mysql经典问题四表查询(教师，学生，成绩，课程表)

---

建立数据库

1.建立一个数据库
create database work;

2.进入数据库work
use work;

3.数据库默认编码可能不支持中文,可以在这里设置下
set names gbk;

4.建立student表
属性有:编号：id (主键，自动增长),姓名:sname,出生年月:sage,性别:ssex(枚举)
create table student(sid int primary key auto_increment,
sname varchar(20),
sage date,
ssex enum(‘男’,’女’));

5.第二个课程表中使用了外键教师标号，因而需要先建立教师表
create table teacher(tid int primary key auto_increment,
tname varchar(20));

6.建立课程表
create table course(cid int primary key auto_increment,
cname varchar(20),
tid int,
foreign key(tid) references teacher(tid));

7.建立成绩表
create table sc(sid int,
cid int,
score int);

8.show tables; //可查看建立的四个表格

9.插入数据，因为里面有主键链接，表格插入数据也要有顺序(注意题目图片上都是字节引号，应该为int，不要单引号)

a，先给student表插入数据
insert into student values(1,'赵雷','1990-01-01','男'),
    (2,'钱电','1990-12-21','男'),
    (3,'孙风','1990-05-20','男'),
    (4,'李云','1990-08-06','男'),
    (5,'周梅','1991-12-01','女'),
    (6,'吴兰','1992-03-01','女'),
    (7,'郑竹','1989-07-01','女'),
    (8,'王菊','1990-01-20','女');

b, 给teacher表插入数据,这里不可以先给course表插入数据,因为course表外链接到teacher的主键
insert into teacher values(1,'张三'),
        (2,'李四'),
        (3,'王五');

c, 给course表插入数据
insert into course values(1,'语文',2),
            (2,'语文',1),
            (3,'语文',3);

d, 最后给sc表插入数据(题目图片少了第1个学生成绩，在这加上 1,1,90;  1,2,80;  1,3,90)
insert into sc values(1,1,90),
            (1,2,80),
            (1,3,90),
            (2,1,70),
            (2,2,60),
            (2,3,80),
            (3,1,80),
            (3,2,80),
            (3,3,80),
            (4,1,50),
            (4,2,30),
            (4,3,20),
            (5,1,76),
            (5,2,87),
            (6,1,31),
            (6,3,34),
            (7,2,89),
            (7,3,98);

———————–数据库建立完成—————————————

---

1、查询”01”课程比”02”课程成绩高的学生的信息及课程分数
select s.sid,s.sname,s.sage,s.ssex,sc1.score,sc2.score from student s,sc sc1,sc sc2 where sc1.cid=1 and sc2.cid=2 and sc1.score>sc2.score and sc1.sid=sc2.sid and s.sid=sc1.sid;

2、查询同时存在”01”课程和”02”课程的情况
select * from course c1,course c2;

3.查询平均成绩大于等于60分的同学的学生编号和学生姓名和平均成绩
select s.sid,s.sname,avg(score) from student s,sc group by s.sid having avg(score)>60;

4、查询名字中含有”风”字的学生信息
select * from student where sname like ‘%风%’;

5、查询课程名称为”数学”，且分数低于60的学生姓名和分数
select s.sname,score from student s,sc where s.sid=sc.sid and cid=2 and score<60;

6、查询所有学生的课程及分数情况；
select cname,score from sc,course where sc.cid=course.cid;

7、查询没学过”张三”老师授课的同学的信息
select s.* from student s where s.sid not in(select sc1.sid from sc sc1,course c,teacher t where t.tid=c.tid and sc1.cid=c.cid and t.tname=’张三’);

8.查询学过”张三”老师授课的同学的信息
select s.* from student s ,sc sc1,course c,teacher t where s.sid=sc1.sid and sc1.cid=c.cid and c.tid=t.tid and t.tname=’张三’;

9、查询学过编号为”01”并且也学过编号为”02”的课程的同学的信息
student(sid) sc(sid cid tid) sc2(sid cid tid) course(cid tid cname)
select s.* from student s,sc sc1,sc sc2 where s.sid=sc1.sid and sc1.sid=sc2.sid and sc1.cid=1 and sc2.cid=2;

10、查询学过编号为”01”但是没有学过编号为”02”的课程的同学的信息
select distinct s.* from student s,sc sc1,sc sc2,sc sc3 where s.sid=sc1.sid and sc1.sid=sc2.sid and sc1.cid=1 and sc2.cid!=2;

11、查询没有学全所有课程的同学的信息
select s.* from student s where s.sid not in(select sc1.sid from sc sc1,sc sc2,sc sc3 where sc1.cid=1 and sc2.cid=2 and sc3.cid =3 and sc1.sid=sc2.sid and sc1.sid=sc3.sid) group by s.sid;

12、查询至少有一门课与学号为”01”的同学所学相同的同学的信息
select distinct s.* from student s,sc sc1 where s.sid=sc1.sid and sc1.cid in(select cid from sc where sid=1) and s.sid<> 1;

13、查询和”01”号的同学学习的课程完全相同的其他同学的信息
select s.* from student s where s.sid in(select distinct sc.sid from sc where sid<>1 and sc.cid in(select distinct cid from sc where sid=1)group by sc.sid having count(1)=(select count(1) from sc where s.sid=1));

14、查询没学过”张三”老师讲授的任一门课程的学生姓名
select s.* from student s where s.sid not in(select sc1.sid from sc sc1,course c,teacher t where sc1.cid=c.cid and c.tid=t.tid and t.tname=’张三’);

15、查询出只有两门课程的全部学生的学号和姓名
select s.* from student s,sc group by sc.sid having count(sc.sid)=2 and s.sid=sc.sid;

16、查询1990年出生的学生名单(注：Student表中Sage列的类型是datetime)
select * from student where sage>=’1900-01-01’ and sage<=’1900-12-31’;
select s.* from student s where s.sage like ‘1900-%’;（方法2）

17、查询每门课程的平均成绩，结果按平均成绩降序排列，平均成绩相同时，按课程编号升序排列
select sc.cid,avg(score) from sc group by sc.cid order by avg(score) DESC , sc.cid;

18、查询任何一门课程成绩在70分以上的姓名、课程名称和分数；
select s.sname,c.cname,score from student s,sc,course c where s.sid=sc.sid and sc.cid=c.cid and score>70;

19、查询平均成绩大于等于85的所有学生的学号、姓名和平均成绩
select s.sname,avg(score) from sc,student s where s.sid=sc.sid group by sc.sid having avg(score)>=85;

20、查询不及格的课程
select s.sname,c.cname,score from student s,sc,course c where s.sid=sc.sid and sc.cid=c.cid and score<60;

21、查询课程编号为01且课程成绩在80分以上的学生的学号和姓名；
select s.sid,s.sname from student s,sc where sc.sid=s.sid and sc.cid=1 and score>80;

22、求每门课程的学生人数
select cid,count(sid) from sc group by sc.cid;

23、统计每门课程的学生选修人数（超过5人的课程才统计）。要求输出课程号和选修人数，查询结果按人数降序排列，若人数相同，按课程号升序排列
select cid,count(sid) from sc group by cid having count(sid)>5 order by count(sid),cid ASC;

24、查询不同课程成绩相同的学生的学生编号、课程编号、学生成绩
select s1.sid,s2.sid,sc1.cid,sc1.score,sc2.score from student s1,student s2,sc sc1,sc sc2 where s1.sid!=s2.sid and s1.sid=sc1.sid and s2.sid=sc2.sid and sc1.cid!=sc2.cid and sc1.score=sc2.score;

25、检索至少选修两门课程的学生学号
select sid from sc group by sid having count(cid)>=2;

26、查询选修了全部课程的学生信息
select s.* from sc,student s where s.sid=sc.sid group by sid having count
(cid)=3;

27、查询各学生的年龄
select s.sname,(TO_DAYS(‘2017-09-07’)-TO_DAYS(s.sage))/365 as age from student s;

28、查询本月过生日的学生
select s.sname from student s where s.sage like ‘_____07%’;

29、查询下月过生日的学生
select s.sname from student s where s.sage like ‘_____08%’;

30、查询学全所有课程的同学的信息
select s.* from student s,sc sc1,sc sc2,sc sc3 where sc1.cid=1 and sc2.cid=2 and sc3.cid=3 and sc1.sid=sc2.sid and sc1.sid=sc3.cid and s.sid =sc1.sid group by s.sid;