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_nosalary

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_not

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_noemp_nosalary

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