SQL窗口函数的移动求和/平均，及相应的Leecode题目

原文链接：https://blog.csdn.net/weixin_45638528/article/details/130229603

移动求和的窗口函数——连续n个数据求和

sum(字段1) over (partition by 字段2 order by 字段3 rows n preceding)

LEECODE 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 |
+---------------+---------+
(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 |
+--------------+--------------+----------------+

来源：力扣（LeetCode）
链接：https://leetcode.cn/problems/restaurant-growth
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

解答
1、首先按visited_on分组，先计算每个日期的总的amount

select visited_on, sum(amount) as amount
from Customer
group by visited_on;

2、对上面得出的表计算移动平均、移动求和

select visited_on,
sum(amount) over (order by visited_on rows 6 preceding) as amount,
avg(amount) over (order by visited_on rows 6 preceding) as average_amount
from (
select visited_on,sum(amount) as amount
from Customer
group by visited_on) as a;

　3、因为题目计算的是以7天为一个时间段的数据，上表我们得出的结果包含了不足7天的数据（如2019-01-01至2019-01-06），因此要进行筛选，筛选思路可以是 每个日期与列表最小间隔日期为6天

select visited_on,amount,round(average_amount,2) as average_amount
from (
select visited_on,
sum(amount) over (order by visited_on rows 6 preceding) as amount,
avg(amount/1.0) over (order by visited_on rows 6 preceding) as average_amount
from (
select visited_on,sum(amount) as amount
from Customer
group by visited_on
) as a
) as b
where datediff(day, (select min(visited_on) from Customer), visited_on)>=6;

原文链接：https://blog.csdn.net/weixin_45638528/article/details/130229603