Question

我有一个列出date，customerID，orderID和orderCost的视图。如何选择每月最高消费客户？该数据仅在2015年持续六个月。

我能够每个月获得每个客户花费的SUM，而不是最大消费客户：

SELECT EXTRACT(YEAR FROM date) AS year, MONTHNAME(date) AS month, customerID, SUM(orderCost) 
FROM CustomerPricedOrder 
GROUP BY MONTH(date), customerID;

*其中CustomerPricedOrder是一个从名为CustomerOrder的表中获取数据的视图

如何通过存储过程找到每月最大消费客户？我需要光标吗？

调用程序时我需要的输出示例：

年........月......... customerID
2015 ........一月....... 4
2015 ........二月...... 21
2015 ........三月......... 6
2015 ........四月......... 11

其中customerID是该月的最高消费客户。

Answer 1

您可以按照降序排序并限制1吗？

SELECT EXTRACT(YEAR FROM date) AS year, MONTHNAME(date) AS month, customerID, SUM(orderCost) as sum_of_order_cost FROM CustomerPricedOrder GROUP BY MONTH(date), customerID order by sum_of_order_cost desc limit 1;

Answer 2

SELECT EXTRACT(YEAR FROM date) AS year, 
MONTHNAME(date) AS month, 
customerID, 
SUM(orderCost) as sum_order_cost, 
MAX(sum_of_order_cost) FROM CustomerPricedOrder 
GROUP BY MONTH(date), customerID 
order by sum_order_cost DESC
LIMIT 1

这可能有效

Answer 3

亚历，

你能试试吗

SELECT EXTRACT(YEAR FROM date) AS year, MONTHNAME(date) AS month, customerID, SUM(orderCost) AS cost 
FROM CustomerPricedOrder 
GROUP BY MONTH(date), customerID ORDER BY cost DESC;

如果我的大脑正确理解这应该可以胜任。

Answer 4

这是一种使用变量来模拟row_number()的方法。该查询对一组中的每个客户（每个组代表一个月）进行排名，从最高消费客户的1开始，并保留排名为1的那些客户：

SELECT * FROM (
    SELECT @rn := if(@prevMonth = month AND @prevYear = year, @rn + 1, 1) rn, 
           @prevMonth := month, @prevYear := year, *
    FROM ( 
        SELECT YEAR(date) AS year, 
            MONTHNAME(date) AS month, 
            customerID, 
            SUM(orderCost) AS sum_order_cost
        FROM CustomerPricedOrder 
        GROUP BY MONTHNAME(date), YEAR(date), customerID
    ) t ORDER BY year, month, sum_order_cost desc
) t WHERE rn = 1

Answer 5

试试这个......

SELECT EXTRACT(YEAR FROM date) AS year, 
MONTHNAME(date) AS month, 
customerID, 
SUM(orderCost) as sum_of_order_cost, 
MAX(sum_of_order_cost) FROM CustomerPricedOrder 
GROUP BY MONTH(date), customerID 
order by orderCost;

Answer 6

似乎是分析函数的直接应用

    select year, month, customer, amount,
    rank() OVER (PARTITION BY year, month ORDER BY amount DESC) rn
    (
          SELECT EXTRACT(YEAR FROM date) AS year, MONTHNAME(date) AS month,customerID, SUM(orderCost) amount
          FROM CustomerPricedOrder 
          GROUP BY MONTH(date), customerID
    ) 

;

这会将每年/每月的金额与指定的最高金额rn -> 1进行排名。使用分析函数时需要注意的一般性问题是您希望如何处理＆＃34;相同的结果＆＃34; （即两个客户在一段时间内的金额相同）。检查是否需要处理这种情况以及Row_Number是否更适合您的需求（即使对于相同的结果也强制使用唯一编号，而rank()将为相同的结果分配相同的订单号），或者您需要添加到分析函数中的排序标准。

如何选择每月最大消费客户

6 个答案: