经典的MySQL试题

1.取得每个部门最高薪水的人员名称

第一步：求出每个部门的最高薪水

seelct
		e.deptno,max(e.sal) as maxsal
from
		emp e
group by
		e.deptno;		

将以上查询结果当成一个临时表t（deptno,maxsal）

select
		e.deptno,e.ename,t.maxsal,e.sal
from
		(seelct
				e.deptno,max(e.sal) as maxsal
		 from
				emp e
		 group by
				e.deptno)t
join
		emp e
on
		t.deptno = d.deptno
where
		t.maxsal = e.sal
order by
		e.deptno;

2.哪些人的薪水在部门平均薪水之上

第一步：求出每个部门的平均薪水

select
				e.deptno,avg(e.sal) as avgsal
from
				emp e
group by
				e.deptno;

将以上查询结果当成临时表t(deptno,avgsal)

select
		t.deptno,e.ename
from
		(select
				e.deptno,avg(e.sal) as avgsal
		from
				emp e
		group by
				e.deptno)t
join
		emp e
on
		e.deptno = t.deptno
where
		e.sal > t.avfsal;

3.取得部门中（所有人的）平均薪水等级

3.1取得部门中所有人的平均薪水等级
第一步：求出部门的平均薪水

select
		e.deptno,avg(e.sal) as avgsal
from
		emp e
group by
		e.deptno;

将以上查询结果当成临时表t(deptno,avgsal)

select
		t.deptno,t.avgsal,s.grade
from
		(select
				e.deptno,avg(e.sal) as avgsal
		from
				emp e
		group by
				e.deptno)t
join
		salgrade s
on
		t.avgsal between s.losal and s.hisal;

3.2取得部门中所有人的平均的薪水等级
第一步：求出每个人的薪水等级

select
		e.deptno,e.enamem,s.grade
from
		emp e
join
		salgrade s
on
		e.sal between s.losal and s.hisal
roder by
		e.deptno;

将以上查询数据当成临时表t(deptno,ename,grade)

select
		t.deptno,avg(t.grade) as avgGrade
from
		(select
				e.deptno,e.enamem,s.grade
		from
				emp e
		join
				salgrade s
		on
				e.sal between s.losal and s.hisal)t
group by
		t.deptno;

4.不用组函数（MAX），取得最高薪水（给出两种解决方案）

第一种：

select
		sal
from
		emp
order by
		sal desc limit 1;

第二种：
首先找到a表中的薪水低于b表中的薪水（去重）

select
		distinct a.sal
from
		emp a
join
		emp b
on
		a.sal < b.sal;

然后：找到不在表中的数据
不在表中的薪水就是最高的薪水

select
		sal
from
		emp
where
		sal not in(select
							distinct a.sal
					from
							emp a
					join
							emp b
					on
							a.sal < b.sal);

5.取得平均薪水的最高部门编号

第一步：求出平均薪水

select
		e.deptno,avg(e.sal) as avgsal
from
		emp e
group by
		e.deptno;

第二步：将以上查询数据结果当成临时表t(deptno,avgsal)
然后求出平均薪水的最大值

select
		max(t.avgsal) as maxAvgSal
from
		(select
				e.deptno,avg(e.sal) as avgsal
		from
				emp e
		group by
				e.deptno)t;

第三步：拿出最高平均薪水进行过滤

select
		e.deptno,avg(e.sal) as avgsal
from
		emp e
group by
		e.deptno
having
		avgsal = (select
							max(t.avgsal) as maxAvgSal
				from
							(select
									e.deptno,avg(e.sal) as avgsal
							from
									emp e
							group by
									e.deptno) t);

6.取得平均薪水最高的部门的部门名称

select
		e.deptno,d.dname,avg(e.sal) as avgsal
from
		emp e
join
		dept d
on
		e.deptno = d.deptno
group by
		e.deptno,d.dname
having
		avgsal = (select
							max(t.avgsal) as maxAvgSal
				from
							(select
									e.deptno,avg(e.sal) as avgsal
							from
									emp e
							group by
									e.deptno) t);

7.求平均薪水的等级最低的部门的部门名称

第一步：求出部门的平均薪水和部门名称

select
		e.deptno,d.dname,avg(e.sal) as avgsal
from
		emp e
join
		dept d
on
		e.deptno = d.deptno
group by
		e.deptno,d.deptno;

第二步：将以上查询结果当成临时表t(deptno,avgsal)与salgrade表连接
从而求出每个部门薪水等级

select
		t.deptno,t.dname,s.grade
from
		(select
				e.deptno,d.dname,avg(e.sal) as avgsal
		from
				emp e
		join
				dept d
		on
				e.deptno = d.deptno
		group by
				e.deptno,d.deptno)t
							join
									salgrade s
							on
									t.avgsal between s.losal and s.hisal;				

第三步：将以上查询结果当成临时表
从而求出最低等级

select min(t.grade) as minGrade from (select
											t.deptno,t.dname,s.grade
									from
											(select
													e.deptno,d.dname,avg(e.sal) as avgsal
											from
													emp e
											join
													dept d
											on
													e.deptno = d.deptno
											group by
													e.deptno,d.deptno)t
																join
																		salgrade s
																on
																		t.avgsal between s.losal and s.hisal)t;

第四步：将最低等级代入部门所对应的编号和名称当中，得出最后结果

select
		t.deptno,t.dname,s.grade
from
		(select
				e.deptno,d.dname,avg(e.sal) as avgsal
		from
				emp e
		join
				dept d
		on
				e.deptno = d.deptno
		group by
				e.deptno,d.deptno)t
							join
									salgrade s
							on
									t.avgsal between s.losal and s.hisal
							where
									s.grade = (select min(t.grade) as minGrade from (select
											t.deptno,t.dname,s.grade
									from
											(select
													e.deptno,d.dname,avg(e.sal) as avgsal
											from
													emp e
											join
													dept d
											on
													e.deptno = d.deptno
											group by
													e.deptno,d.deptno)t
																join
																		salgrade s
																on
																		t.avgsal between s.losal and s.hisal)t);

8.取得比普通员工（员工代码没有在mgr表上出现的）的最高薪水还要高的经理人姓名

第一步：找出普通员工（员工代码没有出现在mgr上的）
先找出mgr有哪些人

select distinct mgr from emp;

注：
not in不会自动忽略控制（会有null参与数学运算）
in会自动会略空值（null不会参与数学运算）

第二步：找出普通员工的最高薪水

select
		max(sal) as maxsal
from
		emp
where
		empno not in(select distinct mgr from emp where mgr is not null);

第三步：求出符合条件的经理的姓名

select ename from emp where sal > (select
											max(sal) as maxsal
									from
											emp
									where
											empno not in(select distinct mgr from emp where mgr is not null));

9.取得薪水最高的前五名员工

select * from emp order by sal desc limit 0,5;

10.取得薪水最高的第六到第十名员工

select * from emp order by sal desc limit 5,5;

11.取得最后入职的5名员工

select * from emp order by hiredate desc limit 5;

12.取得每个薪水等级有多少员工

第一步：查出每个员工的薪水等级

select
		e.ename,s.grade
from
		emp e
join
		salgrade s
on
		e.sal between s.losal and s.hisal
order by
		s.grade;

第二步：将以上查询结果当成临时表t(ename,grade)
从而求出等级的总和

select
		t.grade,count(t.ename) as totalEmp
from
		(select
				e.ename,s.grade
		from
				emp e
		join
				salgrade s
		on
				e.sal between s.losal and s.hisal)t
group by
		t.grade;

**13.有3张表s(学生表),c(课程表),sc(学生选课表)

s(sno,sname) 代表 (学号，姓名)
c(cno,cname,cteacher) 代表 (课号，课名，教师)
sc(sno,cno,scgrade) 代表 (学号，课号，成绩)**

问题：
1.找出没选过“黎明”老师的所有学生姓名
2.列出2门以上（含2门）不及格学生姓名及平均成绩
3.学过1号课程又学过2号课所有学生的姓名

第一步：首先创建表并且添加数据

create table s(
		sno int(4) primary key auto_increment,
		sname varchar(32)		
	);

insert into s(sname) values('zhangsan');
insert into s(sname) values('lisi');
insert into s(sname) values('wangwu');
insert into s(sname) values('zhaoliu');

create table c(
		cno int(4) primary key auto_increment,
		cname varchar(32),
		cteacher varchar(32)	
	);

insert into c(cname,cteacher) values('Java','吴老师');
insert into c(cname,cteacher) values('C++','王老师');
insert into c(cname,cteacher) values('C##','张老师');
insert into c(cname,cteacher) values('MySQL','郭老师');
insert into c(cname,cteacher) values('oRACLE','黎明');

create table sc(
		sno int(4),
		cno int(4),
		scgrade double(3,1),
		constraint sc_sno_pk primary key(sno,cno),
		constraint sc_sno_fk foreign key(sno) references s(sno),
		constraint sc_cno_fk foreign key(cno) references c(cno),
	);

insert into sc(sno,cno,scgrade) values(1,1,30);
insert into sc(sno,cno,scgrade) values(1,2,50);
insert into sc(sno,cno,scgrade) values(1,3,80);
insert into sc(sno,cno,scgrade) values(1,4,90);
insert into sc(sno,cno,scgrade) values(1,5,70);
insert into sc(sno,cno,scgrade) values(2,2,80);
insert into sc(sno,cno,scgrade) values(2,3,50);
insert into sc(sno,cno,scgrade) values(2,4,70);
insert into sc(sno,cno,scgrade) values(2,5,80);
insert into sc(sno,cno,scgrade) values(3,1,60);
insert into sc(sno,cno,scgrade) values(3,2,70);
insert into sc(sno,cno,scgrade) values(3,3,80);
insert into sc(sno,cno,scgrade) values(4,3,50);
insert into sc(sno,cno,scgrade) values(4,4,80);

1.找出没选过“黎明”老师的所有学生姓名
第一步：找出黎明老师授课的编号

select cno from c where cteacher = '黎明';

第二步：找出选过黎明老师的学生编号

select sno from sc where cno = (select cno from c where cteacher = '黎明');

第三步：找出最后结果

select * from s where sno not in(select sno from sc where cno = (select cno from c where cteacher = '黎明'));

2.列出2门以上（含2门）不及格学生姓名及平均成绩
第一步：找出这类学生

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

第二步：找出该类学生平均成绩

select
		sc.sno,avg(sc.scgrade) as scgrade
from
		sc
group by
		sc.sno;

第三步，将上述两张表进行联接

select
		t1.sname,t2.avgscgrade
from
		(select
				sc.sno,sname,count(*) as studentNum
		from
				sc
		join
				s
		on
				sc.sno = s.sno
		where
				scgraade < 60
		group by
				sc.sno,ssname
		having
				studentNum >=2)t1
join
		(select
				sc.sno,avg(sc.scgrade) as scgrade
		from
				sc
		group by
				sc.sno)t2
on
		t1.sno = t2.sno;		

3.学过1号课程又学过2号课所有学生的姓名
第一步：分别找出1号课程和2号课程的学生

select sno from sc where cno = 1;
select sno from sc where cno = 2;

第二步：找出符合条件的学生

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);

14.列出所有员工及领导的名字（所有，运用到外连接）

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

15.列出受雇日期早于其上级的所有员工编号、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;

16.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;

17.列出至少5个员工的所有部门

第一步：先求出每个部门的员工数量并对其分组

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

第二步：分组不满意，having过滤找到大于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;

18.列出薪水比‘SMITH’多的所有员工信息

第一步：找到SMITH的薪水

select sal from emp where ename = 'SMITH';

第二步：找出大于该工资的其他员工

select * from emp where sal > (select sal from emp where ename = 'SMITH');

19.列出所有‘CLERK’(办事员)的姓名及其部门名称，部门人数

第一步：先找出其办事员和所对应的部门，并将查询结果看作t1表

select
		d.deptno,d.dname,e.ename
from
		emp e
join
		dept d
on
		e.deptno = d.deptno
where
		e.job = 'CLERK';

第二步：求出每个部门的员工数量，将查询结果看作t2表

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

第三步：将t1和t2连接

select				
		t1.deptno,t1.dname,t1.ename,t2.totalEmp
from
		(select
				d.deptno,d.dname,e.ename
		from
				emp e
		join
				dept d
		on
				e.deptno = d.deptno
		where
				e.job = 'CLERK')t1
join
		(select
				e.deptno,count(e.ename) as totalEmp
		from
				emp
		group by
				e.deptno)t2
on
		t1.deptno = t2.deptno;

20.列出最低薪水大于1500的各种工作及从事此工作的全部雇员人数

第一步：先求出每种工作岗位的最低薪水

select
		e.job,min(e.sal) as minsal
from
		emp e
group by
		e.job;

第二步：对查询结果不满意，用having过滤，求出薪水大于1500及雇员人数

select
		e.job,min(e.sal) as minsal，count(e.ename) as totalEmp
from
		emp e
group by
		e.job
having
		minsal > 1500;

21.列出在部门‘SALES’<销售部>工作的员工姓名，假定不知道销售部门的部门编号

第一步：找出在部门名称为‘SALES’的部门编号

select deptno from dept where dname = 'SALES';

第二步：将上述查询结果带入此步查询结果中，找出员工姓名

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

22.列出薪水高于公司平均薪水的所有员工，所在部门、上级领导、雇员的工资等级

第一步：求出公司的平均薪水

select avg(sal) as avgsal from emp;

第二步：进行表连接，查询出对应结果

select
		d.dname,
		e.ename,
		b.ename as leadername,
		s.grade
from
		emp e
join
		dept d
on
		e.deptno = d.deptno
left join
		emp b
on
		e.mgr = b.empno
join
		salgrade s
on
		e.sal between s.losal and s.hisal
where
		e.sal > (select avg(sal) as avgsal from emp);

23.列出与‘SCOTT’从事相同工作的所有员工及部门名称

第一步：查询出‘SCOTT’的工作感岗位

select job from emp where ename = 'SCOTT';

第二步：连接表将所有员工及部门名称查询出来

select
		d.dname,
		e.*
from
		emp e
join
		dept d
on
		e.deptno = d.deptno
where
		e.job = (select job from emp where ename = 'SCOTT');

24.列出薪水等于30中员工的薪水的其他员工的姓名和薪水

第一步：找出部门30中薪水有哪些值

select distinct sal from emp where deptno = 30;

第二步：找出其他员工

select
		deptno,ename,sal
from
		emp
where
		sal in(select distinct sal from emp where deptno = 30) and deptno <> 30;

25.列出薪水高于部门30的所有员工的薪水的员工姓名和薪水、部门名称

第一步：先找出部门30的最高薪水

select max(sal) as maxsal from emp where deptno = 30;

第二步：找出这些员工

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 emp where deptno = 30);

26.列出再每个部门工作的员工数量、平均工资和平均服务期限

to_days(日期类型)–> 天数
获取数据库的系统当前时间的函数是：now()
第一步：先求出平均服务期限

select ename,(to_days(now()) - to_days(hiredate))/365 as serveryear from emp;

select avg((to_days(now()) - to_days(hiredate))/365) as avgserveryear from emp;

第二步：求出对应的员工信息

select
		e.deptno,
		count(e.ename) as totalEmp,
		avg(e.sal) as avgsal,
		avg((to_days(now()) - to_days(hiredate))/365) as avgserveryear
from
		emp e
group by
		e.deptno;

27.列出所有员工的姓名、部门名称和工资

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

28.列出所有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;

29.列出各种工作的最低工资及从事此工作的员工姓名

第一步：类出各个工作的最低工资

select
		e.job,min(e.sal) as minsal
from
		emp e
group by
		e.job;

第二步：将上述查询结果当成临时表t（job,minsal）进行连接

select
		e.ename
from
		emp e
join
		(select
				e.job,min(e.sal) as minsal
		from
				emp e
		group by
				e.job) t
on
		e.job = t.job
where
		e.sal = t.minsal;

30.列出各个部门MANAGER的最低薪水

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

31.列出所有员工的年工资，按年薪从低到高排序

select ename,(sal + ifnull (comm,0)) * 12 as yearsal from emp order by yearsal asc;

32.求出员工领导的薪水超过3000的员工名称和领导名称

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

33.求部门名称中带“S”字符的部门员工的工资合计、部门人数

select
		d.dname,
		sum(e.sal) as sumsal,
		count(e.ename) as totalEmp
from
		emp e
join
		dept d
on
		e.deptno = d.deptno
where
		d.dname like '$s%'
group by
		d.dname;

34.给任职日期超过30年的员工加薪10%

update emp_bak1 set sal = sal * 1.1 where (to_days(now()) - to_days(hiredate))/365 > 30;