数据库之经典面试

1，获取各个部门最高薪水的人员

使用连接查询

select e.department,e.`name`,t.dept,t.maxMoney
from 
	(select e.department as dept,max(e.money) as maxMoney from employee e group by e.department)t
	join
		employee e
	on 
		t.dept = e.department
	where 
		e.money = t.maxMoney;

2，哪些人的薪水在部门的平均薪水之上

select
	e.department,e.`name`,t.`平均薪水`,e.money
from
	(select
		e.department as dept,avg(e.money) as 平均薪水
	from
		employee e
	group by 
		e.department)t
join
	employee e
on
	e.department = t.dept
where
	e.money > t.`平均薪水`;

3，求各个部门平均薪水的等级

表：

select 
	t.`部门` as 部门,t.`平均薪水` as 平均薪水,m.grade
from 
	(select e.department as 部门,avg(e.money) as 平均薪水 from employee e group by e.department)t
join
	moneygrade m
on
	t.`平均薪水` between m.low and m.high;

4, 不使用组函数max()，获取最高薪水

方案一：

select
	e.name,e.money
from 
	employee e
order by 
	e.money desc limit 1;

方案二:
第一步

select
	distinct e.name ,e.money
from
	employee e
join
	employee b
on 
	e.money < b.money;

可以发现除了最大值，其他的值都符合要求
第二步

select e.name,e.money from employee e where e.money not in (select
	distinct e.money
from
	employee e
join
	employee b
on 
	e.money < b.money);

5，求平均薪水最低的等级的部门名称

select 
	t.department
from 
	(select e.department ,avg(e.money) as 平均薪水 from employee e group by e.department)t
join
	moneygrade m 
on 
	t.平均薪水 between m.low and m.high
where m.grade = (
	select 
		min(m.grade) 
	from
			(select 
				t.department,t.`平均薪水`,m.grade 
				from 
					(select e.department ,avg(e.money) as 平均薪水 from employee e group by e.department)t
				join moneygrade m 
				on t.平均薪水 between m.low and m.high)m); 

6,三个表，学生表（S），课程表( C)，学生选课表(SC)

课程表：分别对应课号（cno），课名（cname），老师（cteacher）

学生表：学号(sno)，姓名(sname)

课程表：学号(sno)，课号(cno)，成绩(scgrade)
学号，课号分别为外键

向学生表中添加数据

课程表添加数据

选课表添加数据

问题1:找出没选过郑老师的所有学生的姓名

select sname from s where sno not in 
	(select sno from sc where cno = 
		(select cno from c where cteacher = '郑老师'));

问题2：列出两门（含两门）以上不及格学生姓名及平均成绩

#分组先求每个人成绩小于60分的门数

select 
	sc.sno , s.sname ,count(*) as studentSum
from
	sc
join
	s
on
	s.sno = sc.sno
where
	sc.scgrade < 60
group by
	sc.sno
having
	studentSum >= 2;

问题3：即学过1号课程和2号课程所有学生的姓名

select
	s.sname 
from 
	sc
join
	s
on
	sc.sno = s.sno
where
	cno = 1 and sc.sno in(
		select sno from sc where cno = 2
	);

7，新建两个表，分别为dept部门信息表和emp员工表

然后呢插入数据，emp表如下

dept表如下

1，列出所有的员工及领导的名字

select
	e.ename,
	b.ename as leadername
from
	emp e
left join
	emp b
on
	e.mgr = b.empno;

并出现结果

外连接查询的条数永远大于等于内连接

2，列出受雇日期早于其上级的所有的员工编号，姓名，部门名称

#列出受雇日期早于其上级的所有的员工编号，姓名，部门名称
 select 
	d.dname,e.empno,e.ename
from
	emp e
join
	emp b
on
	e.mgr = b.empno
join
	dept d
on
	e.deptno = d.deptno
where
	e.hiredate < b.hiredate;

最终的数据为

3，列出部门名称和这些部门的员工信息，同时列出那些没有员工的部门

select
	d.dname,e.*
from
	emp e
right join
	dept d
on
	e.deptno = d.deptno;

4,列出至少有5个员工的所有部门

select 
	e.deptno,count(e.ename) 
as
	totalEmp
from 
	emp e 
group by 
	e.deptno 
having totalEmp >= 5;

5,列出在部门"SALES"<销售部>工作的员工的姓名，假定不知道销售部门的部门编号

select 
	ename
from
	emp
where
	deptno = (select deptno from dept where dname = 'SALES');

6,列出薪资高于公司平均水平的所有员工，所在部门，上级领导

select 
	e.ename,
	d.dname,
	b.ename
from
	emp e
join
	dept d
on
	e.deptno = d.deptno
left join
	emp b
on
	e.mgr = b.empno
where
	e.sal > (select avg(sal) as angsal from emp);

7,列出薪水高于在部门30狗熊的所有员工的薪金的员工姓名和薪金，部门名称

select 
	d.dname,
	e.ename,
	e.sal
FROM
	emp e
join
	dept d
on
	e.deptno = d.deptno
where
	e.sal > (select max(sal) as maxSal from dept);

8,列出在每个部门工作的员工数量，平均工资

select 
	d.dname,count(*) as '人数',avg(e.sal) as '平均工资' 
from 
	emp e 
join 
	dept d 
on 
	d.deptno = e.deptno 
group by
	e.deptno; 

9,列出所有的员工的姓名，部门名称和工资

select
	e.ename as '员工姓名',
	d.dname as '部门名称',
	e.sal as '工资'
from
	emp e
join
	dept d
on
	e.deptno = d.deptno

10,列出所有的部门的详细信息和人数

select 
	d.deptno,d.dname,d.loc,count(e.ename) as totalEmp
from
	emp e
right join
	dept d
on
	e.deptno = d.deptno
group by
	d.deptno,d.dname,d.loc;

11,列出各种工作的最低工资以及从事此工作的雇员姓名

select distinct
	e.job,t.minSal,e.ename
from 
	emp e
join
		(select 
		e.job,min(e.sal) as minSal
	from
		emp e
	join
		dept d
	group by
		e.job)t
on
	e.sal = t.minSal

12,列出各个部门MANAGER的最低薪金

select 
	w.deptno,min(w.sal) as minSal
from
	emp w
where
	w.job = 'MANAGER'
group by
	w.deptno;

13,求出员工领导超过3000的员工名称和领导名称

select
	e.ename,
	b.ename as leadername
from
	emp e
join
	emp b
on
	e.mgr = b.empno
where
	b.sal > 3000;

14,名字中带S字符的部门的工资合计和部门人数

select 
	d.dname,
	sum(e.sal) as sumMoney,
	count(e.ename) as sumPeople
from
	emp e
join
	dept d
on
	e.deptno = d.deptno
where
	d.dname like '%S%'
group by
	d.dname

15,给任职期间超过30年的员工加薪

update
	emp
set
	sal = sal * 1.1
where
	(to_days(now()) - to_days(hiredate))/365 > 30;