MySQL不使用order by實(shí)現(xiàn)排名的三種思路總結(jié)
假定業(yè)務(wù):
查看在職員工的薪資的第二名的員工信息
創(chuàng)建數(shù)據(jù)庫
drop database if exists emps;
create database emps;
use emps;
create table employees(
empId int primary key,-- 員工編號(hào)
gender char(1) NOT NULL, -- 員工性別
hire_date date NOT NULL -- 員工入職時(shí)間
);
create table salaries(
empId int primary key,
salary double -- 員工薪資
);
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);
題解思路
1、(基礎(chǔ)解法)
先查出salaries表中最高薪資,再以此為條件查出第二高的工資
查詢語句如下:
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)
);
-- ---------------查詢結(jié)果------------ --
+-------+--------+------------+--------+
| empId | gender | hire_date | salary |
+-------+--------+------------+--------+
| 10004 | M | 1986-12-01 | 74057 |
+-------+--------+------------+--------+
2、(自聯(lián)結(jié)查詢)
先對(duì)salaries進(jìn)行自聯(lián)結(jié)查詢,當(dāng)s1<=s2鏈接并以s1.salary分組,此時(shí)count的值,即薪資比他高的人數(shù),用having篩選count=2 的人,就可以得到第二高的薪資了;
查詢語句如下:
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
);
-- ---------------查詢結(jié)果------------ --
+-------+--------+------------+--------+
| empId | gender | hire_date | salary |
+-------+--------+------------+--------+
| 10004 | M | 1986-12-01 | 74057 |
+-------+--------+------------+--------+
3、(自聯(lián)結(jié)查詢優(yōu)化版)
原理和2相同,但是代碼精簡了很多,上面兩種是為了引出最后這種方法,在很多時(shí)候group by和order by都有其局限性,對(duì)于俺們初學(xué)者掌握這種實(shí)用性較廣的思路,還是很有意義的。
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;
-- ---------------查詢結(jié)果------------ --
+-------+--------+------------+--------+
| empId | gender | hire_date | salary |
+-------+--------+------------+--------+
| 10004 | M | 1986-12-01 | 74057 |
+-------+--------+------------+--------+
初淺總結(jié),如有錯(cuò)誤,還望指正。
總結(jié)
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