Find Customer Referee
Problem Statement - link #
SQL Schema
Create table If Not Exists Customer (id int, name varchar(25), referee_id int)
Truncate table Customer
insert into Customer (id, name, referee_id) values ('1', 'Will', 'None')
insert into Customer (id, name, referee_id) values ('2', 'Jane', 'None')
insert into Customer (id, name, referee_id) values ('3', 'Alex', '2')
insert into Customer (id, name, referee_id) values ('4', 'Bill', 'None')
insert into Customer (id, name, referee_id) values ('5', 'Zack', '1')
insert into Customer (id, name, referee_id) values ('6', 'Mark', '2')
Table: Customer
+-------------+---------+
| Column Name | Type |
+-------------+---------+
| id | int |
| name | varchar |
| referee_id | int |
+-------------+---------+
id is the primary key column for this table.
Each row of this table indicates the id of a customer, their name, and the id of the customer who referred them.
Your Task:
Write an SQL query to report the names of the customer that are not referred by the customer with id = 2.
Return the result table in any order.
The query result format is in the following example.
Example #
Input:
Customer table:
+----+------+------------+
| id | name | referee_id |
+----+------+------------+
| 1 | Will | null |
| 2 | Jane | null |
| 3 | Alex | 2 |
| 4 | Bill | null |
| 5 | Zack | 1 |
| 6 | Mark | 2 |
+----+------+------------+
Output:
+------+
| name |
+------+
| Will |
| Jane |
| Bill |
| Zack |
+------+
Solutions #
-- Write your MySQL query statement below
select name from Customer c where c.referee_id!=2 or c.referee_id is null;