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(kù)
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(yǔ)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(yǔ)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(jiǎn)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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