1321.(Medium)餐馆营业额变化增长

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

来源：力扣（LeetCode）

链接：https://leetcode-cn.com/problems/restaurant-growth

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Solution

self join

select a.visited_on, 
    sum(b.amount) amount,
    round(sum(b.amount)/7,2) average_amount
from (select visited_on, sum(amount) amount from Customer GROUP BY visited_on) a, 
    (select visited_on, sum(amount) amount from Customer GROUP BY visited_on) b 
where datediff(a.visited_on,b.visited_on) between 0 and 6
group by a.visited_on
having count(distinct b.visited_on) = 7;

window function

SELECT visited_on,amount,round(average_amount,2)average_amount
from 
    (select 
        visited_on, 
     sum(amount) over (order by visited_on rows 6 preceding) amount,
     avg(amount) over (order by visited_on rows 6 preceding) average_amount
    from (select visited_on, sum(amount) amount from Customer GROUP BY visited_on) a
    ) b
where DATEDIFF(visited_on,(SELECT min(visited_on) from Customer)) >= 6;

Table Schema

Create table If Not Exists Customer (customer_id int, name varchar(20), visited_on date, amount int);
Truncate table Customer;
insert into Customer (customer_id, name, visited_on, amount) values ('1', 'Jhon', '2019-01-01', '100');
insert into Customer (customer_id, name, visited_on, amount) values ('2', 'Daniel', '2019-01-02', '110');
insert into Customer (customer_id, name, visited_on, amount) values ('3', 'Jade', '2019-01-03', '120');
insert into Customer (customer_id, name, visited_on, amount) values ('4', 'Khaled', '2019-01-04', '130');
insert into Customer (customer_id, name, visited_on, amount) values ('5', 'Winston', '2019-01-05', '110');
insert into Customer (customer_id, name, visited_on, amount) values ('6', 'Elvis', '2019-01-06', '140');
insert into Customer (customer_id, name, visited_on, amount) values ('7', 'Anna', '2019-01-07', '150');
insert into Customer (customer_id, name, visited_on, amount) values ('8', 'Maria', '2019-01-08', '80');
insert into Customer (customer_id, name, visited_on, amount) values ('9', 'Jaze', '2019-01-09', '110');
insert into Customer (customer_id, name, visited_on, amount) values ('1', 'Jhon', '2019-01-10', '130');
insert into Customer (customer_id, name, visited_on, amount) values ('3', 'Jade', '2019-01-10', '150');