MySQL不用order by排名 MySQL不使用order by实现排名的三种思路总结

时间:2021-06-02 CodePhage 人气:0

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假定业务：

查看在职员工的薪资的第二名的员工信息

创建数据库

drop database if exists emps;
create database emps;
use emps;

create table employees(
    empId int primary key,-- 员工编号
    gender char(1) NOT NULL, -- 员工性别
	hire_date date NOT NULL -- 员工入职时间
   	);
create table salaries(
    empId int primary key, 
    salary double -- 员工薪资
    );
    
INSERT INTO employees VALUES(10001,'M','1986-06-26');
INSERT INTO employees VALUES(10002,'F','1985-11-21');
INSERT INTO employees VALUES(10003,'M','1986-08-28');
INSERT INTO employees VALUES(10004,'M','1986-12-01');
INSERT INTO salaries VALUES(10001,88958);
INSERT INTO salaries VALUES(10002,72527);
INSERT INTO salaries VALUES(10003,43311);
INSERT INTO salaries VALUES(10004,74057);

题解思路

1、(基础解法)

先查出salaries表中最高薪资,再以此为条件查出第二高的工资

查询语句如下：

select
	E.empId,E.gender,E.hire_date,S.salary
from
	employees E join salaries S 
on 
	E.empId = S.empId
where	
	S.salary=
	(
    select max(salary)from salaries 
    where 
        salary<
        (select max(salary) from salaries)
    );
-- ---------------查询结果------------ --
+-------+--------+------------+--------+
| empId | gender | hire_date  | salary |
+-------+--------+------------+--------+
| 10004 | M      | 1986-12-01 |  74057 |
+-------+--------+------------+--------+

2、(自联结查询)

先对salaries进行自联结查询，当s1<=s2链接并以s1.salary分组，此时count的值，即薪资比他高的人数，用having筛选count=2 的人，就可以得到第二高的薪资了；

查询语句如下：

select
	E.empId,E.gender,E.hire_date,S.salary
from
	employees E join salaries S 
on 
	E.empId = S.empId
where S.salary=
	(
    select 
        s1.salary
    from 
        salaries s1 join salaries s2 
    on 
        s1.salary <= s2.salary
    group by 
        s1.salary              
  	having
  	 count(distinct s2.salary) = 2
    );
-- ---------------查询结果------------ --
+-------+--------+------------+--------+
| empId | gender | hire_date  | salary |
+-------+--------+------------+--------+
| 10004 | M      | 1986-12-01 |  74057 |
+-------+--------+------------+--------+

3、(自联结查询优化版)

原理和2相同，但是代码精简了很多，上面两种是为了引出最后这种方法，在很多时候group by和order by都有其局限性，对于俺们初学者掌握这种实用性较广的思路，还是很有意义的。

select
	E.empId,E.gender,E.hire_date,S.salary
from
	employees E join salaries S 
on
    S.empId =E.empId
where
    (select count(1) from salaries where salary>=S.salary)=2；
-- ---------------查询结果------------ --
+-------+--------+------------+--------+
| empId | gender | hire_date  | salary |
+-------+--------+------------+--------+
| 10004 | M      | 1986-12-01 |  74057 |
+-------+--------+------------+--------+

初浅总结，如有错误，还望指正。

总结

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