数据库多表连接查询练习

数据库多表连接查询练习


一、创建数据库dept、emp

二、插入数据

1.找出销售部门中年纪最大的员工的姓名

mysql> select name,age from emp 
inner join dept on dept.dept1=emp.dept2 
where dept_name='销售' 
order by age desc limit 1;

2.求财务部门最低工资的员工姓名

mysql> select name from emp 
where incoming=(select min(incoming) from emp);

3.列出每个部门收入总和高于9000的部门名称

mysql> select dept_name as 部门名称,sum(incoming) as 总工资 from emp 
inner join dept on emp.dept2=dept.dept1 
group by dept_name having 总工资>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 from emp 
where incoming in (7500,8500) 
order by age desc limit 1;

5.找出销售部门收入最低的员工入职时间

mysql> select worktime_start from emp 
inner join dept on emp.dept2=dept.dept1 
where dept_name='销售' 
order by incoming limit 1;

6.财务部门收入超过2000元的员工姓名

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

7.列出每个部门的平均收入及部门名称

mysql> select dept_name as 部门名称, avg(incoming) as 平均工资 from emp
    -> inner 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
    -> inner 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) as 收入总和 from emp
    -> inner join dept on emp.dept2=dept.dept1
    -> where dept_name='财务';

10.先按部门号大小排序，再依据入职时间由早到晚排序员工信息表

mysql> select * from emp order by dept2 desc,worktime_start;

11.找出哪个部门还没有员工入职；

mysql> select dept_name from emp right join dept on emp.dept2=dept.dept1 where sid is null;

12.列出部门员工收入大于7000的部门编号，部门名称；

mysql> select dept2 as 部门编号,dept_name as 部门名称 from emp
    -> inner join dept on emp.dept2=dept.dept1
    -> where incoming>7000;

13.列出每一个部门的员工总收入及部门名称；

mysql> select dept_name as 部门名称,sum(incoming) as 总收入 from emp                                            -> inner join dept on emp.dept2=dept.dept1                                                                  -> group by dept_name;

14.列出每一个部门中年纪最大的员工姓名，部门名称；

mysql> select emp.name as 姓名, dept.dept_name as 部门名称 from dept
    -> inner join (select max(age) as age,dept2 from emp group by dept2) as em on dept.dept1=em.dept2           
    -> inner join emp on dept.dept1=emp.dept2 and em.age=emp.age;

15.求李四的收入及部门名称

mysql> select incoming as 收入,dept_name from emp
    -> inner join dept on emp.dept2=dept.dept1
    -> where name='李四';

16.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 as 姓名,incoming as 收入,dept_name as 部门名称 from emp
    -> inner join dept on emp.dept2=dept.dept1                                      -> inner join
    -> (select max(incoming) as max_incom, dept2 from emp group by dept2)
    -> as em
    -> on emp.dept2=em.dept2 and emp.incoming=em.max_incom
    -> order by incoming desc;

17.列出部门员工数大于1个的部门名称

mysql> select dept_name as 部门名称 from dept
    -> inner join (select count(*) as 员工数,dept2 from emp group by dept2)         -> as de on dept.dept1=de.dept2
    -> where de.员工数>1;

19.查找张三所在的部门名称

mysql> select dept_name as 部门名称 from emp
    -> inner join dept on emp.dept2=dept.dept1
    -> where name='张三';