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<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;
题目:
代码:
#include<iostream>
using namespace std;
int main(){//一、分析问题//已知:10 个苹果到地面的高度a[10],陶陶把手伸直的时候能够达到的最大高度height//未知:陶陶能够摘到的苹果的数目sum。//关系ÿ…
目录结构 注:提前言明 本文借鉴了以下博主、书籍或网站的内容,其列表如下: 1、参考书籍:《Oracle Database SQL Language Reference》 2、参考书籍:《PostgreSQL中文手册》 3、EDB Postgres Advanced Server User Gui…
//轨道炮 #include<iostream>
using namespace std;
#include<algorithm>
int logs[100010];
int main()
{int n;cin >> n;for (int i 1;i < n;i){cin >> logs[i];}sort(logs 1, logs n 1);int ans 1000000000;for (int i 2;i < n;i){if (…