MySQL不使用order by實現(xiàn)排名的三種思路總結
假定業(yè)務:
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 gender char(1) NOT NULL, -- 員工性別 hire_date date NOT NULL -- 員工入職時間 ); 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) ); -- ---------------查詢結果------------ -- +-------+--------+------------+--------+ | empId | gender | hire_date | salary | +-------+--------+------------+--------+ | 10004 | M | 1986-12-01 | 74057 | +-------+--------+------------+--------+
2、(自聯(lián)結查詢)
先對salaries進行自聯(lián)結查詢,當s1<=s2鏈接并以s1.salary分組,此時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 ); -- ---------------查詢結果------------ -- +-------+--------+------------+--------+ | empId | gender | hire_date | salary | +-------+--------+------------+--------+ | 10004 | M | 1986-12-01 | 74057 | +-------+--------+------------+--------+
3、(自聯(lián)結查詢優(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,在很多時候group by和order by都有其局限性,對于俺們初學者掌握這種實用性較廣的思路,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; -- ---------------查詢結果------------ -- +-------+--------+------------+--------+ | empId | gender | hire_date | salary | +-------+--------+------------+--------+ | 10004 | M | 1986-12-01 | 74057 | +-------+--------+------------+--------+
初淺總結,如有錯誤,還望指正。
總結
到此這篇關于MySQL不使用order by實現(xiàn)排名的三種思路的文章就介紹到這了,更多相關MySQL不用order by排名內(nèi)容請搜索腳本之家以前的文章或繼續(xù)瀏覽下面的相關文章希望大家以后多多支持腳本之家!