表:Customers
+---------------+---------+ | Column Name | Type | +---------------+---------+ | customer_id | int | | name | varchar | +---------------+---------+ customer_id 是该表主键 该表包含消费者的信息
表:Orders
+---------------+---------+ | Column Name | Type | +---------------+---------+ | order_id | int | | order_date | date | | customer_id | int | | cost | int | +---------------+---------+ order_id 是该表主键 该表包含id为customer_id的消费者的订单信息 每一个消费者 每天一笔订单
写一个 SQL 语句,找到每个用户的最近三笔订单。如果用户的订单少于 3 笔,则返回他的全部订单。
返回的结果按照 customer_name
升序排列。如果排名有相同,则继续按照 customer_id
升序排列。如果排名还有相同,则继续按照 order_date
降序排列。
查询结果格式如下例所示:
Customers
+-------------+-----------+ | customer_id | name | +-------------+-----------+ | 1 | Winston | | 2 | Jonathan | | 3 | Annabelle | | 4 | Marwan | | 5 | Khaled | +-------------+-----------+Orders
+----------+------------+-------------+------+ | order_id | order_date | customer_id | cost | +----------+------------+-------------+------+ | 1 | 2020-07-31 | 1 | 30 | | 2 | 2020-07-30 | 2 | 40 | | 3 | 2020-07-31 | 3 | 70 | | 4 | 2020-07-29 | 4 | 100 | | 5 | 2020-06-10 | 1 | 1010 | | 6 | 2020-08-01 | 2 | 102 | | 7 | 2020-08-01 | 3 | 111 | | 8 | 2020-08-03 | 1 | 99 | | 9 | 2020-08-07 | 2 | 32 | | 10 | 2020-07-15 | 1 | 2 | +----------+------------+-------------+------+ Result table: +---------------+-------------+----------+------------+ | customer_name | customer_id | order_id | order_date | +---------------+-------------+----------+------------+ | Annabelle | 3 | 7 | 2020-08-01 | | Annabelle | 3 | 3 | 2020-07-31 | | Jonathan | 2 | 9 | 2020-08-07 | | Jonathan | 2 | 6 | 2020-08-01 | | Jonathan | 2 | 2 | 2020-07-30 | | Marwan | 4 | 4 | 2020-07-29 | | Winston | 1 | 8 | 2020-08-03 | | Winston | 1 | 1 | 2020-07-31 | | Winston | 1 | 10 | 2020-07-15 | +---------------+-------------+----------+------------+ Winston 有 4 笔订单, 排除了 "2020-06-10" 的订单, 因为它是最老的订单。 Annabelle 只有 2 笔订单, 全部返回。 Jonathan 恰好有 3 笔订单。 Marwan 只有 1 笔订单。 结果表我们按照 customer_name 升序排列,customer_id 升序排列,order_date 降序排列。
进阶:
- 你能写出来最近
n
笔订单的通用解决方案吗?