一、題目描述

獲取所有員工當前的(dept_manager.to_date='9999-01-01')manager，如果員工是manager的話不顯示(也就是如果當前的manager是自己的話結果不顯示)。輸出結果第一列給出當前員工的emp_no,第二列給出其manager對應的emp_no。
CREATE TABLE `dept_emp` (
`emp_no` int(11) NOT NULL, -- '所有的員工編號'
`dept_no` char(4) NOT NULL, --'部門編號'
`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, --'部門編號'
`emp_no` int(11) NOT NULL,-- '經理編號'
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`dept_no`)); 如插入： 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_emp VALUES(10003,'d004','1995-12-03','9999-01-01');
INSERT INTO dept_emp VALUES(10004,'d004','1986-12-01','9999-01-01');
INSERT INTO dept_emp VALUES(10005,'d003','1989-09-12','9999-01-01');
INSERT INTO dept_emp VALUES(10006,'d002','1990-08-05','9999-01-01');
INSERT INTO dept_emp VALUES(10007,'d005','1989-02-10','9999-01-01');
INSERT INTO dept_emp VALUES(10008,'d005','1998-03-11','2000-07-31');
INSERT INTO dept_emp VALUES(10009,'d006','1985-02-18','9999-01-01');
INSERT INTO dept_emp VALUES(10010,'d005','1996-11-24','2000-06-26');
INSERT INTO dept_emp VALUES(10010,'d006','2000-06-26','9999-01-01');
INSERT INTO dept_manager VALUES('d001',10002,'1996-08-03','9999-01-01');
INSERT INTO dept_manager VALUES('d002',10006,'1990-08-05','9999-01-01');
INSERT INTO dept_manager VALUES('d003',10005,'1989-09-12','9999-01-01');
INSERT INTO dept_manager VALUES('d004',10004,'1986-12-01','9999-01-01');
INSERT INTO dept_manager VALUES('d005',10010,'1996-11-24','2000-06-26');
INSERT INTO dept_manager VALUES('d006',10010,'2000-06-26','9999-01-01');

答案

SELECT dept_emp.emp_no,dept_manager.emp_no manager_no FROM dept_emp 
inner join dept_manager on dept_emp.dept_no=dept_manager.dept_no
WHERE dept_manager.to_date='9999-01-01' 
and not dept_emp.emp_no=dept_manager.emp_no

二、題目描述

獲取所有部門中當前(dept_emp.to_date = '9999-01-01')員工當前(salaries.to_date='9999-01-01')薪水最高的相關資訊，給出dept_no, emp_no以及其對應的salary，按照部門編號升序排列。
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 `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`)); 如插入： 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_emp VALUES(10003,'d001','1996-08-03','1997-08-03');

INSERT INTO salaries VALUES(10001,90000,'1986-06-26','1987-06-26');
INSERT INTO salaries VALUES(10001,88958,'2002-06-22','9999-01-01');
INSERT INTO salaries VALUES(10002,72527,'1996-08-03','1997-08-03');
INSERT INTO salaries VALUES(10002,72527,'2000-08-02','2001-08-02');
INSERT INTO salaries VALUES(10002,72527,'2001-08-02','9999-01-01');
INSERT INTO salaries VALUES(10003,90000,'1996-08-03','1997-08-03');

答案

select dept_no,salaries.emp_no,max(salary) from salaries
inner join dept_emp on salaries.emp_no=dept_emp.emp_no
where dept_emp.to_date = '9999-01-01' and salaries.to_date='9999-01-01'
group by dept_emp.dept_no