LeetCode MySQL-607. Sales Person

    AllenLi
    LeetCode MySQL

    Write an SQL query to report the names of all the salespersons who did not have any orders related to the company with the name “RED”.
    Return the result table in any order.

    寫一個 SQL Query來報告沒有在 "RED" 公司的訂單銷售人員。
可任意排序結果。

    Table: SalesPerson

    +-----------------+---------+
| Column Name     | Type    |
+-----------------+---------+
| sales_id        | int     |
| name            | varchar |
| salary          | int     |
| commission_rate | int     |
| hire_date       | date    |
+-----------------+---------+
sales_id is the primary key column for this table.
Each row of this table indicates the name and the ID of a salesperson alongside their salary, commission rate, and hire date.

    Table: Company

    +-------------+---------+
| Column Name | Type    |
+-------------+---------+
| com_id      | int     |
| name        | varchar |
| city        | varchar |
+-------------+---------+
com_id is the primary key column for this table.
Each row of this table indicates the name and the ID of a company and the city in which the company is located.

    Table: Orders

    +-------------+------+
| Column Name | Type |
+-------------+------+
| order_id    | int  |
| order_date  | date |
| com_id      | int  |
| sales_id    | int  |
| amount      | int  |
+-------------+------+
order_id is the primary key column for this table.
com_id is a foreign key to com_id from the Company table.
sales_id is a foreign key to sales_id from the SalesPerson table.
Each row of this table contains information about one order. This includes the ID of the company, the ID of the salesperson, the date of the order, and the amount paid.

    Example 1:

    Input: 
SalesPerson table:
+----------+------+--------+-----------------+------------+
| sales_id | name | salary | commission_rate | hire_date  |
+----------+------+--------+-----------------+------------+
| 1        | John | 100000 | 6               | 4/1/2006   |
| 2        | Amy  | 12000  | 5               | 5/1/2010   |
| 3        | Mark | 65000  | 12              | 12/25/2008 |
| 4        | Pam  | 25000  | 25              | 1/1/2005   |
| 5        | Alex | 5000   | 10              | 2/3/2007   |
+----------+------+--------+-----------------+------------+
Company table:
+--------+--------+----------+
| com_id | name   | city     |
+--------+--------+----------+
| 1      | RED    | Boston   |
| 2      | ORANGE | New York |
| 3      | YELLOW | Boston   |
| 4      | GREEN  | Austin   |
+--------+--------+----------+
Orders table:
+----------+------------+--------+----------+--------+
| order_id | order_date | com_id | sales_id | amount |
+----------+------------+--------+----------+--------+
| 1        | 1/1/2014   | 3      | 4        | 10000  |
| 2        | 2/1/2014   | 4      | 5        | 5000   |
| 3        | 3/1/2014   | 1      | 1        | 50000  |
| 4        | 4/1/2014   | 1      | 4        | 25000  |
+----------+------------+--------+----------+--------+
Output: 
+------+
| name |
+------+
| Amy  |
| Mark |
| Alex |
+------+
Explanation: 
According to orders 3 and 4 in the Orders table, it is easy to tell that only salesperson John and Pam have sales to company RED, so we report all the other names in the table salesperson.

    Solution:
    1. 選擇標題 SalesPerson 中的 name 為 name。
    2. 且 SalePerson 的 sales_id 不存在，以下：
    3. (Orders 的 sales_id 中，含有在 Company 的 ‘RED’ 紀錄)。 

    //orders.sales_id data
sales_id = (1, 4)
com_id   = (1, 1)

    Code1:

    SELECT s.name AS name
    FROM SalesPerson AS s
WHERE s.sales_id NOT IN 
	(SELECT o.sales_id
		FROM orders AS o
			LEFT JOIN Company AS c
				ON o.com_id = c.com_id
		WHERE c.name = 'RED');

    newSalesPerson

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