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Database

2051. The Category of Each Member in the Store 🔒

中文文档

Description

Table: Members

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| member_id   | int     |
| name        | varchar |
+-------------+---------+
member_id is the column with unique values for this table.
Each row of this table indicates the name and the ID of a member.

 

Table: Visits

+-------------+------+
| Column Name | Type |
+-------------+------+
| visit_id    | int  |
| member_id   | int  |
| visit_date  | date |
+-------------+------+
visit_id is the column with unique values for this table.
member_id is a foreign key (reference column) to member_id from the Members table.
Each row of this table contains information about the date of a visit to the store and the member who visited it.

 

Table: Purchases

+----------------+------+
| Column Name    | Type |
+----------------+------+
| visit_id       | int  |
| charged_amount | int  |
+----------------+------+
visit_id is the column with unique values for this table.
visit_id is a foreign key (reference column) to visit_id from the Visits table.
Each row of this table contains information about the amount charged in a visit to the store.

 

A store wants to categorize its members. There are three tiers:

  • "Diamond": if the conversion rate is greater than or equal to 80.
  • "Gold": if the conversion rate is greater than or equal to 50 and less than 80.
  • "Silver": if the conversion rate is less than 50.
  • "Bronze": if the member never visited the store.

The conversion rate of a member is (100 * total number of purchases for the member) / total number of visits for the member.

Write a solution to report the id, the name, and the category of each member.

Return the result table in any order.

The result format is in the following example.

 

Example 1:

Input: 
Members table:
+-----------+---------+
| member_id | name    |
+-----------+---------+
| 9         | Alice   |
| 11        | Bob     |
| 3         | Winston |
| 8         | Hercy   |
| 1         | Narihan |
+-----------+---------+
Visits table:
+----------+-----------+------------+
| visit_id | member_id | visit_date |
+----------+-----------+------------+
| 22       | 11        | 2021-10-28 |
| 16       | 11        | 2021-01-12 |
| 18       | 9         | 2021-12-10 |
| 19       | 3         | 2021-10-19 |
| 12       | 11        | 2021-03-01 |
| 17       | 8         | 2021-05-07 |
| 21       | 9         | 2021-05-12 |
+----------+-----------+------------+
Purchases table:
+----------+----------------+
| visit_id | charged_amount |
+----------+----------------+
| 12       | 2000           |
| 18       | 9000           |
| 17       | 7000           |
+----------+----------------+
Output: 
+-----------+---------+----------+
| member_id | name    | category |
+-----------+---------+----------+
| 1         | Narihan | Bronze   |
| 3         | Winston | Silver   |
| 8         | Hercy   | Diamond  |
| 9         | Alice   | Gold     |
| 11        | Bob     | Silver   |
+-----------+---------+----------+
Explanation: 
- User Narihan with id = 1 did not make any visits to the store. She gets a Bronze category.
- User Winston with id = 3 visited the store one time and did not purchase anything. The conversion rate = (100 * 0) / 1 = 0. He gets a Silver category.
- User Hercy with id = 8 visited the store one time and purchased one time. The conversion rate = (100 * 1) / 1 = 1. He gets a Diamond category.
- User Alice with id = 9 visited the store two times and purchased one time. The conversion rate = (100 * 1) / 2 = 50. She gets a Gold category.
- User Bob with id = 11 visited the store three times and purchased one time. The conversion rate = (100 * 1) / 3 = 33.33. He gets a Silver category.

Solutions

Solution 1

MySQL

# Write your MySQL query statement below
SELECT
    m.member_id,
    name,
    CASE
        WHEN COUNT(v.visit_id) = 0 THEN 'Bronze'
        WHEN 100 * COUNT(charged_amount) / COUNT(v.visit_id) >= 80 THEN 'Diamond'
        WHEN 100 * COUNT(charged_amount) / COUNT(v.visit_id) >= 50 THEN 'Gold'
        ELSE 'Silver'
    END AS category
FROM
    Members AS m
    LEFT JOIN Visits AS v ON m.member_id = v.member_id
    LEFT JOIN Purchases AS p ON v.visit_id = p.visit_id
GROUP BY member_id;