获取所有非manager的员工emp

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	1985-11-21
10003	1959-12-03	Parto	Bamford	M	1986-08-28

有一个部门领导表dept_manager简况如下:

dept_no	emp_no	from_date	to_date
d001	10002	1996-08-03	9999-01-01
d002	10003	1990-08-05	9999-01-01

请你找出所有非部门领导的员工emp_no，以上例子输出:

emp_no

10001

示例1

输入:

drop table if exists  `dept_manager` ; 
drop table if exists  `employees` ; 
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`));
INSERT INTO dept_manager VALUES('d001',10002,'1996-08-03','9999-01-01');
INSERT INTO dept_manager VALUES('d002',10003,'1990-08-05','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');

输出:

原站题解

上次编辑到这里，代码来自缓存点击恢复默认模板

Sqlite 解法, 执行用时: 9ms, 内存消耗: 3364KB, 提交时间: 2021-09-09

select emp_no
from employees
where emp_no not in(
select emp_no
from dept_manager);

Sqlite 解法, 执行用时: 10ms, 内存消耗: 3188KB, 提交时间: 2020-07-06

select emp_no
from employees where emp_no 
not in (select emp_no from dept_manager)

Sqlite 解法, 执行用时: 10ms, 内存消耗: 3196KB, 提交时间: 2020-10-30

select emp_no from employees 
where emp_no not in (select emp_no from dept_manager)

Sqlite 解法, 执行用时: 10ms, 内存消耗: 3196KB, 提交时间: 2020-07-07

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

Sqlite 解法, 执行用时: 10ms, 内存消耗: 3232KB, 提交时间: 2021-08-03

select emp_no from employees where emp_no not in 
(select emp_no from dept_manager)

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