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Commit 4769d32

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feat: update sql solutions to lc problems: No.0607,1112,1407,1607 (doocs#1781)
* No.0607.Sales Person * No.1112.Highest Grade For Each Student * No.1407.Top Travellers * No.1607.Sellers With No Sales
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‎solution/0100-0199/0175.Combine Two Tables/README_EN.md‎

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@@ -76,6 +76,10 @@ addressId = 1 contains information about the address of personId = 2.
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## Solutions
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**Solution 1: LEFT JOIN**
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We can use a left join to join the `Person` table with the `Address` table on the condition `Person.personId = Address.personId`, which will give us the first name, last name, city, and state of each person. If the address of a `personId` is not in the `Address` table, it will be reported as `null`.
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### **SQL**

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

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<!-- 这里可写通用的实现逻辑 -->
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<!-- tabs:start -->
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**方法一:左连接 + 分组统计**
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### **SQL**
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我们可以使用左连接将 `SalesPerson` 表与 `Orders` 表连接起来,再与 `Company` 表连接起来,然后按照 `sales_id` 分组,每组统计有多少个公司的名字为 `RED` 的订单,最后筛选出没有这样的订单的销售人员的姓名。
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```sql
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SELECT name
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FROM salesperson
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WHERE
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sales_id NOT IN (
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SELECT s.sales_id
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FROM
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orders AS o
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INNER JOIN salesperson AS s ON o.sales_id = s.sales_id
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INNER JOIN company AS 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:start -->
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```sql
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SELECT
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name
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FROM SalesPerson AS s
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WHERE
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0 = (
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SELECT
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COUNT(*)
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FROM
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Orders AS o
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JOIN Company AS c ON o.com_id = c.com_id
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WHERE o.sales_id = s.sales_id AND c.name = 'RED'
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);
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```
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### **SQL**
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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
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sales_id NOT IN (
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SELECT sales_id
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FROM
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Company AS c
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JOIN Orders AS o USING (com_id)
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GROUP BY sales_id
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HAVING sum(name = 'RED') > 0
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);
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SELECT s.name
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FROM
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SalesPerson AS s
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LEFT JOIN Orders USING (sales_id)
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LEFT JOIN Company AS c USING (com_id)
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GROUP BY sales_id
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HAVING ifnull(sum(c.name = 'RED'), 0) = 0;
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```
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<!-- tabs:end -->

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

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## Solutions
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<!-- tabs:start -->
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**Solution 1: LEFT JOIN + GROUP BY**
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### **SQL**
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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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```sql
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SELECT name
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FROM salesperson
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WHERE
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sales_id NOT IN (
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SELECT s.sales_id
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FROM
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orders AS o
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INNER JOIN salesperson AS s ON o.sales_id = s.sales_id
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INNER JOIN company AS 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:start -->
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```sql
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SELECT
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name
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FROM SalesPerson AS s
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WHERE
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0 = (
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SELECT
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COUNT(*)
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FROM
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Orders AS o
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JOIN Company AS c ON o.com_id = c.com_id
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WHERE o.sales_id = s.sales_id AND c.name = 'RED'
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);
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```
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### **SQL**
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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
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sales_id NOT IN (
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SELECT sales_id
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FROM
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Company AS c
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JOIN Orders AS o USING (com_id)
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GROUP BY sales_id
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HAVING sum(name = 'RED') > 0
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);
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SELECT s.name
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FROM
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SalesPerson AS s
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LEFT JOIN Orders USING (sales_id)
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LEFT JOIN Company AS c USING (com_id)
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GROUP BY sales_id
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HAVING ifnull(sum(c.name = 'RED'), 0) = 0;
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```
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SELECT name
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FROM salesperson
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WHERE
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sales_id NOT IN (
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SELECT s.sales_id
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FROM
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orders AS o
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INNER JOIN salesperson AS s ON o.sales_id = s.sales_id
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INNER JOIN company AS c ON o.com_id = c.com_id
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WHERE c.name = 'RED'
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);
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# Write your MySQL query statement below
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SELECT s.name
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FROM
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SalesPerson AS s
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LEFT JOIN Orders USING (sales_id)
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LEFT JOIN Company AS c USING (com_id)
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GROUP BY sales_id
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HAVING ifnull(sum(c.name = 'RED'), 0) = 0;

‎solution/1100-1199/1112.Highest Grade For Each Student/README.md‎

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<!-- 这里可写通用的实现逻辑 -->
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**方法一:RANK() OVER() 窗口函数**
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我们可以使用 `RANK() OVER()` 窗口函数,按照每个学生的成绩降序排列,如果成绩相同,按照课程号升序排列,然后取每个学生排名为 1ドル$ 的记录。
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**方法二:子查询**
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我们可以先查询每个学生的最高成绩,然后再查询每个学生的最高成绩对应的最小课程号。
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### **SQL**
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ORDER BY student_id;
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```
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```sql
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# Write your MySQL query statement below
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SELECT student_id, min(course_id) AS course_id, grade
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FROM Enrollments
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WHERE
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(student_id, grade) IN (
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SELECT student_id, max(grade) AS grade
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FROM Enrollments
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GROUP BY 1
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)
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GROUP BY 1
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ORDER BY 1;
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```
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‎solution/1100-1199/1112.Highest Grade For Each Student/README_EN.md‎

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## Solutions
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**Solution 1: RANK() OVER() Window Function**
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We can use the `RANK() OVER()` window function to sort the grades of each student in descending order. If the grades are the same, we sort them in ascending order by course number, and then select the record with a rank of 1ドル$ for each student.
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**Solution 2: Subquery**
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We can first query the highest grade of each student, and then query the minimum course number corresponding to the highest grade of each student.
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### **SQL**
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ORDER BY student_id;
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```
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```sql
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# Write your MySQL query statement below
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SELECT student_id, min(course_id) AS course_id, grade
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FROM Enrollments
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WHERE
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(student_id, grade) IN (
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SELECT student_id, max(grade) AS grade
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FROM Enrollments
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GROUP BY 1
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)
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GROUP BY 1
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ORDER BY 1;
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```
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‎solution/1400-1499/1407.Top Travellers/README.md‎

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<!-- 这里可写通用的实现逻辑 -->
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**方法一:左连接 + 分组统计**
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我们可以使用左连接,将 `Users` 表与 `Rides` 表按照用户 id 连接,然后按照用户 id 分组,统计每个用户的旅行距离。注意,如果用户没有旅行记录,那么旅行距离为 0ドル$。
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### **SQL**

‎solution/1400-1499/1407.Top Travellers/README_EN.md‎

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## Solutions
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**Solution 1: LEFT JOIN + GROUP BY**
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We can use a left join to join the `Users` table with the `Rides` table on the condition of user id, and then group by user id to calculate the travel distance for each user. Note that if a user has no travel records, the travel distance is 0ドル$.
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### **SQL**

‎solution/1600-1699/1607.Sellers With No Sales/README.md‎

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<!-- 这里可写通用的实现逻辑 -->
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**方法一:左连接 + 分组 + 筛选**
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我们可以使用左连接,将 `Seller` 表与 `Orders` 表按照字段 `seller_id` 连接,然后按照 `seller_id` 分组,统计每个卖家在 2020ドル$ 年的卖出次数,最后筛选出卖出次数为 0ドル$ 的卖家。
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### **SQL**
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```sql
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# Write your MySQL query statement below
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SELECT
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seller_name
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SELECT seller_name
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FROM
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seller AS s
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LEFT JOIN orders AS o ON s.seller_id = o.seller_id AND year(sale_date) = '2020'
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WHERE o.seller_id IS NULL
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ORDER BY seller_name;
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Seller
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LEFT JOIN Orders USING (seller_id)
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GROUP BY seller_id
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HAVING ifnull(sum(year(sale_date) = 2020), 0) = 0
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ORDER BY 1;
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```
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‎solution/1600-1699/1607.Sellers With No Sales/README_EN.md‎

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## Solutions
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**Solution 1: LEFT JOIN + GROUP BY + FILTER**
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We can use a left join to join the `Seller` table with the `Orders` table on the condition `seller_id`, and then group by `seller_id` to count the number of sales for each seller in the year 2020ドル$. Finally, we can filter out the sellers with zero sales.
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### **SQL**
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```sql
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# Write your MySQL query statement below
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SELECT
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seller_name
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SELECT seller_name
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FROM
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seller AS s
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LEFT JOIN orders AS o ON s.seller_id = o.seller_id AND year(sale_date) = '2020'
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WHERE o.seller_id IS NULL
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ORDER BY seller_name;
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Seller
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LEFT JOIN Orders USING (seller_id)
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GROUP BY seller_id
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HAVING ifnull(sum(year(sale_date) = 2020), 0) = 0
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ORDER BY 1;
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```
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<!-- tabs:end -->

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