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获取所有非manager员工当前的薪水情况

[编程题]获取所有非manager员工当前的薪水情况

热度指数：339640 时间限制：C/C++ 1秒，其他语言2秒空间限制：C/C++ 32M，其他语言64M
算法知识视频讲解

有一个员工表employees简况如下:

emp_no	birth_date	first_name	last_name	gender	hire_date
10001	1953-09-02	Georgi	Facello	M	1986-06-26
10002	1964-06-02	Bezalel	Simmel	F	1996-08-03

有一个，部门员工关系表dept_emp简况如下:

emp_no	dept_no	from_date	to_date
10001	d001	1986-06-26	9999-01-01
10002	d001	1996-08-03	9999-01-01

有一个部门经理表dept_manager简况如下:

dept_no	emp_no	from_date	to_date
d001	10002	1996-08-03	9999-01-01

有一个薪水表salaries简况如下:

emp_no	salary	from_date	to_date
10001	88958	1986-06-26	9999-01-01
10002	72527	1996-08-03	9999-01-01

获取所有非manager员工薪水情况，给出dept_no、emp_no以及salary，以上例子输出:

dept_no	emp_no	salary
d001	10001	88958

示例1

输入

drop table if exists  `dept_emp` ; 
drop table if exists  `dept_manager` ; 
drop table if exists  `employees` ; 
drop table if exists  `salaries` ; 
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`));
INSERT INTO dept_emp VALUES(10001,'d001','1986-06-26','9999-01-01');
INSERT INTO dept_emp VALUES(10002,'d001','1996-08-03','9999-01-01');
INSERT INTO dept_manager VALUES('d001',10002,'1996-08-03','9999-01-01');
INSERT INTO employees VALUES(10001,'1953-09-02','Georgi','Facello','M','1986-06-26');
INSERT INTO employees VALUES(10002,'1964-06-02','Bezalel','Simmel','F','1996-08-03');
INSERT INTO salaries VALUES(10001,88958,'1986-06-26','9999-01-01');
INSERT INTO salaries VALUES(10002,72527,'1996-08-03','9999-01-01');