【LeetCode】1321. 餐馆营业额变化增长
表: Customer
+---------------+---------+ | Column Name | Type | +---------------+---------+ | customer_id | int | | name | varchar | | visited_on | date | | amount | int | +---------------+---------+ 在 SQL 中，(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）。
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。average_amount 要保留两位小数。
结果按 visited_on 升序排序。
返回结果格式的例子如下。
示例 1:
输入： 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 | +-------------+--------------+--------------+-------------+ 输出： +--------------+--------------+----------------+ | 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 | +--------------+--------------+----------------+ 解释： 第一个七天消费平均值从 2019-01-01 到 2019-01-07 是restaurant-growth/restaurant-growth/ (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
方法一：使用窗口函数筛选日期
with cte1 as(select distinct visited_onfrom Customerwhere datediff(visited_on, (select min(visited_on) from Customer))>=6
)select
visited_on,sum(amount) as amount,round((sum(amount)/7),2) as average_amount
from(selecta.visited_on as visited_on,b.visited_on as old_visited_on,amount,dense_rank() over(partition by a.visited_on order by b.visited_on desc) as rnkfromcte1 a,Customer bwhere a.visited_on>=b.visited_on)tmp
where rnk<=7
group by visited_on方法二：使用datediff()函数筛选日期（datediff函数返回date1 - date2的计算结果）
with cte1 as(select distinct visited_onfrom Customerwhere datediff(visited_on, (select min(visited_on) from Customer))>=6
)select
cte1.visited_on,
sum(amount) as amount,
round((sum(amount)/7),2) as average_amount
from
cte1,Customer
where cte1.visited_on>=Customer.visited_on
and datediff(cte1.visited_on, Customer.visited_on)<=6
group by cte1.visited_on