Question

1321. Restaurant Growth

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Description

Table: Customer

+---------------+---------+
| Column Name   | Type  |
+---------------+---------+
| customer_id   | int   |
| name      | varchar |
| visited_on  | date  |
| amount    | int   |
+---------------+---------+
In SQL,(customer_id, visited_on) is the primary key for this table.
This table contains data about customer transactions in a restaurant.
visited_on is the date on which the customer with ID (customer_id) has visited the restaurant.
amount is the total paid by a customer.

 

You are the restaurant owner and you want to analyze a possible expansion (there will be at least one customer every day).

Compute the moving average of how much the customer paid in a seven days window (i.e., current day + 6 days before). average_amount should be rounded to two decimal places.

Return the result table ordered by visited_on in ascending order.

The result format is in the following example.

 

Example 1:

Input: 
Customer table:
+-------------+--------------+--------------+-------------+
| 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     | 
+-------------+--------------+--------------+-------------+
Output: 
+--------------+--------------+----------------+
| 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     |
+--------------+--------------+----------------+
Explanation: 
1st moving average from 2019-01-01 to 2019-01-07 has an average_amount of (100 + 110 + 120 + 130 + 110 + 140 + 150)/7 = 122.86
2nd moving average from 2019-01-02 to 2019-01-08 has an average_amount of (110 + 120 + 130 + 110 + 140 + 150 + 80)/7 = 120
3rd moving average from 2019-01-03 to 2019-01-09 has an average_amount of (120 + 130 + 110 + 140 + 150 + 80 + 110)/7 = 120
4th moving average from 2019-01-04 to 2019-01-10 has an average_amount of (130 + 110 + 140 + 150 + 80 + 110 + 130 + 150)/7 = 142.86

Solutions

Solution 1

# 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;

Solution 2

# 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;