一、题目描述
查找employees表所有emp_no为奇数，且last_name不为Mary(注意大小写)的员工信息，并按照hire_date逆序排列
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：
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');
答案
select * from employees where last_name!="Mary" and emp_no%2=1 order by hire_date desc
二、题目描述
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,
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');
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');
答案
select title, avg(salary) from salaries inner join titles on salaries.emp_no=titles.emp_no where titles.to_date='9999-01-01' and salaries.to_date='9999-01-01' group by title