[LeetCode] 584. Find Customer Referee_Easy tag: SQL

转载于 2018-08-28 23:04:00 发布 · 113 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/Johnsonxiong/p/9551557.html

博客围绕数据库查询展开，要求从存储客户信息和推荐人信息的表中，查询出未被ID为2的人推荐的客户列表，提示使用is Null和!= 进行查询，代码转载自相关博客。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

转载于:https://www.cnblogs.com/Johnsonxiong/p/9551557.html