数据库第五次作业----多表查询
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);
1.找出销售部门中年纪最大的员工的姓名
2.求财务部门最低工资的员工姓名
3.列出每个部门收入总和高于9000的部门名称
4.求工资在7500到8500元之间，年龄最大的人的姓名及部门
5.找出销售部门收入最低的员工入职时间
6.财务部门收入超过2000元的员工姓名
7.列出每个部门的平均收入及部门名称
8.IT技术部入职员工的员工号
9.财务部门的收入总和；
10.找出哪个部门还没有员工入职；
11.列出部门员工收入大于7000的部门编号，部门名称；
12.列出每一个部门的员工总收入及部门名称；
13.列出每一个部门中年纪最大的员工姓名，部门名称；
14.求李四的收入及部门名称
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
1、mysql> select name from emp,dept-> where emp.dept2=dept.dept1-> and age>all(select age from emp where emp.dept2=102);
2、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、mysql> select dept_name from emp left join dept-> on emp.dept2=dept.dept1-> group by dept_name-> having sum(incoming)>9000;
4、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、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、mysql> select name from emp left join dept-> on emp.dept2=dept.dept1-> where incoming>2000 and dept_name='财务';
7、mysql> select avg(incoming),dept_name from emp left join dept-> on emp.dept2=dept.dept1-> group by dept_name;
8、mysql> select sid from emp left join dept-> on emp.dept2=dept.dept1-> where dept_name='IT技术';
9、mysql> select sum(incoming) from emp left join dept-> on emp.dept2=dept.dept1-> where dept_name='财务';
10、mysql> select distinct dept_name from emp,dept-> where dept.dept1 not in(select dept2 from emp);
11、mysql> select dept1,dept_name from emp,dept-> where dept.dept1=emp.dept2-> and incoming>7000;
12、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、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、mysql> select incoming,dept_name from emp,dept-> where dept.dept1=emp.dept2-> and name='李四';
15、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;
