SQL练习10：获取所有非manager的员工emp_no

题目链接：牛客网

题目描述
获取所有非manager的员工emp_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`));

如插入为:
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');

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','1985-11-21');
INSERT INTO employees VALUES(10003,'1959-12-03','Parto','Bamford','M','1986-08-28');
INSERT INTO employees VALUES(10004,'1954-05-01','Chirstian','Koblick','M','1986-12-01');
INSERT INTO employees VALUES(10005,'1955-01-21','Kyoichi','Maliniak','M','1989-09-12');
INSERT INTO employees VALUES(10006,'1953-04-20','Anneke','Preusig','F','1989-06-02');
INSERT INTO employees VALUES(10007,'1957-05-23','Tzvetan','Zielinski','F','1989-02-10');
INSERT INTO employees VALUES(10008,'1958-02-19','Saniya','Kalloufi','M','1994-09-15');
INSERT INTO employees VALUES(10009,'1952-04-19','Sumant','Peac','F','1985-02-18');
INSERT INTO employees VALUES(10010,'1963-06-01','Duangkaew','Piveteau','F','1989-08-24');
INSERT INTO employees VALUES(10011,'1953-11-07','Mary','Sluis','F','1990-01-22');

解法一
查询的要求是不在dept_manager表中的emp_no，使用not in + 子查询。MYSQL官方文档中提到in使用时效率不高，推荐使用exists或join。

SELECT emp_no 
FROM employees 
WHERE emp_no NOT IN(SELECT DISTINCT emp_no
                   FROM dept_manager)

解法二
使用left join来连接两个表，再筛选非manager的员工。

SELECT e.emp_no
FROM employees e LEFT JOIN dept_manager d
ON e.emp_no = d.emp_no 
WHERE dept_no IS NULL

解法三
使用not exists + 子查询的方式。

SELECT emp_no
FROM employees e
WHERE NOT EXISTS(SELECT DISTINCT d.emp_no 
                FROM dept_manager d
                WHERE e.emp_no = d.emp_no)