SQL_基础查找练习：limit、distinct、max()、avg()、exists

1、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));

答案1：order by , limit

select * from employees 
order by hire_date desc limit 0,1

答案2：max( )

select * from employees
where hire_date = (select max(hire_date) from employees)

2、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));

答案1：limit a,b

select * from employees where hire_date=
(select hire_date from employees order by hire_date desc limit 2,1)

答案2： distinct

select * from employees
where hire_date=
(select distinct hire_date from employees order by hire_date desc limit 2,1)

3、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));

答案：

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

4、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));

答案：inner join

select e.last_name, e.first_name,d.dept_no from employees as e 
inner join dept_emp as d 
on e.emp_no = d.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，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));

答案：left join

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，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`));

答案：inner join

select e.emp_no,s.salary
from salaries as s
inner join employees as e
on s.emp_no=e.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
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));

答案：count 、group by、having

select emp_no,count(*) as t
from salaries
group by emp_no
having t>15

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));

答案1：distinct

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

答案2：group by

select salary from salaries
where to_date='9999-01-01'
group by salary
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’
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));

答案：

select d.dept_no, d.emp_no, s.salary
from salaries as s
inner join dept_manager as d
on s.emp_no = d.emp_no 
and d.to_date=s.to_date
and d.to_date='9999-01-01' 

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));

答案：not in

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

11、获取所有员工当前的manager，如果当前的manager是自己的话结果不显示，当前表示to_date=’9999-01-01’。
结果第一列给出当前员工的emp_no,第二列给出其manager对应的manager_no。
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 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));

答案：not in

select e.emp_no, m.emp_no as manager_no 
from dept_emp as e , dept_manager as m
where e.dept_no=m.dept_no 
and e.emp_no not in (select emp_no from dept_manager)
and e.to_date='9999-01-01' 
and m.to_date='9999-01-01'

12、获取所有部门中当前员工薪水最高的相关信息，给出dept_no, emp_no以及其对应的salary
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 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));

答案：group by、having、max()

select d.dept_no, d.emp_no, s.salary
from dept_emp as d inner join salaries as s
where d.emp_no=s.emp_no 
and d.to_date=s.to_date
and d.to_date='9999-01-01'
group by d.dept_no having max(s.salary)

13、从titles表获取按照title进行分组，每组个数大于等于2，给出title以及对应的数目t。
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);

答案：count(*)、group by、having

select title,count(*) as t from titles
group by title
having t>=2

14、从titles表获取按照title进行分组，每组个数大于等于2，给出title以及对应的数目t。
注意对于重复的emp_no进行忽略。
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);

答案：

select title ,count(distinct emp_no) as t
from titles
group by title
having t>=2

15、查找employees表所有emp_no为奇数，且last_name不为Mary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));

答案：%、<>

select * from employees
where emp_no%2 == 1
and last_name <> 'Mary'
order by hire_date desc

16、统计出当前各个title类型对应的员工当前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);

答案：avg

select t.title, avg(s.salary) as avg
from titles as t,salaries as s
where t.emp_no=s.emp_no and t.to_date=s.to_date
and t.to_date='9999-01-01'
group by t.title

17、获取当前（to_date=’9999-01-01’）薪水第二多的员工的emp_no以及其对应的薪水salary
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));

答案：

select emp_no ,salary from salaries
where to_date='9999-01-01'
order by salary desc limit 1,1

18、查找当前薪水(to_date=’9999-01-01’)排名第二多的员工编号emp_no、薪水salary、last_name以及first_name，不准使用order by
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));

答案：max()、not in、！=、<>

select e.emp_no, max(s.salary), e.last_name, e.first_name 
from employees as e , salaries as s
where e.emp_no=s.emp_no
and salary !=(select max(salary) from salaries where to_date='9999-01-01')

19、查找员工编号emp_no为10001其自入职以来的薪水salary涨幅值growth
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));

答案：

select
(select salary from salaries where emp_no='10001' order by to_date desc limit 1)-
(select salary from salaries where emp_no='10001' order by to_date limit 1)
as growth

20、查找排除当前最大、最小salary之后的员工的平均工资avg_salary。
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));

答案：

select avg(salary) as avg_salary from salaries
where to_date='9999-01-01'
and salary not in (select max(salary)from salaries)
and salary not in (select min(salary)from salaries)

21、使用含有关键字exists查找未分配具体部门的员工的所有信息。
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));
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));

答案：

select * from employees where not exists
(select emp_no from dept_emp where emp_no=employees.emp_no)

22、