update README.md

Author: int32bit

database/SecondHighestSalary/README.md

@@ -0,0 +1,47 @@
+# Second Highest Salary
+
+## Problem
+Write a SQL query to get the second highest salary from the Employee table.
+```
++----+--------+
+| Id | Salary |
++----+--------+
+| 1  | 100    |
+| 2  | 200    |
+| 3  | 300    |
++----+--------+
+```
+For example, given the above Employee table, the second highest salary is 200. If there is no second highest salary, then the query should return null.
+
+## Solution 1
+
+可以使用子查询的方式,即
+
+```sql
+select max(Salary) as SecondHighestSalary from Employee where Salary < (select max(Salary) from Employee);
+```
+
+这种方式更通用，但使用了子查询（不相关子查询），效率会低点。
+
+## Solution 2
+
+使用mysql特有`LIMIT` 和 `IFNULL`语法，下面简单介绍下:
+
+### LIMIT
+
+LIMIT可以限制select语句的输出行数，可以有一个参数或者两个参数。
+
+如果只有一个参数，则表示限制的最大行数，如果不足最大行数，则输出所有行。
+
+如果两个参数，前一个参数表示与第一行的偏移量，偏移量为0表示从第一行开始输出，第二个参数表示最大行数。如果偏移量超出行数范围，输出为空
+
+
+### IFNULL
+
+两个参数，如果第一个参数为空集合，则输出第二个参数。
+
+综合以上两个函数，可以实现为:
+
+```sql
+select ifnull((select distinct Salary from Employee order by Salary desc limit 1,1), null) as SecondHighestSalary;
+```

database/SecondHighestSalary/solve.sql

@@ -5,4 +5,7 @@ insert into Employee values(1, 100);
 insert into Employee values(2, 200);
 insert into Employee values(3, 300);
 */
-select max(Salary) from Employee where Salary < (select max(Salary) from Employee);
+/*
+select max(Salary) as SecondHighestSalary from Employee where Salary < (select max(Salary) from Employee);
+*/
+select ifnull((select distinct Salary from Employee order by Salary desc limit 1,1), null) as SecondHighestSalary;