feat: update sql solutions to lc problems: No.0183,1821,1873

.prettierignore

@@ -21,6 +21,7 @@ node_modules/
 /solution/1600-1699/1635.Hopper Company Queries I/Solution.sql
 /solution/1600-1699/1651.Hopper Company Queries III/Solution.sql
 /solution/1700-1799/1767.Find the Subtasks That Did Not Execute/Solution.sql
+/solution/1800-1899/1873.Calculate Special Bonus/Solution.sql
 /solution/2100-2199/2118.Build the Equation/Solution.sql
 /solution/2100-2199/2153.The Number of Passengers in Each Bus II/Solution.sql
 /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

@@ -73,6 +73,14 @@ Orders table:
 
 ## Solutions
 
+**Solution 1: NOT IN**
+
+List all customer IDs of existing orders, and use `NOT IN` to find customers who are not in the list.
+
+**Solution 2: LEFT JOIN**
+
+Use `LEFT JOIN` to join the tables and return the data where `CustomerId` is `NULL`.
+
 <!-- tabs:start -->
 
 ### **SQL**

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

@@ -65,6 +65,10 @@ Customers
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：WHERE 子句**
+
+我们可以直接使用 `WHERE` 子句来筛选出 `year` 为 `2021` 且 `revenue` 大于 $0$ 的客户。
+
 <!-- tabs:start -->
 
 ### **SQL**

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

@@ -61,6 +61,10 @@ Thus only customers 1 and 4 have positive revenue in the year 2021.
 
 ## Solutions
 
+**Solution 1: WHERE Clause**
+
+We can directly use the `WHERE` clause to filter out the customers whose `year` is `2021` and `revenue` is greater than $0$.
+
 <!-- tabs:start -->
 
 ### **SQL**

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

@@ -63,27 +63,18 @@ Employees 表:
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：IF 语句 + ORDER BY 子句**
+
+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。
+
 <!-- tabs:start -->
 
 ### **SQL**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```sql
-SELECT
-    employee_id,
-    CASE
-        WHEN employee_id % 2 = 0
-        OR LEFT(name, 1) = 'M' THEN 0
-        ELSE salary
-    END AS bonus
-FROM
-    employees;
-```
-
-MySQL
-
-```sql
+# Write your MySQL query statement below
 SELECT
     employee_id,
     IF(
@@ -93,19 +84,8 @@ SELECT
         salary
     ) AS bonus
 FROM
-    employees;
-```
-
-```sql
-# Write your MySQL query statement below
-SELECT
-    employee_id,
-    CASE
-        WHEN (employee_id % 2 = 1 AND NAME NOT LIKE "M%") THEN salary
-        ELSE 0
-    END AS bonus
-FROM Employees
-ORDER BY employee_id;
+    employees
+ORDER BY 1;
 ```
 
 <!-- tabs:end -->

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

@@ -59,25 +59,16 @@ The rest of the employees get a 100% bonus.
 
 ## Solutions
 
-<!-- tabs:start -->
+**Solution 1: IF Statement + ORDER BY Clause**
 
-### **SQL**
+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`.
 
-```sql
-SELECT
-    employee_id,
-    CASE
-        WHEN employee_id % 2 = 0
-        OR LEFT(name, 1) = 'M' THEN 0
-        ELSE salary
-    END AS bonus
-FROM
-    employees;
-```
+<!-- tabs:start -->
 
-MySQL
+### **SQL**
 
 ```sql
+# Write your MySQL query statement below
 SELECT
     employee_id,
     IF(
@@ -87,19 +78,8 @@ SELECT
         salary
     ) AS bonus
 FROM
-    employees;
-```
-
-```sql
-# Write your MySQL query statement below
-SELECT
-    employee_id,
-    CASE
-        WHEN (employee_id % 2 = 1 AND NAME NOT LIKE "M%") THEN salary
-        ELSE 0
-    END AS bonus
-FROM Employees
-ORDER BY employee_id;
+    employees
+ORDER BY 1;
 ```
 
 <!-- tabs:end -->

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

@@ -0,0 +1,12 @@
+# Write your MySQL query statement below
+SELECT
+    employee_id,
+    IF(
+        employee_id % 2 = 0
+        OR LEFT(name, 1) = 'M',
+        0,
+        salary
+    ) AS bonus
+FROM
+    employees
+ORDER BY 1;