表: Customer
+ --------- + --------------- +
| Column the Name | Type |
+ --------------- + + ---------
| CUSTOMER_ID | int |
| name | VARCHAR |
| visited_on | DATE |
| AMOUNT | int |
+ --------------- + --- + ------
(CUSTOMER_ID, visited_on) is the primary key for the table
of the customer transaction data table contains a restaurant
customer visited_on representation (customer_id) in the day visited the restaurant visited_on
amount is the total spending one day a customer of
You are the restaurant owner, and now you want to analyze possible changes in revenue growth (at least one customer a day)
Write a SQL query to calculate the average customer spending seven days (a + 6 days before the date of that date) is a period of time
Examples of query results format is as follows:
Sort query results by visited_on
average_amount to two decimal places, the format of date data for ( 'YYYY-MM-DD' )
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 | 01/09/2019 | 110 |
| 1 | Jhon | 10/01/2019 | 130 |
| 3 | jade | 10/01/2019 | 150 |
+ -------------- + ------------- + -------------- + ----- -------- +
Results Table:
+ ---------------- + -------------- -------------- + +
| 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 |
+ -------------- + ---------------- + -------------- +
The first seven days the average consumer from 2019-01-01 to 2019-01-07 (100 + 110 + 120 + 130 + 110 + 140 + 150) / 7 = 122.86
second seven days the average consumer from 2019-01 2019-01-08 -02 to (110 + 120 + 130 + 110 + 140 + 150 + 80) / 120 = 7
third seven days the average consumption is from 2019-01-03 to 2019-01-09 (120 + 130 + 110 + 140 + 150 + 80 + 110) / 120 = 7
fourth seven days from 2019-01-04 to 2019-01-10 average consumption is (130 + 110 + 140 + 150 + 80 + 110 + 130 + 150) / 7 = 142.86
Source: stay button (LeetCode)
link: https: //leetcode-cn.com/problems/restaurant-growth
copyrighted by deduction from all networks. Commercial reprint please contact the authorized official, non-commercial reprint please indicate the source.
Moderation: Write a SQL query to calculate the average customer spending seven days (a + 6 days before the date of that date) is a period of time
Thoughts: query the current average of six days before this date.
Problem solving:
Method One: first select the range, and then statistics
SELECT t1.visited_on, sum(t2.amount) amount, ROUND(SUM(t2.amount)/7.0, 2) average_amount
FROM
(SELECT DISTINCT visited_on FROM customer
WHERE visited_on >= DATE_ADD((SELECT min(visited_on) FROM customer), INTERVAL 6 DAY)) t1
LEFT JOIN
customer t2
ON t1.visited_on <= DATE_ADD(t2.visited_on, INTERVAL 6 DAY) AND t1.visited_on >= t2.visited_on
GROUP BY t1.visited_on;
Method Two:
First to go heavy sum for each table by date, and then do the Cartesian product, then screened to meet the conditions, and then after ordering. Get answers
select c1.visited_on,sum(c2.amount) amount,round(sum(c2.amount)/7,2) average_amount
from
(select customer_id,name,visited_on,sum(amount) amount from Customer group by visited_on)c1 cross join
(select customer_id,name,visited_on,sum(amount) amount from Customer group by visited_on) c2
where datediff(c1.visited_on,c2.visited_on)<7 and datediff(c1.visited_on,c2.visited_on)>=0
group by c1.visited_on
having count(*)>6
order by c1.visited_onKnowledge points:
DATE_ADD () function to add to the date specified time interval.
DATE_ADD (DATE, INTERVAL expr of the type) DATE parameter is valid date expression. expr parameter is the time interval you want to add.
