创建课程表c
⑴ SQL serever 数据库 学生管理 三个表学生表S，学生选课表SC，课程表C
1.SELECT COUNT(DISTINCT(学号）） FROM SC
2.SELECT 姓名 FROM S,SC WHERE S.学号=SC.学号 AND S.学号=1号
3.SELECT S.学号，姓名，课程名，成绩 FROM S,SC,C WHERE S.学号=SC.学号 AND SC.课程号=C.课程 AND 课程号=1号
4.SELECT S.学号，姓名 FROM S,SC WHERE S.学号=SC.学号 AND SC.课程=‘人工智能’
5.SELECT MAX(成绩） FROM S,SC WHERE S.学号=SC.学号 AND S.学号=1号
6.SELECT 学号，姓名 FROM S WHERE SUBSTRING(姓名，3,2）=’月’
7.SELECT 课程号，COUNT(*） FROM C,SC WHERE C.课程号=SC.课程号
8.SELECT S.学号，姓名 FROM S,SC,C WHERE S.学号=SC.学号 AND SC.课程号=C.课程号 AND C.成绩>80
9.SELECT DISTINCT（省区） FROM S WHERE S.系别=‘物理’
10.SELECT * FROM S ORDER BY 系别 ASC，年龄 DSC
11.SELECT AVG(成绩 ） FROM C WHERE 课程号=2号
12.SELECT 学号，姓名 FROM S WHERE SUBSTRING(姓名，3,2）=’阳’
⑵ 求一份数据结构课设报告（C语言）：个人课程表 设计要求：1.建立个人课程表信息
可以帮你写一个
⑶ 如果课程表c是使用以下sql语句创建的：create table c(cno c(2) not null ,cname c(20) ,cfen() n(1)):下
正确答案是D？
⑷ 对于教学数据库的3个基表(S表示学生表,C表示课程表,SC表示学生选课表,假设基表中
1.create table S
(SNO char(10) primary key,
SNAME varchar(20),
AGE int,
SEX char(2)default '男')
2.SELECT CNO,CNAME FROM C WHERE TEACHER='LI'
3.insert into C values('C01','高等数学','LI')
4.SELECT SNO,SNAME FROM S WHERE AGE>21 AND SEX='男'
5.select CNAME,TEACHER FROM C INNER JOIN SC ON C.CNO=SC.CNO WHERE SC.SNO='S1'
6.SELECT SNAME FROM S
WHERE SEX='男' and SNO in(select SNO from SC inner join
C on SC.CNO=C.CNO WHERE C.TEACHER='LI')
7.select CNO,CNAME FROM C
WHERE CNO in(select CNO from SC inner join
S on SC.SNO=S.SNO WHERE S.SNAME='WANG')
8.SELECT CNO 课程号,SEX 性别,count(*) 总人数,AVG(GRADE) 平均成绩 FROM SC INNER JOIN
S ON SC.SNO=S.SNO group by CNO,SEX
9.select SNAME,SNO from S
where SNO in (select SNO FROM SC group by SNO having count(SNO)>=2)
⑸ 一个面试题 oracle:学生表s（sno、sname)，课程表c（cno、cname,cteacher）选课表sc（sno、cno、score）
--1
select * from s where s.sno not in(select sno from c,sc where c.cno=sc.cno and c.cteacher='张三')
--2
select (select sname from s where s.sno=main.sno),avg(sc.score)
from s main,sc
where main.sno=sc.sno
and main.sno in(select sno from s s1,sc sc1 where s1.sno=sc1.sno and sc1.score<60 group by s1.sno having count(sno)>=2)
⑹ 大学数据库题 ，建立一个课程表C,有课程号Cno,课程名称Cn,学分xf,课时ks,课程类型
create table t_C (cno int primary key auto increment,
cn varchar(20),
xf int(11),
ks int(11),
kl varchar(30));
cno int primary key auto是设置cno为主键并且自增
⑺ 用SQL语言创建三张表 ， 学生表S，学生修课表SC，课程表C
哎 孩子这个还真像帮你做
不过你这分数和题量不大合适
其实还是懒了呵呵
不过你打了这么多字 要是不是复制的话 好好看书其实做更快了
SQL语句还是挺简单的 都不难 期末老师还是会考的
⑻ 求救！！！用C语言编一个课程表
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
intmain()
{
char*p[8][5],*a;
inti=0,j=0,n;
a=(char*)malloc(sizeof(char));
for(j=0;j<5;j++)
for(i=0;i<8;i++)
{
printf("周%d第%d节",j+1,i+1);
scanf("%s",a);
p[j][i]=a;
a=(char*)malloc(sizeof(char));
}
printf("周一 周二 周三 周四版权 周五 ");
for(j=0;j<5;j++){
for(i=0;i<8;i++)
{
printf("%s ",p[j][i]);
}
printf(" ");
}
}
⑼ 设在学生数据库中有三张表，表结构如下所示： Student(学生表): Student表的主键：sno(学号) Course(课程
(1) create table S_C
(sno char(10) not null
,cno char(8) not null
,score int null
,constraint PK_SC primary key (sno,cno)
)
(2)insert into Student (sno,sname,ssex)
values('1010','李小丽','女')
(3)create index IND_CName on Course (cname)
(4)update student set sage=23 where sno='1005'
(5)delete from course where cname='管理信息系统'
(6)select cno,cname,ctime
from course
where teacher='李元'
order by cno ASC
(7)select sno,sum(score) as score
from S_C
group by sno
(8)create view V_Student
as
select *
from student
where ssex='男' and sage>=18 and sage<=24
(9)select sno,sname
from student
where sno in (select sno from S_C where cno='001')
(10)select A.sno,A.sname
from student A
left join S_C B on A.sno=B.sno
left join Course C on B.cno=C.cno
where C.cname='关系数据库'
(11)select sno,sname from
(select A.sno,A.sname,count(1) as count_
from student A
left join S_C B on A.sno=B.sno
group by A.sno,A.sname) A
where count_>3
(12)select C.sname
from
(select * from S_C where cno='002') A
inner join
(select * from S_C where cno='004' on) B on A.sno=B.sno
left join student C on A.sno=C.sno
⑽ 数据库查询，有三张表学生表STU，课程表Course,SC学生课程关系表（s_id,c_id）
SELECTSTU.s_nameFROMSTU,(SELECTSC.s_idFROMSCGROUPBYs_id
HAVINGCOUNT(*)=(SELECTCOUNT(*)FROMCourse)AStmpWHERESTU.s_id=tmp.s_id