题目链接：https://leetcode-cn.com/problems/second-highest-salary/
题目
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 |
+---------------------+
解答
第一次解答报错。。。
---- oracle ----
/* Write your PL/SQL query statement below */
select Salary as SecondHighestSalary
from
(
select Salary,
rownum as rn
from
(
select Salary
from Employee
order by Salary desc
) b
) c
where c.rn = 2 ---- 执行报错 当数据只有1行时 执行为空 而不是null第二次解答。。。依旧报错
---- oracle ----
/* Write your PL/SQL query statement below */
select Salary as SecondHighestSalary
from
(
select Salary,
row_number() over(order by Salary desc) as rn
from Employee
) c
where c.rn = 2 ---- 依旧报错参考评论之后，再改进。。
解答一
---- oracle ----
/* Write your PL/SQL query statement below */
select max(Salary) as SecondHighestSalary
from Empolyee
where Salary <> (select max(Salary) from Employee) ---- 672ms解答二
修改第一次出错的版本，添加判断后再次尝试。。。
---- oracle ----
select Salary as SecondHighestSalary
from
(
select Salary
from
(
select Salary,
rownum as rn
from
(
select distinct(Salary)
from Employee
order by Salary desc
) b
) c
where c.rn = 2
union all
select null from dual
)
where rownum = 1 -- 未针对薪水进行去重操作 增加distinct
-- 不添加distinct还执行报错 添加之后通过
---- 861ms解答三
好久没用过MySQL，用法都忘光了。。
使用子查询和LIMIT子句
---- MySQL ----
select
(
select distinct Salary
from Employee
order by Salary desc
limit 1 offset 1
) as SecondHighestSalary; ---- 112ms 好快解答四
为了解决NULL的问题，可以使用IFNULL函数
---- MySQL ----
select
ifnull(
(
select distinct Salary
from Employee
order by Salary desc
limit 1 offset 1
),
NULL) as SecondHighestSalary ---- 108ms思考
去重、排序、获取第二个、返回为NULL时设置返回NULL
注意
oracle中rownum的使用必须要包含第一条记录，也就是类似rownum <= 10，所以不能使用rownum = 2提取第2行数据，必须利用嵌套查询实现。
group by的速度比distinct速度要快。