607. Sales Person

Description

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 with unique values) 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 with unique values) 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 with unique values) for this table.
com_id is a foreign key (reference column) to com_id from the Company table.
sales_id is a foreign key (reference column) 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.

Write a solution to find 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.

The result format is in the following example.

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.

Solutions

Solution 1: LEFT JOIN + GROUP BY

We can use a left join to join the SalesPerson table with the Orders table on the condition of sales id, and then join the result with the Company table on the condition of company id. After that, we can group by sales_id and count the number of orders with the company name RED. Finally, we can filter out the salespersons who do not have any orders with the company name RED.

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# Write your MySQL query statement below
SELECT s.name
FROM
    SalesPerson AS s
    LEFT JOIN Orders USING (sales_id)
    LEFT JOIN Company AS c USING (com_id)
GROUP BY sales_id
HAVING IFNULL(SUM(c.name = 'RED'), 0) = 0;