SQL面试题

有3个表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，即学过1号课程又学过2号课所有学生的姓名。

创建表和导入表数据
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_cno_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(3,4,60);
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。
先找出选过黎明老师的学生编号-> 黎明老师的授课的编号
select cno from c where cteacher = '黎明';

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

select * from s where sno not in(select sno from sc where cno =(select cno from c where cteacher = '黎明'));
+-----+---------+
| sno | sname   |
+-----+---------+
|   3 | wangwu  |
|   4 | zhaoliu |
+-----+---------+

2.列出2门以上（含2门）不及格学生姓名及平均成绩。
select
	t1.sname,t2.avgscgrade
from
	(select 
	sc.sno,s.sname,count(*) as studentNum
from 
	sc
join
	s
on
	sc.sno = s.sno
where 
	scgrade <60
group by
	sc.sno,s.sname
having
	studentNum >=2)t1
join
	(select
	sc.sno,avg(sc.scgrade) as avgscgrade
from
	sc
group by
	sc.sno)t2
on
	t1.sno = t2.sno；
+----------+------------+
| sname    | avgscgrade |
+----------+------------+
| zhangsan |   64.00000 |
+----------+------------+
3.即学过1号课程又学过2号课所有学生的姓名
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);
+----------+
| sname    |
+----------+
| zhangsan |
| wangwu   |
+----------+