Given a table customer holding customers information and the referee.
+------+------+-----------+ | id | name | referee_id| +------+------+-----------+ | 1 | Will | NULL | | 2 | Jane | NULL | | 3 | Alex | 2 | | 4 | Bill | NULL | | 5 | Zack | 1 | | 6 | Mark | 2 | +------+------+-----------+
Write a query to return the list of customers NOT referred by the person with id ‘2‘.
For the sample data above, the result is:
+------+ | name | +------+ | Will | | Jane | | Bill | | Zack | +------+
note: 要用 is Null 和 != .
Code
SELECT name FROM customer WHERE referee_id != 2 OR referee_id IS NULL
[LeetCode] 584. Find Customer Referee_Easy tag: SQL
原文：https://www.cnblogs.com/Johnsonxiong/p/9551557.html