当前位置:Gxlcms > 数据库问题 > mysql内连接的复杂应用

mysql内连接的复杂应用

时间:2021-07-01 10:21:17 帮助过:17人阅读

查找所有员工自入职以来的薪水涨幅情况,给出员工编号emp_noy以及其对应的薪水涨幅growth,并按照growth进行升序
CREATE TABLE `employees` (
`emp_no` int(11) NOT NULL,
`birth_date` date NOT NULL,
`first_name` varchar(14) NOT NULL,
`last_name` varchar(16) NOT NULL,
`gender` char(1) NOT NULL,
`hire_date` date NOT NULL,
PRIMARY KEY (`emp_no`));
CREATE TABLE `salaries` (
`emp_no` int(11) NOT NULL,
`salary` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`from_date`));

 

用内连接实现:

select a.emp_no, (b.salary - c.salary) as growth
from
    employees as a
    inner join salaries as b
    on a.emp_no = b.emp_no and b.to_date = ‘9999-01-01‘
    inner join salaries as c
    on a.emp_no = c.emp_no and a.hire_date = c.from_date
order by growth asc

mysql内连接的复杂应用

标签:null   primary   var   create   char   mysql   编号   入职   tab   

人气教程排行