*18查找当前薪水排名第二多的员工编号emp_no薪水salarylast_name以及first_name，不准使用order by

Posted 2020-11-25 guoyu1

1、题目描述：

查找当前薪水(to_date=‘9999-01-01‘)排名第二多的员工编号emp_no、薪水salary、last_name以及first_name，不准使用order by
CREATE TABLE `employees` (
`emp_no` int(11) NOT NULL,
`birth_date` date NOT NULL,
`first_name` varchar(14) NOT NULL,
`last_name` varchar(16) NOT NULL,
`gender` char(1) NOT NULL,
`hire_date` date NOT NULL,
PRIMARY KEY (`emp_no`));
CREATE TABLE `salaries` (
`emp_no` int(11) NOT NULL,
`salary` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`from_date`));

输入描述:

无

输出描述:

emp_no	salary	last_name	first_name
10009	94409	Peac	Sumant

2、代码：

（1）解法1：除了最高工资就是它最高，先找到最高工资，然后找到第二高工资

select e.emp_no,max(s.salary),e.last_name,e.first_name from employees e, salaries s
where s.salary < (select max(salary) from salaries)
and s.to_date="9999-01-01" and e.emp_no=s.emp_no;

 

（2）解法2：表内连接求第二高工资

select e.emp_no,s.salary,e.last_name,e.first_name
from employees e join salaries s on e.emp_no=s.emp_no 
where  s.to_date=‘9999-01-01‘ and s.salary = 
(
     select s1.salary
     from salaries s1 join salaries s2 on s1.salary<=s2.salary 
     where s1.to_date=‘9999-01-01‘ and s2.to_date=‘9999-01-01‘
     group by s1.salary having count(distinct s2.salary)=2
 );