数据库SQL实战题目详解（全61题）---（1-20）部分

题目来源：牛客网–《数据库SQL实战》
https://www.nowcoder.com/ta/sql?page=0
题目答案为博主自写已通过运行，题目难度近似于阶梯上升，可根据自身情况分部分作答。

其他题目链接：
21-40题链接：
https://blog.csdn.net/weixin_41744624/article/details/104412955
41-60题链接：
https://blog.csdn.net/weixin_41744624/article/details/104413757
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另附可以使用的在线编程工具
sqlfiddle在线工具
网页版方便个人在闲暇时间的小题目练习
也可以从牛客网参考本答案直接作答
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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));

select A.* from employees A 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));

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));

select s.* ,d.dept_no
from salaries as s 
join dept_manager as d 
on s.emp_no=d.emp_no
where 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));

select A.last_name,A.first_name ,B.dept_no from employees A
join dept_emp B 
on A.emp_no =B.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));

select A.last_name, A.first_name,B.dept_no from employees A
left join dept_emp B on A.emp_no = B.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));

select A.emp_no , B.salary from employees A
join salaries B
on A.emp_no = B.emp_no and A.hire_date = B.from_date
order by A.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));

select emp_no ,count(salary) as t from salaries
group by emp_no
having count(salary)>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));

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’
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));

select A.dept_no , A.emp_no,B.salary from dept_manager A
join salaries B
on A.emp_no=B.emp_no 
where A.to_date='9999-01-01' and  B.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));

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));

select A.emp_no,B.emp_no as manager_no from dept_emp A
left join dept_manager B 
on A.dept_no = B.dept_no
where A.emp_no != B.emp_no and A.to_date='9999-01-01' and B.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));

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

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);

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

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

select * from employees 
where (emp_no%2 !=0) 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（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,
to_date date DEFAULT NULL);

select A.title ,avg(B.salary)as avg from titles A
join salaries B
on A.emp_no = B.emp_no and A.to_date='9999-01-01'  and B.to_date='9999-01-01'
group by A.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));

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));

elect A.emp_no , max(B.salary) , A.last_name, A.first_name from employees A
join salaries B
on A.emp_no = B.emp_no
where B.to_date='9999-01-01'
and B.salary <(select max(salary) from salaries)

19、

查找所有员工的last_name和first_name以及对应的dept_name，也包括暂时没有分配部门的员工
CREATE TABLE departments (
dept_no char(4) NOT NULL,
dept_name varchar(40) NOT NULL,
PRIMARY KEY (dept_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 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));

select A.last_name , A.first_name, C.dept_name from employees A
left join dept_emp B on A.emp_no=B.emp_no
left join departments C on B.dept_no=C.dept_no

20、

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));

select (max(salary)-min(salary)) as growth from salaries
where emp_no = '10001'