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Commit bfd80c4

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feat: update sql solutions to lc problems: No.0183,1821,1873 (#1779)
* No.0183.Customers Who Never Order * No.1821.Find Customers With Positive Revenue this Year * No.1873.Calculate Special Bonus
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‎.prettierignore‎

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/solution/1600-1699/1635.Hopper Company Queries I/Solution.sql
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/solution/1600-1699/1651.Hopper Company Queries III/Solution.sql
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/solution/1700-1799/1767.Find the Subtasks That Did Not Execute/Solution.sql
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/solution/1800-1899/1873.Calculate Special Bonus/Solution.sql
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/solution/2100-2199/2118.Build the Equation/Solution.sql
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/solution/2100-2199/2153.The Number of Passengers in Each Bus II/Solution.sql
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/solution/2200-2299/2205.The Number of Users That Are Eligible for Discount/Solution.sql

‎solution/0100-0199/0183.Customers Who Never Order/README_EN.md‎

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## Solutions
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**Solution 1: NOT IN**
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List all customer IDs of existing orders, and use `NOT IN` to find customers who are not in the list.
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**Solution 2: LEFT JOIN**
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Use `LEFT JOIN` to join the tables and return the data where `CustomerId` is `NULL`.
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### **SQL**

‎solution/1800-1899/1821.Find Customers With Positive Revenue this Year/README.md‎

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<!-- 这里可写通用的实现逻辑 -->
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**方法一:WHERE 子句**
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我们可以直接使用 `WHERE` 子句来筛选出 `year` 为 `2021` 且 `revenue` 大于 0ドル$ 的客户。
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### **SQL**

‎solution/1800-1899/1821.Find Customers With Positive Revenue this Year/README_EN.md‎

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## Solutions
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**Solution 1: WHERE Clause**
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We can directly use the `WHERE` clause to filter out the customers whose `year` is `2021` and `revenue` is greater than 0ドル$.
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### **SQL**

‎solution/1800-1899/1873.Calculate Special Bonus/README.md‎

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<!-- 这里可写通用的实现逻辑 -->
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**方法一:IF 语句 + ORDER BY 子句**
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我们可以使用 `IF` 语句来判断奖金的计算方式,然后使用 `ORDER BY` 将结果按照 `employee_id` 排序。
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### **SQL**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```sql
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SELECT
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employee_id,
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CASE
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WHEN employee_id % 2 = 0
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OR LEFT(name, 1) = 'M' THEN 0
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ELSE salary
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END AS bonus
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FROM
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employees;
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```
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MySQL
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```sql
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# Write your MySQL query statement below
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SELECT
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employee_id,
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IF(
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salary
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) AS bonus
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FROM
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employees;
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```
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```sql
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# Write your MySQL query statement below
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SELECT
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employee_id,
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CASE
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WHEN (employee_id % 2 = 1 AND NAME NOT LIKE "M%") THEN salary
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ELSE 0
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END AS bonus
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FROM Employees
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ORDER BY employee_id;
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employees
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ORDER BY 1;
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```
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‎solution/1800-1899/1873.Calculate Special Bonus/README_EN.md‎

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## Solutions
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**Solution 1: IF Statement + ORDER BY Clause**
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### **SQL**
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We can use the `IF` statement to determine the calculation method of the bonus, and then use `ORDER BY` to sort the results by `employee_id`.
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```sql
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SELECT
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employee_id,
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CASE
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WHEN employee_id % 2 = 0
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OR LEFT(name, 1) = 'M' THEN 0
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ELSE salary
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END AS bonus
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FROM
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employees;
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```
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MySQL
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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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employee_id,
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IF(
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salary
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) AS bonus
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FROM
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employees;
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```
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```sql
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# Write your MySQL query statement below
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SELECT
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employee_id,
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CASE
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WHEN (employee_id % 2 = 1 AND NAME NOT LIKE "M%") THEN salary
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ELSE 0
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END AS bonus
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FROM Employees
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ORDER BY employee_id;
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employees
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ORDER BY 1;
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```
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<!-- tabs:end -->

‎solution/1800-1899/1873.Calculate Special Bonus/Solution.SQL‎

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# Write your MySQL query statement below
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SELECT
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employee_id,
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IF(
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employee_id % 2 = 0
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OR LEFT(name, 1) = 'M',
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0,
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salary
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) AS bonus
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FROM
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employees
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ORDER BY 1;

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