LeetCode Solution, Medium, 607. Sales Person
607. Sales Person
題目敘述
SQL Schema >
Create table If Not Exists SalesPerson (sales_id int, name varchar(255), salary int, commission_rate int, hire_date date) Create table If Not Exists Company (com_id int, name varchar(255), city varchar(255)) Create table If Not Exists Orders (order_id int, order_date date, com_id int, sales_id int, amount int) Truncate table SalesPerson insert into SalesPerson (sales_id, name, salary, commission_rate, hire_date) values ('1', 'John', '100000', '6', '4/1/2006') insert into SalesPerson (sales_id, name, salary, commission_rate, hire_date) values ('2', 'Amy', '12000', '5', '5/1/2010') insert into SalesPerson (sales_id, name, salary, commission_rate, hire_date) values ('3', 'Mark', '65000', '12', '12/25/2008') insert into SalesPerson (sales_id, name, salary, commission_rate, hire_date) values ('4', 'Pam', '25000', '25', '1/1/2005') insert into SalesPerson (sales_id, name, salary, commission_rate, hire_date) values ('5', 'Alex', '5000', '10', '2/3/2007') Truncate table Company insert into Company (com_id, name, city) values ('1', 'RED', 'Boston') insert into Company (com_id, name, city) values ('2', 'ORANGE', 'New York') insert into Company (com_id, name, city) values ('3', 'YELLOW', 'Boston') insert into Company (com_id, name, city) values ('4', 'GREEN', 'Austin') Truncate table Orders insert into Orders (order_id, order_date, com_id, sales_id, amount) values ('1', '1/1/2014', '3', '4', '10000') insert into Orders (order_id, order_date, com_id, sales_id, amount) values ('2', '2/1/2014', '4', '5', '5000') insert into Orders (order_id, order_date, com_id, sales_id, amount) values ('3', '3/1/2014', '1', '1', '50000') insert into Orders (order_id, order_date, com_id, sales_id, amount) values ('4', '4/1/2014', '1', '4', '25000')
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.
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.
The query 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.
題目翻譯
解法解析
這邊使用了 SubQuery 來篩選條件,我們可以先思考子條件
解法範例
MySQL
# Write your MySQL query statement below
SELECT s.name
FROM salesperson s
WHERE s.sales_id NOT IN (
SELECT o.sales_id
FROM orders o
LEFT JOIN company c ON o.com_id = c.com_id
WHERE c.name = 'RED'
);