数据库作业MySQL

新增员工表emp和部门表dept

create table dept (dept1 int ,dept_name varchar(11)) charset=utf8;
create table emp (sid int ,name varchar(11),age int,worktime_start date,incoming int,dept2 int) charset=utf8;

 insert into dept values
(101,'财务'),
(102,'销售'),
(103,'IT技术'),
(104,'行政');
 
 insert into emp values
(1789,'张三',35,'1980/1/1',4000,101),
(1674,'李四',32,'1983/4/1',3500,101),
(1776,'王五',24,'1990/7/1',2000,101),
(1568,'赵六',57,'1970/10/11',7500,102),
(1564,'荣七',64,'1963/10/11',8500,102),
(1879,'牛八',55,'1971/10/20',7300,103);
(1668, '钱九', 64, '1963/5/4', 8000, 102);
(1724, '武十', 22, '2023/5/8', 1500, 103);
(1770, '孙二', 65, '1986/8/12', 9500, 101);
(18400, '苟一', 65, '1986/8/12', 1500, 101);

目的
1.找出销售部门中年纪最大的员工的姓名

mysql> select name from emp,dept
    -> where emp.dept2=dept.dept1
    -> and age>all(select age from emp where emp.dept2=102);

2.求财务部门最低工资的员工姓名

mysql> select name from emp,dept
    -> where emp.dept2=dept.dept1
    -> and incoming in (select min(incoming) from emp
    -> group by dept.dept_name
    -> having dept_name='财务');

3.列出每个部门收入总和高于9000的部门名称

mysql> select dept_name from emp left join dept
    -> on emp.dept2=dept.dept1
    -> group by dept_name
    -> having sum(incoming)>9000;

4.求工资在7500到8500元之间，年龄最大的人的姓名及部门

mysql> select name,dept_name from emp,dept
    -> where emp.dept2=dept.dept1
    -> and age=(select max(age) from emp where incoming between 7500 and 8500);

5.找出销售部门收入最低的员工入职时间

mysql>  select worktime_start from emp,dept
    -> where emp.dept2=dept.dept1
    ->  and incoming =(select min(incoming) from emp left join dept on emp.dept2=dept.dept1
    -> where dept_name='销售');

6.财务部门收入超过2000元的员工姓名

mysql> select name from emp left join dept
    -> on emp.dept2=dept.dept1
    -> where incoming>2000 and dept_name='财务';

7.列出每个部门的平均收入及部门名称

mysql> select avg(incoming),dept_name from emp left join dept
    -> on emp.dept2=dept.dept1
    -> group by dept_name;

8.IT技术部入职员工的员工号

mysql> select sid from emp left join dept
    -> on emp.dept2=dept.dept1
    -> where dept_name='IT技术';

9.财务部门的收入总和；

mysql> select sum(incoming) from emp left join dept
    -> on emp.dept2=dept.dept1
    -> where dept_name='财务';

10.找出哪个部门还没有员工入职；

mysql> select distinct dept_name from emp,dept
    -> where dept.dept1 not in(select dept2 from emp);

11.列出部门员工收入大于7000的部门编号，部门名称；

mysql> select dept1,dept_name from emp,dept
    -> where dept.dept1=emp.dept2
    -> and incoming>7000;

12.列出每一个部门的员工总收入及部门名称；

mysql>  select sum,dept_name from dept left join
    -> (select sum(incoming) sum,dept2 from emp group by dept2) e
    -> on dept.dept1=e.dept2;

13.列出每一个部门中年纪最大的员工姓名，部门名称；

mysql> select name,dept_name from dept left join emp
    -> on dept.dept1=emp.dept2
    -> where age in(select max(age) from emp group by dept2);

14.求李四的收入及部门名称

mysql> select incoming,dept_name from emp,dept
    -> where dept.dept1=emp.dept2
    -> and name='李四';

15.列出每个部门中收入最高的员工姓名，部门名称，收入，并按照收入降序

mysql> select name,dept_name,incoming from emp join dept
    -> on dept.dept1=emp.dept2
    -> where(dept2,incoming)
    -> in (select max(incoming),dept2 from emp group by dept2,dept.dept_name)
    ->  order by incoming desc;