mysql的一些常用操作二

紧跟上一节，我们创建了四个表：

Student、Teacher、Course、Score

接下来就是实际的一些操作了：
1.求每门课程的学生人数。

select course.cname '课程名称',count(*) '人数' from score,course where score.CId=course.CId group by score.CId

 2.查询课程编号为 01 且课程成绩在 80 分及以上的学生的学号和姓名

select a.sid,a.sname from Student a,Score b where a.sid=b.sid and b.cid='01' and b.scoreore >=80;

3.统计每门课程的学生选修人数（超过 5 人的课程才统计）

select b.cname '课程',count(*) from course a,score b where a.cid=b.cid group by a.cid having count(*)>5;

 

4.检索至少选修两门课程的学生学号

select sid from score group by sid having count(cid)>=2;

5.选修了全部课程的学生信息

select a.* from student a,score b where a.sid=b.sid group by a.sid having count(cid)=(select count(*) from course);

 6 .查询存在不及格的课程

select  distinct a.* from course a,score b where a.cid=b.cid and b.scoreore<60;

 7.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

select a.sname,b.cname,c.scoreore from student a,course b,score c where a.sid=c.sid and b.cid=c.cid and c.scoreore>70;

8.查询所有学生的课程及分数情况（存在学生没成绩，没选课的情况

select a.sname,c.scoreore from student a left join score c on a.sid=c.sid left join course b on b.cid=c.cid;

 9.查询课程名称为「数学」，且分数低于 60 的学生姓名和分数

select a.sname,score c where a.sid=c.sid and b.cid=c.cid and b.cname="数学" and c.scoreore<60;

 

 10.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 a.sid,a.sname,avg(b.scoreore) from student a,score b where a.sid=b.sid group by a.sid having avg(b.scoreore)>=85;

 11.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 cid,avg(scoreore) from score group by cid order by avg(scoreore) desc,cid asc;

12.查询各科成绩最高分、最低分和平均分

以如下形式显示：课程 ID，课程 name，最高分，最低分，平均分，及格率，中等率，优良率，优秀率

及格为>=60，中等为：70-80，优良为：80-90，优秀为：>=90

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 score.cid,Course.Cname,max(score.scoreore),min(score.scoreore),AVG(score.scoreore),count(score.SId),sum(case when scoreore <60 then 1 else 0 end)/count(score.cid) '不合格率',sum(case when scoreore <80 and scoreore >=60 then 1 else 0 end)/count(score.cid) '合格率',sum(case when scoreore <90 and scoreore >=80 then 1 else 0 end)/count(score.cid) '优良率',sum(case when scoreore >= 90  then 1 else 0 end)/count(score.cid) '优秀率'
from score,Course
where score.cid =  Course.cid
GROUP BY score.cid;

 13.查询男生、女生人数

select ssex,count(*) from student group by ssex;

 14.检索" 01 "课程分数小于 60，按分数降序排列的学生信息

select a.*,b.scoreore from student a,score b where a.sid=b.sid and b.cid="01" and b.scoreore<60 order by scoreore desc;

 15.按平均成绩从高到低显示所有学生的所有课程的成绩以及平均成绩

select a.sid,a.scoreore,b.`平均成绩` from score a right join (select sid,avg(scoreore) '平均成绩'from score group by sid) b

on a.sid = b .sid

ORDER BY b.`平均成绩` desc;

 16.查询没学过"张三"老师讲授的任一门课程的学生姓名

select distinct student.sname from student where student.sid not in

(select c.sid from teacher a,score c where a.tid=b.tid and b.cid=c.cid and a.tname="张三") ;

17.成绩不重复，查询选修「张三」老师所授课程的学生中，成绩最高的学生信息及其成绩

select  d.*,c.scoreore from teacher a,score c,student d

where a.tid=b.tid and b.cid=c.cid and d.sid=c.sid and tname="张三" 

order by c.scoreore desc limit 1;

 18.成绩有重复的情况下，查询选修「张三」老师所授课程的学生中，成绩最高的学生信息及其成绩

目前我们表中没有重复的成绩，先修改一下:update score set score=90 where skey=17;

若是取出张三老师下的所有学生信息和成绩，即：

select  d.*,student d

where a.tid=b.tid and b.cid=c.cid and d.sid=c.sid and tname="张三" 

order by c.scoreore desc;则有:

select  d.*,student d

where a.tid=b.tid and b.cid=c.cid and d.sid=c.sid and tname="张三" and scoreore =

(select max(scoreore) from teacher a,score c where a.tid=b.tid and b.cid=c.cid and a.tname="张三");

 

19.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 any_value(a.sid),any_value(a.cid),any_value(a.scoreore) from score a

inner join score b on a.sid=b.sid and a.cid!=b.cid and a.scoreore=b.scoreore group by a.cid,b.cid;

20.查询每门功成绩最好的前两名

select any_value(a.sid),any_value(a.scoreore) from score a left join score as b 
on a.cid = b.cid
and a.scoreore < b.scoreore
group by a.cid,a.sid
having count(b.scoreore)<2
order by a.cid;

 21.查询每门课程被选修的学生数

select cid,count(*) from score group by cid;

 22.查询出只选修两门课程的学生学号和姓名

select a.sid,a.sname from student a,score b where a.sid=b.sid group by a.sid having count(*)=2;

 23.查询同名学生名单，并统计同名人数

select sname,count(*) from student group by sname having count(*)>1;

 24.查询 1990 年出生的学生名单

select sname from student where year(sage)=1990;

 25.查询各学生的年龄.

select sid,sname,timestampdiff(year,sage,curdate()) from student;

 26.查询本周过生日的学生

select sname from student where week(curdate())=week(sage);

没有学生；

27.查询本月过生日的学生

select sname from student where month(curdate())=month(sage);

没有学生;

28.查询「李」姓老师的数量

select count(*) from teacher where tname like "李%";

 29.查有成绩的学生信息

select * from student where sid in (select sid from score);

 30.查询所有同学的学生编号、学生姓名、选课总数、所有课程的成绩总和

select a.SId,a.Sname,count(b.cid),sum(b.scoreore) from student a left join score b on a.sid = b.sid GROUP BY a.sid;

 31.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 a.sid,score b where a.sid=b.sid having avg(b.scoreore)>60;

 32.查询不存在" 01 "课程但存在" 02 "课程的情况

select * from score where cid = '02' and sid not in (select sid from score where cid='01');

 33.查询存在" 01 "课程但可能不存在" 02 "课程的情况

select * from score where cid="01";

 34.按各科成绩进行排序，并显示排名， Score 重复时保留名次空缺

select a.cid,a.sid,any_value(a.scoreore),count(b.sid)+1 from score a left join score b 
on a.cid=b.cid 
and a.scoreore<b.scoreore 
GROUP BY  a.cid,a.sid
order by a.cid,count(b.sid)+1; 

 35.查询" 01 "课程比" 02 "课程成绩高的学生的信息及课程分数

select c.*,b.scoreore from student c,(select scoreore,sid from score where cid = '01')a,sid from score where cid = '02')b
where a.sid = b.sid 
and c.sid=b.sid
and a.scoreore >b.scoreore;

 36.查询没有学全所有课程的同学的信息

select a.* from Student a left join score b 
on a.sid = b.sid 
group by a.SId
having count(b.cid)<(select count(cname) from Course);

37.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 a.* from student a,score b where a.sid = b.sid and b.cid in 
(select cid from score where sid ='01')
group by a.sid;

38.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 * from Student where sid in
(select  sid from score where sid not in 
(select  sid from score where  cid not in 
(select cid from score where sid ='01'))
GROUP BY sid 
having count(*)=(select count(cid) from score where sid ='01')and sid<>'01');

 首先，筛选出01同学的课程——选出有谁没有选择01同学课程的学生sid——再对上一层逻辑做否定，选择了01号同学的子集或者含有01号同学所选课程之外的含有其他课程的 同学。因此最后再加上判断条件，课程数量相等。就可以判断是和01号同学学习相同的课程。