Question

我有一个非常基本的订单表：



+----------+---------+------------+---------+
| order_id | cust_id | order_date | book_id |
+----------+---------+------------+---------+
| 1        | 1       | 10/10/2014 | 1       |
+----------+---------+------------+---------+
| 2        | 1       | 10/10/2014 | 2       |
+----------+---------+------------+---------+
| 3        | 1       | 10/12/2014 | 1       |
+----------+---------+------------+---------+
| 4        | 2       | 10/18/2014 | 6       |
+----------+---------+------------+---------+
| 5        | 2       | 10/18/2014 | 77      |
+----------+---------+------------+---------+
| 6        | 2       | 10/18/2014 | 103     |
+----------+---------+------------+---------+
| 7        | 2       | 10/10/2014 | 13      |
+----------+---------+------------+---------+
| 8        | 3       | 10/09/2014 | 1       |
+----------+---------+------------+---------+
| 9        | 3       | 10/11/2014 | 2       |
+----------+---------+------------+---------+
| 10       | 3       | 10/12/2014 | 3       |
+----------+---------+------------+---------+

查询应该：

- 对于每个客户，显示order_date，其中最大订单数量。示例：客户1在10/10/2014有2个订单

- 特殊情况：当客户有大多数订单的多天时，显示最早的日期。示例：客户3在不同日期有3个单项订单。需要显示10/09/2014。

我应该能够在不使用临时表的情况下解决这个问题。

基拉

Answer 1

您需要两个条款：1）查找每个客户每天的最大订单数量; 2）找到最大订单数量发生的日期。第二步是消除具有相同订单数量的多天。


SELECT * FROM 
    (SELECT cust_id, 
        order_date, 
        count() cnt 
    FROM 
        orders 
    GROUP BY cust_id, order_date) x 
JOIN 
    (SELECT cust_id, 
            MAX(cnt) max_cnt 
    FROM 
        (SELECT cust_id, 
                order_date, 
                count() AS cnt 
         FROM 
            orders 
        GROUP BY cust_id, order_date) n 
GROUP BY cust_id) y 
ON 
    y.max_cnt = x.cnt 
AND 
    y.cust_id = x.cust_id 

JOIN  (SELECT   x.cust_id, 
                min(x.order_date) AS dd 
        FROM 
            (SELECT cust_id, 
                    order_date, 
                    count() cnt 
            FROM orders 
            GROUP BY by cust_id, order_date) x 
        JOIN 
            (SELECT cust_id, 
                    MAX(cnt) max_cnt 
            FROM 
            (SELECT cust_id, 
                    order_date, 
                    count() as cnt 
            FROM 
                orders 
            GROUP BY 
                cust_id, order_date) n 
            GROUP BY cust_id) y 
        ON 
            y.max_cnt = x.cnt 
        AND 
            y.cust_id = x.cust_id 
        GROUP BY cust_id) AS t2 
    ON 
        t2.cust_id=x.cust_id 
    AND 
        t2.dd=x.order_date;

MySQL：查找最大订单日

1 个答案: