)
GROUP BY Emp_ID, First_Name, Last_Name, Store_Num, Store_Name;

Emp. Name            Store                Num Rentals
-------------------- -------------------- -----------
104: Victoria Lee    20: SQ Hill's DVD              4
102: Steve Baker     10: Oakland's DVD              3
103: Lily Evans      10: Oakland's DVD              3
105: Gloria Tremblay 20: SQ Hill's DVD              4
101: Jim Smith       10: Oakland's DVD              5
106: Emma Wesley     30: DVD Hub                    3

6 rows selected.

这可以获得每个员工处理的租金数量以及他们在哪个商店工作。我只想在每家商店找到最优秀的员工。结果表应该是......

Emp. Name            Store                Num Rentals
-------------------- -------------------- -----------
104: Victoria Lee    20: SQ Hill's DVD              4
105: Gloria Tremblay 20: SQ Hill's DVD              4
101: Jim Smith       10: Oakland's DVD              5
106: Emma Wesley     30: DVD Hub                    3

提前感谢您的所有帮助！

Answer 1

您可以使用分析函数RANK：

select "Emp. Name", "Store", "Num Rentals" FROM
    (SELECT Emp_ID || ': ' || First_Name || ' ' || Last_Name AS "Emp. Name",
            Store_Num || ': ' || Store_Name AS "Store",
            COUNT(Emp_ID) AS "Num Rentals",
            rank() over(partition by Store_Num, Store_Name order by COUNT(Emp_ID) desc) rw
    FROM Employee JOIN rental USING (Emp_ID)
            JOIN store USING (Store_Num)
    GROUP BY Emp_ID, First_Name, Last_Name, Store_Num, Store_Name)
WHERE rw = 1;

RANK为PARTITION BY中表示的每个组计算符合ORDER BY expr（几行可以具有相同排名）的每一行的等级。

分析函数在GROUP BY

之后计算

Answer 2

select t."Store", max("Emp. Name") keep (dense_rank first order by "Num Rentals" desc)
  from (SELECT Emp_ID || ': ' || First_Name || ' ' || Last_Name AS "Emp. Name",
            Store_Num || ': ' || Store_Name AS "Store",
            COUNT(Emp_ID) AS "Num Rentals"
    FROM Employee JOIN rental USING (Emp_ID)
            JOIN store USING (Store_Num)
    GROUP BY Emp_ID, First_Name, Last_Name, Store_Num, Store_Name) t;

在SQL查询中使用MAX

2 个答案: