牛客网—数据库SQL实战(1)

1. 查找最晚入职员工的所有信息

题目描述

查找最晚入职员工的所有信息
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`));

输入描述:

输出描述:

emp_no birth_date first_name last_name gender hire_date
10008 1958-02-19 Saniya Kalloufi M 1994-09-15
select *
from employees
order by hire_date desc limit 0,1

2. 查找入职员工时间排名倒数第三的员工所有信息

题目描述

查找入职员工时间排名倒数第三的员工所有信息
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`));

输入描述:

输出描述:

emp_no birth_date first_name last_name gender hire_date
10005 1955-01-21 Kyoichi Maliniak M 1989-09-12
select * 
from employees 
order by hire_date desc limit 2,1

3. 查找当前薪水详情以及部门编号dept_no

题目描述

查找各个部门当前(to_date='9999-01-01')领导当前薪水详情以及其对应部门编号dept_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 `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`));

输入描述:

输出描述:

emp_no salary from_date to_date dept_no
10002 72527 2001-08-02 9999-01-01 d001
10004 74057 2001-11-27 9999-01-01 d004
10005 94692 2001-09-09 9999-01-01 d003
10006 43311 2001-08-02 9999-01-01 d002
10010 94409 2001-11-23 9999-01-01 d006
select s.*, d.dept_no
from salaries as s, dept_manager as d
where s.to_date = '9999-01-01'
    and d.to_date = '9999-01-01'
    and s.emp_no = d.emp_no

4. 查找所有已经分配部门的员工的last_name和first_name

题目描述

查找所有已经分配部门的员工的last_name和first_name
CREATE TABLE `dept_emp` (
`emp_no` int(11) NOT NULL,
`dept_no` char(4) 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`));

输入描述:

输出描述:

last_name first_name dept_no
Facello Georgi d001
省略 省略 省略
Piveteau Duangkaew d006
select e.last_name, e.first_name, d.dept_no
from employees as e, dept_emp as d
where d.emp_no = e.emp_no

5. 查找所有员工的last_name和first_name以及对应部门编号dept_no

题目描述

查找所有员工的last_name和first_name以及对应部门编号dept_no,也包括展示没有分配具体部门的员工
CREATE TABLE `dept_emp` (
`emp_no` int(11) NOT NULL,
`dept_no` char(4) 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`));

输入描述:

输出描述:

last_name first_name dept_no
Facello Georgi d001
省略 省略 省略
Sluis Mary NULL(在sqlite中此处为空,MySQL为NULL)
select e.last_name,e.first_name,d.dept_no
from employees as e
    left join dept_emp as d
    on e.emp_no = d.emp_no

6. 查找所有员工入职时候的薪水情况

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`));

输入描述:

输出描述:

emp_no salary
10011 25828
省略 省略
10001 60117
select e.emp_no, s.salary
from employees e, salaries s
where e.emp_no = s.emp_no
    and e.hire_date = s.from_date
order by e.emp_no desc

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

题目描述

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`));

输入描述:

输出描述:

emp_no t
10001 17
10004 16
10009 18
select emp_no, count(emp_no) as t
from salaries
group by emp_no
having t > 15

8. 找出所有员工当前薪水salary情况

题目描述

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`));

输入描述:

输出描述:

salary
94692
94409
88958
88070
74057
72527
59755
43311
25828
select distinct salary
from salaries
where to_date = "9999-01-01"
order by salary desc

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'

题目描述

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`));

输入描述:

输出描述:

dept_no emp_no salary
d001 10002 72527
d004 10004 74057
d003 10005 94692
d002 10006 43311
d006 10010 94409
select d.dept_no, d.emp_no, s.salary 
from  salaries as s
    join dept_manager as d
    on d.emp_no = s.emp_no 
where  d.to_date = '9999-01-01' 
    and  s.to_date = '9999-01-01'
select d.dept_no, d.emp_no, s.salary 
from  salaries as s, dept_manager as d
where  d.to_date = '9999-01-01' 
    and  s.to_date = '9999-01-01'
    and d.emp_no = s.emp_no 

10. 获取所有非manager的员工emp_no

题目描述

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`));

输入描述:

输出描述:

emp_no
10001
10003
10007
10008
10009
10011
select emp_no
from employees
where emp_no not in (select emp_no from dept_manager)

 

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