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,'数学',1),(3,'英语',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”课程的学生情况
SELECT a.*, b.score, c.score FROM student a, sc b,sc c where 1 =1
and a.sid = b.sid and a.sid = c.sid and b.cid = 1 and c.cid = 2
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 t.sid,t.avgnum, (select sname from student where sid = t.sid) as snamefrom (select sid, sum(score)/3 as avgnum from sc group by sid having sum(score)/3 >= 60) as t
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 s.sid, s.sname, ame,sc1.scoreFROM sc sc1,course c,student sWHERE sc1.cid=c.cid AND s.sid = sc1.sid;
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
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;
select t1.*from student t1where t1.sid not in (select sid from sc where cid =2)and t1.sid in (select sid from sc where cid =1)
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;
select t.sid,(select sname from student where sid = t.sid) as name,t.count from(select sid, count(sid) as count from sc group by sid having count(sid) = 2) as t;
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,ame,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,ame,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(‘-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;
