24、获取所有非manager员工当前的薪水情况

1、题目描述

获取所有非manager员工当前的薪水情况，给出dept_no、emp_no以及salary ，当前表示to_date='9999-01-01'
CREATE TABLE `dept_emp` (
`emp_no` int(11) NOT NULL,
`dept_no` char(4) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`dept_no`));
CREATE TABLE `dept_manager` (
`dept_no` char(4) NOT NULL,
`emp_no` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`dept_no`));
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`));

输入描述:

无

输出描述:

dept_no	emp_no	salary
d001	10001	88958
d004	10003	43311
d005	10007	88070
d006	10009	95409

2、代码

select t1.dept_no,t1.emp_no,s.salary
from (
select d.dept_no,d.emp_no,d.from_date,d.to_date
from dept_emp d
where d.emp_no not in
(select m.emp_no 
 from dept_manager m 
 where m.to_date='9999-01-01')
) t1  inner join salaries s 
on t1.emp_no=s.emp_no and t1.to_date='9999-01-01' and s.to_date='9999-01-01';