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
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
|
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：
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
|
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：
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
|
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。
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
|
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
到此这篇关于MySQL不使用order by实现排名的三种思路的文章就介绍到这了,更多相关MySQL不用order by排名内容请搜索服务器之家以前的文章或继续浏览下面的相关文章希望大家以后多多支持服务器之家！
原文链接：https://blog.csdn.net/Tinwares/article/details/117425956