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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
# Write your MySQL query statement below
WITH
t AS (
SELECT
visited_on,
SUM(amount) OVER (
ORDER BY visited_on
ROWS 6 PRECEDING
) AS amount,
RANK() OVER (
ORDER BY visited_on
ROWS 6 PRECEDING
) AS rk
FROM
(
SELECT visited_on, SUM(amount) AS amount
FROM Customer
GROUP BY visited_on
) AS tt
)
SELECT visited_on, amount, ROUND(amount / 7, 2) AS average_amount
FROM t
WHERE rk > 6;
# Write your MySQL query statement below
SELECT
a.visited_on,
SUM(b.amount) AS amount,
ROUND(SUM(b.amount) / 7, 2) AS average_amount
FROM
(SELECT DISTINCT visited_on FROM customer) AS a
JOIN customer AS b ON DATEDIFF(a.visited_on, b.visited_on) BETWEEN 0 AND 6
WHERE a.visited_on >= (SELECT MIN(visited_on) FROM customer) + 6
GROUP BY 1
ORDER BY 1;