动力节点 mysql 郭鑫 34道经典的面试题二

13.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。

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(4,3,50);
insert into sc(sno,cno,scgrade) values(4,4,80);

首先分析下上面表的设计

一个学生可以选修多门课程

同一门课程可以被多个学生选择

学生和课程之间是多对多的关系

所以就要引入第三张中间表来解决学生和课程之间的关系

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)
);

第三张表的sno的值必须来自学生表，使用外键约束

cno必须来自课程表，使用外键约束

然后建立一个sno 和cno的一个复合主键

接下来我们就可以做题了

/*
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。
*/

第一题的第一种做法：

--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'));

第一题的第二种做法：

第一步：找到黎明老师所上课对应的课程对应的课程编号
select cno from c where  cteacher = '黎明';

第二步：求出那些学生选修了黎明老师的课程
select
    sno
from
    sc
join
   (
        select cno from c where  cteacher = '黎明'

   )    t
 on
    sc.cno = t.cno;

第三步：求出那些学生没有选择黎明老师的课
select
  sno,sname
from
   s
where
   sno not  in
                 (

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

                        )    t
                   on
                      sc.cno = t.cno

                 );   

+-----+---------+
| sno | sname   |
+-----+---------+
|   3 | wangwu  |
|   4 | zhaoliu |
+-----+---------+

第二题：

2、列出2门以上（含2门）不及格学生姓名及平均成绩。

思路一 ：在sc表中首先按照学生编号进行分组，得到哪些学生的有两门以上的成绩低于60分

select
     sc.sno ,count(*) as studentNum
from
     sc
where
      scgrade < 60
group by
      sc.sno
having
      studentNum >= 2;

第二步：查询出该学生对应的编号

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

   ) as b
on
   a.sno = b.sno;
     

+-----+----------+
| sno | sname    |
+-----+----------+
|   1 | zhangsan |
+-----+----------+
1 row in set

接下来需要获得该学生的平均成绩，我们得到该学生的sno的值是1，我们需要在sc表中求出该学生的平均成绩，首先需要将上面的表和sc表关联起来形成一个临时表，然后对这个临时表按照sno进行group by

我们来看下下面的代码

select
     sc.sno ,count(*) as studentNum
from
     sc
where
      scgrade < 60
group by
      sc.sno
having
      studentNum >= 2;

第二步：查询出该学生对应的编号

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

   ) as b
on
   a.sno = b.sno;   

第三步得到该学生的平均成绩，把上面的表当成临时表m

select
   m.sno,m.sname,avg(d.scgrade)
from
   sc as d
join
   (
        select
            a.sno , a.sname
        from
            s as a
        join
           (
              select
                    sc.sno ,count(*) as studentNum
                from
                     sc
                where
                      scgrade < 60
                group by
                      sc.sno
                having
                      studentNum >= 2

           ) as b
        on
           a.sno = b.sno

   )  as m
on
  m.sno = d.sno
group by
   d.sno ;

+-----+----------+----------------+
| sno | sname    | avg(d.scgrade) |
+-----+----------+----------------+
|   1 | zhangsan |       64.00000 |
+-----+----------+----------------+
1 row in set

第三题：

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 |
+----------+

不能写成下面的形式会存在错误