SQL选择每个部门的最高销售日期

Question

我很难编写棘手的查询。

我有下表：

我要为每个部门打印利润最大的日期；

我尝试自己提出这样的查询：

Select DISTINCT(Name), Date_sale, MAX(A) as B FROM (SELECT 
 Departments.Name, SALES.Date_sale, SUM(GOODS.Price * SALES.Quantity) 
 AS A FROM DEPARTMENTS, GOODS, SALES
 WHERE DEPARTMENTS.Dept_id = GOODS.Dept_id AND GOODS.Good_id = 
 SALES.Good_id GROUP BY DEPARTMENTs.Name, SALES.Date_sale) 
 GROUP BY Name, Date_sale;

但是由于我按名称和日期分组，因此部门打印了几次的问题。

我应该如何解决？

Answer 1

您可以尝试以下方式-

with cte as 
(
 SELECT 
 Departments.Name, SALES.Date_sale, SUM(GOODS.Price * SALES.Quantity) 
 AS profit FROM DEPARTMENTS inner join GOODS on DEPARTMENTS.Dept_id = GOODS.Dept_id
 inner join SALES on GOODS.Good_id = SALES.Good_id
 GROUP BY DEPARTMENTs.Name, SALES.Date_sale
)A

select * from cte a
where profit =
     (select max(profit) from cte b on a.department=b.department)

或者您可以使用row_number()

select * from
(
select *, row_number() over(partition by department oder by profit desc) as rn
from cte
)A where rn=1

Answer 2

您可以使用ROW_NUMBER进行书写，这将为部门按日期分组的每个日期的总计数提供一个数字，如下所示，然后您可以使用rn = 1来获得最高的销售日期

SELECT NAME, DATE_SALE, A
FROM
    (
        SELECT
            DEPARTMENTS.NAME, SALES.DATE_SALE,
            ROW_NUMBER() OVER(
                PARTITION BY DEPARTMENTS.NAME
                ORDER BY SUM(GOODS.PRICE * SALES.QUANTITY) DESC NULLS LAST
            ) AS RN,
            SUM(GOODS.PRICE * SALES.QUANTITY) AS A
        FROM DEPARTMENTS
            JOIN GOODS ON ( DEPARTMENTS.DEPT_ID = GOODS.DEPT_ID )
            JOIN SALES ON ( GOODS.GOOD_ID = SALES.GOOD_ID )
        GROUP BY DEPARTMENTS.NAME,
            SALES.DATE_SALE
    )
WHERE RN = 1;

重要，请使用标准的ANSI联接。

干杯！

Answer 3

我将在此处使用join-s，因为需要从通过第三个表链接的2个表中提取信息。

类似这样的东西（但是我没有测试这个查询，只是建议一种方法）：

Select department.name as dept, MAX(sales.quantity) as max_sales, sales.date_sale 
  from goods
Left outer join departments on departments.dept_id = goods.dept_id
Left outer join sales on sales.good_id = goods.good_id
Group by dept