1321. 餐馆营业额变化增长
SQL架构表: 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。average_amount
要 保留两位小数。
查询结果按 visited_on
排序。
查询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
select a.visited_on, sum(b.amount) amount, round(avg(b.amount),2) average_amount from # 一天可能有多个人购买东西 (select visited_on, sum(amount) amount from customer group by visited_on order by visited_on) a left join (select visited_on, sum(amount) amount from customer group by visited_on order by visited_on) b on datediff(a.visited_on,b.visited_on) between 0 and 6 group by a.visited_on having count(*) = 7
标签:营业额,1321,2019,120,amount,01,餐馆,110,visited From: https://www.cnblogs.com/fulaien/p/17519938.html