工具：SQL Developer数据库学习的精髓就在于 练俗话说的好，孰能生巧，加油吧，SQL君表的结构首先建立以下几张表学生表# 建表
create table student(
sno varchar2(10) primary key,
sname varchar2(20),
sage number(2),
ssex varchar2(5)
);
# 插入数据
insert into student values ('s001','张三',23,'男');
insert into student values ('s002','李四',23,'男');
insert into student values ('s003','吴鹏',25,'男');
insert into student values ('s004','琴沁',20,'女');
insert into student values ('s005','王丽',20,'女');
insert into student values ('s006','李波',21,'男');
insert into student values ('s007','刘玉',21,'男');
insert into student values ('s008','萧蓉',21,'女');
insert into student values ('s009','陈萧晓',23,'女');
insert into student values ('s010','陈美',22,'女');
commit;
教师表#建表
create table teacher(
tno varchar2(10) primary key,
tname varchar2(20)
);
#插入数据
insert into teacher values ('t001', '刘阳');
insert into teacher values ('t002', '谌燕');
insert into teacher values ('t003', '胡明星');
commit;
课程表#建表
create table course(
cno varchar2(10),
cname varchar2(20),
tno varchar2(20),
constraint pk_course primary key (cno,tno)
);
# 插入数据
insert into course values ('c001','J2SE','t002');
insert into course values ('c002','Java Web','t002');
insert into course values ('c003','SSH','t001');
insert into course values ('c004','Oracle','t001');
insert into course values ('c005','SQL SERVER 2005','t003');
insert into course values ('c006','C#','t003');
insert into course values ('c007','JavaScript','t002');
insert into course values ('c008','DIV+CSS','t001');
insert into course values ('c009','PHP','t003');
insert into course values ('c010','EJB3.0','t002');
commit;
选课表# 建表
create table sc(
sno varchar2(10),
cno varchar2(10),
score number(4,2),
constraint pk_sc primary key (sno,cno)
);
# 插入数据
insert into sc values ('s001','c001',78.9);
insert into sc values ('s002','c001',80.9);
insert into sc values ('s003','c001',81.9);
insert into sc values ('s004','c001',60.9);
insert into sc values ('s001','c002',58.5);
insert into sc values ('s002','c002',10);
insert into sc values ('s003','c002',58);
insert into sc values ('s001','c003','59');
insert into sc values ('s001','c007','70');
insert into sc values ('s001','c010','71.50');
insert into sc values ('s005','c002','75.50');
insert into sc values('s006','c004',55);
insert into sc values('s007','c004',99);
commit;
问题！！！注意：以下代码复制粘贴时不要复制注释部分，否则语句执行会出现错误1、c001课程比c002课程成绩高的所有学生的学号自连接select x.sno from sc x join sc y on x.sno=y.sno
where (x.cno='c001' and y.cno='c002' and x.score>y.score);
2、查询平均成绩大于60 分的同学的学号和平均成绩select sno,avg(score) from sc
group by sno having avg(score)>60;
3、查询所有同学的学号、姓名、选课数、总成绩select s.sno,s.sname,count(*),sum(score) from student s
join sc on s.sno=sc.sno;
4、查询姓“刘”的老师的个数select count(*) from teacher
where tname like '刘%';
5、查询没学过“谌燕”老师课的同学的学号、姓名1、子查询法select sno,sname from student s where s.sno not in (
select diatinct sno from sc where cno in
(select cno from course where tno=
(select tno from teacher where sname='谌燕')
)
);
2、内连接法select distinct s.sno,s.sname from sc join student s on sc.sno=s.sno
where cno not in
(select c.cno from course c join teacher t
on c.tno=t.tno where t.tname='谌燕');
6、查询学过“c001”并且也学过编号“c002”课程的同学的学号、姓名自连接select s.sno,s.sname from
(select x.sno from sc x join sc y on x.sno=y.sno
where x.cno='c001' and y.cno='c002') e
join student s on e.sno=s.sno;
7、查询学过“谌燕”老师所教的所有课的同学的学号、姓名思路：（1）找出谌燕老师教的课程（2）找出所有学习了谌燕老师课程的学生（3）学生所学课程数量等于谌燕老师所教课程的数量select s.sno,s.sname from student s join sc on sc.sno=s.sno
join course c on c.cno=sc.cno
join teacher t on t.tno=c.tno
where t.tname='谌燕'
group by s.sno,s.sname
having count(*)=
# 按学号划分学习谌燕老师课程的组并计算学习课程数
(select count(*) from teacher t join course c on t.tno=c.tno
where tname='谌燕') # 统计谌燕老师所教课程的总数
8、查询课程编号“c002”的成绩比课程编号“c001”课程低的所有同学的学号、姓名自连接select sno,sname from student where sno in
(select a.sno from sc a join sc b on a.sno=b.sno
where a.cno='c002' and b.cno='c001' and a.score<b.score)
9、查询所有课程成绩小于60 分的同学的学号、姓名该题目有点歧义，在这里我理解为只要有一门成绩小于60分就被记录子查询法select sno,sname from student where sno in
(select distinct sno from sc where sc.score<60)
内连接法select distinct s.sno,s.sname from sc join student s
on sc.sno=s.sno where score<60;
10、查询没有学全所有课的同学的学号、姓名该题的所有课理解为在sc表中出现的所有课，而不是course表中的所有课select sno,sname from student where sno in
(select sno from sc group by sno
having count(cno)<
(select count(distinct(cno)) from sc));
11、查询至少有一门课与学号为“s001”的同学所学相同的同学的学号和姓名解法1select sno,sname from student where sno in
(select distinct a.sno from
(select * from sc) a join (select * from sc where sc.sno='s001') b
on a.cno=b.cno where a.sno<>b.sno);
解法2select distinct s.sno,s.sname from student s join
(select sno from sc where cno in
(select cno from sc where sno='s001')
and sno<>'s001'
) e on s.sno=e.sno;
12、查询至少学过学号为“s001”同学所有门课的其他同学学号和姓名题意的意思是其他同学学的课程最少要等于s001同学学习的所有课程感觉目前解的有点问题select s.sno,s.sname from sc join student s
on sc.sno=s.sno
where sc.sno<>'s001' and sc.cno in (select sno from sc where sno='s001');
13、把“SC”表中“谌燕”老师教的课的成绩都更改为此课程的平均成绩未完成update sc
set score=
(select avg(c.score) from course c join teacher t
on t.tno=c.tno where t.tname='谌燕' and c.cno=sc.cno)
14、查询和“s001”号的同学学习的课程完全相同的其他同学学号和姓名多次反选select sno,sname from student where sno in
(select sc.sno fromsc where sc.sno not in
(select sno from sc where sc.sno not in #学的课和s001没有一门相同或至少有一门不同的
(select cno from sc where sno='s001')) # s001学过的课程
and sno!='s001' group by sc.sno having count(sc.cno)=
(select count(cno) from sc where sno='s001'));
15、删除学习“谌燕”老师课的SC 表记录delete from sc where sc.cno in
(select cno from course where tno in
(select tno from teacher where tname='谌燕')
);
16、向SC 表中插入一些记录，这些记录要求符合以下条件：没有上过编号“c002”课程的同学学号、“c002”号课的平均成绩insert into sc(sno,cno,score)
select distinct s.sno,sc.cno,
(select avg(score) from sc where cno='c002')
from student s join sc on s.sno=sc.sno
where not exists
(select * from sc where cno='c002')
and sc.cno='c002';
17、查询各科成绩最高和最低的分：以如下形式显示：课程ID，最高分，最低分select cno,max(score),min(score) from sc group by cno;
18、按各科平均成绩从低到高和及格率的百分数从高到低顺序用case when then endselect cno,avg(score),
sum(case when score>=60 then 1 else 0 end)/count(*) as jg
from sc group by cno order by avg(score),jg desc;
case when score>=60 then 1 else 0 end如果一条记录满足score>=60则记为1，否则记为0，使用sum进行计算总和19、查询不同老师所教不同课程平均分从高到低显示select t.tname,c.cname,avg(score) from sc join course c on sc.cno=c.cno
join teacher t on c.tno=t.tno
group by c.cno,t.tname,c.cname
order by avg(score) desc
20、查询各科成绩前三名的记录:(不考虑成绩并列情况)解法1：使用分区select * from (select sno,cno,score,
row_number()over(partition by cno order by score desc) rn from sc)
where rn<4;
partition 是分区关键字，将数据按照cno进行分区over前的分区函数有以下三种row_number(）正常排序，同名次情况不考虑，按照顺序分配名次rank() 跳跃排序，当有两个第二名时，接下来的就是第四名dense_rank() 连续排序，有两个第二名，那么接下来的就是第三名row_number的使用rank()的使用dense_rank()的使用解法2:相关子查询select cno,sno,score from sc s1 where
(select count(1) from sc s2 where s1.score<=s2.score and s1.cno=s2.cno)
<4;
相关子查询的子查询与主查询相关，运行时主查询和子查询相互依赖21、查询各课程被选修的学生数select cno,count(*) from sc group by cno;
22、查询出只选修了一门课程的全部学生的学号和姓名select sno,sname from student where sno in
(select sno from sc group by sno having count(*)=1);
23、查询男生、女生人数select ssex,count(*) from student group by ssex;
24、查询姓“张”的学生名单select * from stuent where sname like '张%';
25、查询同名同姓学生名单，并统计同名人数思路：姓名分组，统计数量超过1select sname,count(*) from student group by sname having count(*)>1;
26、1995 年出生的学生名单select * from student where sage=to_char(sysdate,'YYYY')-1995;
27、查询每门课程的平均成绩，结果按平均成绩升序排列，平均成绩相同时，按课程号降序排列select avg(score) avgs from sc group by cno order by avgs asc,cno desc;
28、查询平均成绩大于65 的所有学生的学号、姓名和平均成绩select s.sno,s.sname,avg(score) from student s left join sc
on s.sno=sc.sno
group by s.sno,s.sname
having avg(score)>65;
29、查询课程名称为“oracle”，且分数低于60 的学生姓名和分数select s.sname,score from student s join sc on s.sno=sc.sno
where cno =
(select cno from course where cname='oracle')
and sc.score<60;
30、查询所有学生的选课情况，包括学生姓名，课程名称select s.sno,s.sname,c.cname from student
s join sc on s.sno=sc.sno
join course c on sc.cno=c.cno ;
31、查询任何一门课程成绩在70 分以上的学生姓名、课程名称和分数问题歧义，按照任何一个人，一门课程在70分以上就进行记录，语句如下select s.sname,c.cname,score from sc join student s on sc.sno=s.sno
join course c on sc.cno=c.cno
where score>=70;
按照一个同学，所有的课程都在70分以上才进行记录，使用分区函数32、查询不及格的课程，并按课程号从大到小排列,将课程号、课程名称、学生号打印出来select sc.cno,c.cname,sc.score from sc join course
33、查询课程编号为c001 且课程成绩在80 分以上的学生的学号和姓名34、求选了课程的学生人数select count(distinct sno) from sc;
35、查询选修“谌燕”老师所授课程的学生中，成绩最高的学生姓名及其成绩思路1、先查谌燕老师所教的所有课程；2、再将包含于这些课程的最高成绩按课程分组查询出来；3、根据cno和最高成绩两个匹配条件连接sc表，找出sno；4、关联学生表，找出sname；select s.sname,sc.score from sc join student s on sc.sno=s.sno
join (select sc.cno,max(score) maxs from sc join course c on sc.cno=c.cno
join teacher t on t.tno=c.tno
where tname='谌燕' group by sc.cno) e
on sc.cno=e.cno
where sc.score=e.maxs;
36、查询各个课程及相应的选修人数select cno,count(sno) from sc group by cno;
37、查询不同课程成绩相同的学生的学号、课程号、学生成绩select
from sc s1 join sc s2 on s1.score=s2.score
where s1.cno<>s2.cno;
38、查询每门功课成绩最好的前两名select cno,sno,score from
(select sno,cno,score,
row_number()over(partition by cno order by score desc)rn from sc)
where rn<=2;
39、统计每门课程的学生选修人数（超过10 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 cno,count(sno) from sc
group by cno having(sno)>10
order by count(sno) desc,cno asc
40、检索至少选修两门课程的学生学号1、根据学生分组，找出学生所学的课程2、统计学生课程的数量，将数量超过1的学生学号找出来select sno from sc group by sno having count(cno)>1;
41、查询学生选修的全部课程的课程号和课程名1、先从sc表中找出所有cno2、根据cno关联找出cnameselect cno,cname from course where cno in
(select distinct cno from sc);
42、查询没学过“谌燕”老师讲授的任一门课程的学生姓名1、找出谌燕老师教授的课程；2、学生学的cno in 1中的课程，将学生sno找出，取反，并关联student表查出姓名select sno,sname from student where sno in
(select sno from sc where cno not in
(select cno from course where tno in
(select tno from teacher where tname='谌燕')));
43、查询两门以上不及格课程的同学的学号及其平均成绩思路:1、先按学生分组，统计课程<60的学号，2、学号in (不及格的sno)，给出查询要求结果；select sno,avg(score) from sc where score<60
group by sno having count(*)>=2
44、检索“c002”课程分数小于90，按分数降序排列的同学学号select sno from sc where cno='c004'
and score<90
order by score desc;
45、删除“s002”同学的“c001”课程的成绩delete from sc where sno='s002' and cno='c001';

输入10个学生姓名、学号和成绩，将其中不及格者的姓名、学号和成绩输出#include<iostream> #include<iomanip>
#include<string>
//用N代表输入次数（学生人数），用name存储姓名，用num存储学号，用grade存储分数
//首先输入姓名，学号，分数；
//循环判断分数小于60的输出相应信息。
using namespace std;
#define N 10
//N代表输入学生数目
string name[10];
string num[10];
int grade[10];
int main()
{
int i;
for(i=0;i<N;i++)
cin>>name[i]>>num[i]>>grade[i];
for(i=0;i<N;i++)
if(grade[i]<60)
cout<<setw(5)<<name[i]<<setw(5)<<num[i]<<setw(5)<<grade[i]<<endl;
return 0;
}