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  • SQL练习题16:统计出当前(titles.to_date='9999-01-01')各个title类型对应的员工当前(salaries.to_date='9999-01-01')薪水对应的平均工资。结果给出title以及平均工资avg。

    统计出当前(titles.to_date='9999-01-01')各个title类型对应的员工当前(salaries.to_date='9999-01-01')薪水对应的平均工资。结果给出title以及平均工资avg。
    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`));
    CREATE TABLE IF NOT EXISTS "titles" (
    `emp_no` int(11) NOT NULL,
    `title` varchar(50) NOT NULL,
    `from_date` date NOT NULL,

    `to_date` date DEFAULT NULL);

    如插入:
    INSERT INTO salaries VALUES(10001,88958,'1986-06-26','9999-01-01');
    INSERT INTO salaries VALUES(10003,43311,'2001-12-01','9999-01-01');
    INSERT INTO salaries VALUES(10004,70698,'1986-12-01','1995-12-01');
    INSERT INTO salaries VALUES(10004,74057,'1995-12-01','9999-01-01');
    INSERT INTO salaries VALUES(10006,43311,'2001-08-02','9999-01-01');
    INSERT INTO salaries VALUES(10007,88070,'2002-02-07','9999-01-01');

    INSERT INTO titles VALUES(10001,'Senior Engineer','1986-06-26','9999-01-01');
    INSERT INTO titles VALUES(10003,'Senior Engineer','2001-12-01','9999-01-01');
    INSERT INTO titles VALUES(10004,'Engineer','1986-12-01','1995-12-01');
    INSERT INTO titles VALUES(10004,'Senior Engineer','1995-12-01','9999-01-01');
    INSERT INTO titles VALUES(10006,'Senior Engineer','2001-08-02','9999-01-01');
    INSERT INTO titles VALUES(10007,'Senior Staff','1996-02-11','9999-01-01');
     
    输出:
    title
    avg
    Senior Engineer
    62409.25
    Senior Staff
    88070.0
     
    程序:
    SELECT a.title,AVG(b.salary) AS avg
    FROM titles AS a, salaries AS b
    where a.emp_no = b.emp_no
    and a.to_date='9999-01-01'
    and b.to_date='9999-01-01'
    GROUP BY a.title
    
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  • 原文地址:https://www.cnblogs.com/zhuozige/p/13715053.html
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