Leetcode1321. 餐馆营业额变化增长（中等）

题目
表: Customer

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| customer_id   | int     |
| name          | varchar |
| visited_on    | date    |
| amount        | int     |
+---------------+---------+

(customer_id, visited_on) 是该表的主键
该表包含一家餐馆的顾客交易数据
visited_on 表示 (customer_id) 的顾客在 visited_on 那天访问了餐馆
amount 是一个顾客某一天的消费总额

你是餐馆的老板，现在你想分析一下可能的营业额变化增长（每天至少有一位顾客）

写一条 SQL 查询计算以 7 天（某日期 + 该日期前的 6 天）为一个时间段的顾客消费平均值

查询结果格式的例子如下：

查询结果按 visited_on 排序
average_amount 要 保留两位小数，日期数据的格式为 ('YYYY-MM-DD')

Customer 表:

+-------------+--------------+--------------+-------------+
| customer_id | name         | visited_on   | amount      |
+-------------+--------------+--------------+-------------+
| 1           | Jhon         | 2019-01-01   | 100         |
| 2           | Daniel       | 2019-01-02   | 110         |
| 3           | Jade         | 2019-01-03   | 120         |
| 4           | Khaled       | 2019-01-04   | 130         |
| 5           | Winston      | 2019-01-05   | 110         | 
| 6           | Elvis        | 2019-01-06   | 140         | 
| 7           | Anna         | 2019-01-07   | 150         |
| 8           | Maria        | 2019-01-08   | 80          |
| 9           | Jaze         | 2019-01-09   | 110         | 
| 1           | Jhon         | 2019-01-10   | 130         | 
| 3           | Jade         | 2019-01-10   | 150         | 
+-------------+--------------+--------------+-------------+

结果表:

+--------------+--------------+----------------+
| visited_on   | amount       | average_amount |
+--------------+--------------+----------------+
| 2019-01-07   | 860          | 122.86         |
| 2019-01-08   | 840          | 120            |
| 2019-01-09   | 840          | 120            |
| 2019-01-10   | 1000         | 142.86         |
+--------------+--------------+----------------+

第一个七天消费平均值从 2019-01-01 到 2019-01-07 是 (100 + 110 + 120 + 130 + 110 + 140 + 150)/7 = 122.86
第二个七天消费平均值从 2019-01-02 到 2019-01-08 是 (110 + 120 + 130 + 110 + 140 + 150 + 80)/7 = 120
第三个七天消费平均值从 2019-01-03 到 2019-01-09 是 (120 + 130 + 110 + 140 + 150 + 80 + 110)/7 = 120
第四个七天消费平均值从 2019-01-04 到 2019-01-10 是 (130 + 110 + 140 + 150 + 80 + 110 + 130 + 150)/7 = 142.86

解答

这不就很像窗口函数么
先选出最终表的visited_on

select C.visited_on
from Customer as C
where C.visited_on >= DATE_ADD((SELECT min(visited_on) FROM customer), INTERVAL 6 DAY);

然后与主表左连接
连接条件就是左表的时间要大于右表的时间 而且 右表的时间+6要大于左表的时间

select tmp.visited_on, sum(CC.amount) as amount,
avg(CC.amount) as average_amount
from (select C.visited_on
from Customer as C
where C.visited_on >= DATE_ADD((SELECT min(visited_on) FROM customer), INTERVAL 6 DAY)) as tmp
left join Customer as CC
on tmp.visited_on >= CC.visited_on and DATE_ADD(tmp.visited_on, INTERVAL 6 DAY) >= tmp.visited_on
group by tmp.visited_on

别的解答

先对每个表按日期进行去重求和，然后做笛卡尔积，然后筛选出符合条件的，再之后排序。得到答案

select c1.visited_on,sum(c2.amount) amount,round(sum(c2.amount)/7,2) average_amount
from 
(select customer_id,name,visited_on,sum(amount) amount from Customer group by visited_on)c1 cross join 
(select customer_id,name,visited_on,sum(amount) amount from Customer group by visited_on) c2 
where datediff(c1.visited_on,c2.visited_on)<7  and datediff(c1.visited_on,c2.visited_on)>=0
group by c1.visited_on
having count(*)>6
order by c1.visited_on