mysql经典问题四表查询(教师，学生，成绩，课程表)

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)

-----------------------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;

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(sc.score) from student s,sc group by s.sid having avg(sc.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=‘1900-01-01’ and sage70;

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 score80;

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;