# LeetCode Solution, Easy, 584. Find Customer Referee

# [584. Find Customer Referee](https://leetcode.com/problems/find-customer-referee/)

## 題目敘述

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.

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 1:

    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 |
    +------+

### 題目翻譯

現在有一張資料表 `Customer`，編寫一個 SQL Query 在不要求排序的情況下找出，沒有連接到 `id = 2` 的 customer。

## 解法解析

基本上就是簡單的用 `WHERE` 關鍵字來篩選即可，需要注意的就是其中 `referee_id` 是有 `NULL` 的值，如果只單純用 `<> 2` 或 `!= 2` 會沒有辦法比對出來。所以需要再加上 `IS NULL` 的條件。

### 解法範例

#### SQL

```sql
# Write your MySQL query statement below
SELECT name
FROM `Customer`
WHERE referee_id <> 2
    OR referee_id IS NULL;
```