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feat: add sql solution to leetcode problem: No.0607
See https://leetcode-cn.com/problems/game-play-analysis-i
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‎solution/0600-0699/0607.Sales Person/README.md

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<!-- 这里写题目描述 -->
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None
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<p>给定 3 个表：&nbsp;<code>salesperson</code>，&nbsp;<code>company</code>，&nbsp;<code>orders</code>。<br>
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输出所有表&nbsp;<code>salesperson</code>&nbsp;中，没有向公司 &#39;RED&#39; 销售任何东西的销售员。</p>
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<p><strong>示例：</strong><br>
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<strong>输入</strong></p>
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<p>表：&nbsp;<code>salesperson</code></p>
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<pre>+----------+------+--------+-----------------+-----------+
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| sales_id | name | salary | commission_rate | hire_date |
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+----------+------+--------+-----------------+-----------+
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| 1 | John | 100000 | 6 | 4/1/2006 |
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| 2 | Amy | 120000 | 5 | 5/1/2010 |
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| 3 | Mark | 65000 | 12 | 12/25/2008|
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| 4 | Pam | 25000 | 25 | 1/1/2005 |
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| 5 | Alex | 50000 | 10 | 2/3/2007 |
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+----------+------+--------+-----------------+-----------+
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</pre>
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<p>表&nbsp;<code>salesperson</code> 存储了所有销售员的信息。每个销售员都有一个销售员编号&nbsp;<strong>sales_id</strong> 和他的名字&nbsp;<strong>name&nbsp;</strong>。</p>
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<p>表：&nbsp;<code>company</code></p>
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<pre>+---------+--------+------------+
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| com_id | name | city |
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+---------+--------+------------+
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| 1 | RED | Boston |
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| 2 | ORANGE | New York |
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| 3 | YELLOW | Boston |
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| 4 | GREEN | Austin |
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+---------+--------+------------+
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</pre>
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<p>表&nbsp;<code>company</code>&nbsp;存储了所有公司的信息。每个公司都有一个公司编号&nbsp;<strong>com_id</strong>&nbsp;和它的名字 <strong>name</strong>&nbsp;。</p>
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<p>表：&nbsp;<code>orders</code></p>
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<pre>+----------+------------+---------+----------+--------+
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| order_id | order_date | com_id | sales_id | amount |
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+----------+------------+---------+----------+--------+
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| 1 | 1/1/2014 | 3 | 4 | 100000 |
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| 2 | 2/1/2014 | 4 | 5 | 5000 |
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| 3 | 3/1/2014 | 1 | 1 | 50000 |
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| 4 | 4/1/2014 | 1 | 4 | 25000 |
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+----------+----------+---------+----------+--------+
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</pre>
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<p>表&nbsp;<code>orders</code>&nbsp;存储了所有的销售数据，包括销售员编号 <strong>sales_id </strong>和公司编号 <strong>com_id</strong>&nbsp;。</p>
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<p><strong>输出</strong></p>
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<pre>+------+
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| name |
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+------+
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| Amy |
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| Mark |
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| Alex |
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+------+
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</pre>
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<p><strong>解释</strong></p>
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<p>根据表&nbsp;<code>orders</code>&nbsp;中的订单 &#39;3&#39; 和 &#39;4&#39; ，容易看出只有 &#39;John&#39; 和 &#39;Pam&#39; 两个销售员曾经向公司 &#39;RED&#39; 销售过。</p>
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<p>所以我们需要输出表&nbsp;<code>salesperson</code>&nbsp;中所有其他人的名字。</p>
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## 解法
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### **SQL**
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```
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```sql
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# Write your MySQL query statement below
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SELECT name
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FROM salesperson
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WHERE sales_id NOT IN
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(SELECT sales_id
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FROM orders
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WHERE com_id =
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(SELECT com_id
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FROM company
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WHERE name = 'RED'))
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WHERE sales_id
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NOT IN (
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SELECT s.sales_id FROM orders o
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INNER JOIN salesperson s ON o.sales_id = s.sales_id
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INNER JOIN company c ON o.com_id = c.com_id
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WHERE c.name = 'RED'
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);
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```
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<!-- tabs:end -->

‎solution/0600-0699/0607.Sales Person/README_EN.md

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## Description
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None
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<p>Given three tables: <code>salesperson</code>, <code>company</code>, <code>orders</code>.<br />
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Output all the <b>names</b> in the table <code>salesperson</code>, who didn&rsquo;t have sales to company &#39;RED&#39;.</p>
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<p><b>Example</b><br />
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<b>Input</b></p>
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<p>Table: <code>salesperson</code></p>
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<pre>
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+----------+------+--------+-----------------+-----------+
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| sales_id | name | salary | commission_rate | hire_date |
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+----------+------+--------+-----------------+-----------+
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| 1 | John | 100000 | 6 | 4/1/2006 |
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| 2 | Amy | 120000 | 5 | 5/1/2010 |
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| 3 | Mark | 65000 | 12 | 12/25/2008|
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| 4 | Pam | 25000 | 25 | 1/1/2005 |
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| 5 | Alex | 50000 | 10 | 2/3/2007 |
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+----------+------+--------+-----------------+-----------+
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</pre>
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The table <code>salesperson</code> holds the salesperson information. Every salesperson has a <b>sales_id</b> and a <b>name</b>.
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<p>Table: <code>company</code></p>
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<pre>
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+---------+--------+------------+
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| com_id | name | city |
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+---------+--------+------------+
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| 1 | RED | Boston |
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| 2 | ORANGE | New York |
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| 3 | YELLOW | Boston |
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| 4 | GREEN | Austin |
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+---------+--------+------------+
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</pre>
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The table <code>company</code> holds the company information. Every company has a <b>com_id</b> and a <b>name</b>.
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<p>Table: <code>orders</code></p>
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<pre>
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+----------+------------+---------+----------+--------+
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| order_id | order_date | com_id | sales_id | amount |
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+----------+------------+---------+----------+--------+
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| 1 | 1/1/2014 | 3 | 4 | 100000 |
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| 2 | 2/1/2014 | 4 | 5 | 5000 |
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| 3 | 3/1/2014 | 1 | 1 | 50000 |
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| 4 | 4/1/2014 | 1 | 4 | 25000 |
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+----------+----------+---------+----------+--------+
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</pre>
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The table <code>orders</code> holds the sales record information, salesperson and customer company are represented by <b>sales_id</b> and <b>com_id</b>.
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<p><b>output</b></p>
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<pre>
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+------+
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| name |
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+------+
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| Amy |
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| Mark |
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| Alex |
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+------+
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</pre>
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<p><b>Explanation</b></p>
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<p>According to order &#39;3&#39; and &#39;4&#39; in table <code>orders</code>, it is easy to tell only salesperson &#39;John&#39; and &#39;Pam&#39; have sales to company &#39;RED&#39;,<br />
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so we need to output all the other <b>names</b> in the table <code>salesperson</code>.</p>
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## Solutions
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<!-- tabs:start -->
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### **SQL**
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```
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```sql
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# Write your MySQL query statement below
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SELECT name
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FROM salesperson
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WHERE sales_id NOT IN
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(SELECT sales_id
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FROM orders
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WHERE com_id =
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(SELECT com_id
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FROM company
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WHERE name = 'RED'))
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WHERE sales_id
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NOT IN (
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SELECT s.sales_id FROM orders o
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INNER JOIN salesperson s ON o.sales_id = s.sales_id
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INNER JOIN company c ON o.com_id = c.com_id
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WHERE c.name = 'RED'
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);
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```
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<!-- tabs:end -->

‎solution/0600-0699/0607.Sales Person/Solution.sql

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# Write your MySQL query statement below
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SELECT name
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FROM salesperson
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WHERE sales_id NOT IN
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(SELECT sales_id
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FROM orders
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WHERE com_id =
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(SELECT com_id
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FROM company
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WHERE name = 'RED'))
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WHERE sales_id
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NOT IN (
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SELECT s.sales_id FROM orders o
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INNER JOIN salesperson s ON o.sales_id = s.sales_id
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INNER JOIN company c ON o.com_id = c.com_id
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WHERE c.name = 'RED'
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);

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