MySQL第二高的薪水

Posted 2021-09-01 willem_chen

mysql第二高的薪水

• SQL架构
• 题
• 题解
• • 答1
  • 答2
  • 答3

SQL架构

Create table If Not Exists Employee (Id int, Salary int);
Truncate table Employee
insert into Employee (Id, Salary) values ('1', '100');
insert into Employee (Id, Salary) values ('2', '200');
insert into Employee (Id, Salary) values ('3', '300');

题

编写一个 SQL 查询，获取 Employee 表中第二高的薪水（Salary） 。

+----+--------+
| Id | Salary |
+----+--------+
| 1  | 100    |
| 2  | 200    |
| 3  | 300    |
+----+--------+

例如上述 Employee 表，SQL查询应该返回 200 作为第二高的薪水。如果不存在第二高的薪水，那么查询应返回 null。

+---------------------+
| SecondHighestSalary |
+---------------------+
| 200                 |
+---------------------+

题解

方法一：使用子查询和 LIMIT 子句

算法

将不同的薪资按降序排序，然后使用 LIMIT 子句获得第二高的薪资。

答1

SELECT
    (SELECT DISTINCT
            Salary
        FROM
            Employee
        ORDER BY Salary DESC
        LIMIT 1 OFFSET 1) AS SecondHighestSalary;
        
 +---------------------+
| SecondHighestSalary |
+---------------------+
| 200                 |
+---------------------+

然而，如果没有这样的第二最高工资，这个解决方案将被判断为 “错误答案”，因为本表可能只有一项记录。为了克服这个问题，我们可以将其作为临时表。

答2

方法二：使用 IFNULL 和 LIMIT 子句

解决 “NULL” 问题的另一种方法是使用 “IFNULL” 函数，如下所示。

SELECT
    IFNULL(
      (SELECT DISTINCT Salary
       FROM Employee
       ORDER BY Salary DESC
        LIMIT 1 OFFSET 1),
    NULL) AS SecondHighestSalary;

答3

mysql> select max(Salary) as SecondHighestSalary from Employee where Salary < (select max(Salary) from Employee);
+---------------------+
| SecondHighestSalary |
+---------------------+
|                 200 |
+---------------------+
1 row in set (0.00 sec)

mysql> select distinct(Salary) as SecondHighestSalary  from Employee order by Salary desc limit 1,1;
+---------------------+
| SecondHighestSalary |
+---------------------+
|                 200 |
+---------------------+
1 row in set (0.00 sec)