mysql数据库面试题（2）

 

题目描述

6.查找所有员工入职时候的薪水情况，给出emp_no以及salary， 并按照emp_no进行逆序
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`));
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`));

 

输入

select emp_no,salary from salaries
group by emp_no having(min(from_date))
order by emp_no desc

 

p_no	salary
10011	25828
省略	省略
10001	60117

 

 

 

 

 

 

 

7.查找薪水涨幅超过15次的员工号emp_no以及其对应的涨幅次数t
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`));

 

输入

select emp_no, count(emp_no)
from salaries
group by emp_no
having count(emp_no)>15;

 

emp_no	t
10001	17
10004	16
10009	18

 

 

 

 

 

 

 

 

 

8.找出所有员工当前(to_date='9999-01-01')具体的薪水salary情况，对于相同的薪水只显示一次,并按照逆序显示
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`));

 

输入

select salary from salaries
where to_date='9999-01-01'
group by salary
order by salary desc;

输出

salary

94692

94409

88958

88070

74057

72527

59755

43311

25828

 

 

 

 

 

 

 

 

 

 

 

 

 

9.获取所有部门当前manager的当前薪水情况，给出dept_no, emp_no以及salary，当前表示to_date='9999-01-01'
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 `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`));

 

输入

select d.dept_no,d.emp_no,s.salary
from dept_manager d ,salaries s
where d.emp_no = s.emp_no
and d.to_date = '9999-01-01'
and s.to_date = '9999-01-01';

 

输出

 

dept_no	emp_no	salary
d001	10002	72527
d004	10004	74057
d003	10005	94692
d002	10006	43311
d006	10010	94409

 

 

 

 

 

 

 

 

 

10.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`));

输入

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

输出

 

emp_no

10001

10003

10007

10008

10009

10011