sql不用order by查找第二的思路
例：查找當前薪水(to_date=‘9999-01-01‘)排名第二多的員工編號emp_no、薪水salary、last_name以及first_name，不準使用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`));
解答思路：
此題方法較多
方法一：
利用兩次MAX，連同select嵌套
主要思想為多層SELECT嵌套與MAX()函數結合 1、先利用MAX()函數找出salaries中當前薪水最高者，即SELECT MAX(salary) FROM salaries WHERE to_date = ‘9999-01-01‘ 2、再利用INNER JOIN連接employees與salaries表，限定條件為【同一員工】e.emp_no = s.emp_no、【當前】s.to_date = ‘9999-01-01‘與【非薪水最高】s.salary NOT IN (SELECT MAX(salary) FROM salaries WHERE to_date = ‘9999-01-01‘) 3、在以上限制條件下找薪水最高者，即為所有員工薪水的次高者SELECT
FROM employees AS e INNER JOIN salaries AS s
ON e.emp_no = s.emp_no
WHERE s.to_date = ‘9999-01-01‘
AND s.salary NOT IN (SELECT MAX(salary) FROM salaries WHERE to_date = ‘9999-01-01‘)
方法二:
利用max選出最大，用where 限定小於最大值的，然後再取最大的那個
select a.emp_no,max(b.salary),a.last_name,a.first_name
from employees a inner join salaries b
on a.emp_no=b.emp_no
where b.salary<(select max(salary) from salaries)
and b.to_date = ‘9999-01-01‘
sql不用order by查找第二的思路