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  • SQL练习题10:获取所有非manager的员工emp_no

    题目:

    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');

    程序:

    SELECT a.emp_no
    FROM employees AS a LEFT JOIN dept_manager AS b
    ON a.emp_no = b.emp_no
    WHERE b.emp_no IS NULL
    
     
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  • 原文地址:https://www.cnblogs.com/zhuozige/p/13688173.html
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