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MySQL 练习题 附答案，未完
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综合练习题
表结构
整合一下方便查看
teacher
student
course
scors
练习题
1、自行创建测试数据
create table student(
sid int primary key auto_increment,
sname char(32),
gender enum("女","男"),
class_id int,
constraint fk_class_id foreign key (class_id) references class(cid)
on DELETE cascade
on update cascade
);
create table teacher(
tid int primary key auto_increment,
tname char(32)
)；
create table class(
cid int primary key auto_increment,
caption char(32)
)；
create table course(
cid int primary key auto_increment,
cname char(32),
teacher_id int,
constraint fk_teacher_id foreign key (teacher_id) references teacher(tid)
on delete cascade
on update cascade
)；
create table score(
sid int primary key auto_increment,
student_id int,
corse_id int,
number int,
constraint fk_student_id foreign key (student_id) references student(sid)
on delete cascade
on update cascade,
constraint fk_corse_id foreign key (corse_id) references score(sid)
on delete cascade
on update cascade
);
View Code
2、查询“体育”课程比“数学”课程成绩高的所有学生的学号；
select a.student_id
from (select * from score where corse_id = (select cid from course where cname = '数学'))a
inner join (select * from score where corse_id = (select cid from course where cname = '体育'))b
on a.student_id = b.student_id and a.number > b.number;
View Code
select a.student_id
from
(select score.student_id, score.number
from score
left join course c on score.corse_id = c.cid
where c.cname = '体育') a
inner join
(select score.student_id, score.number
from score
left join course c on score.corse_id = c.cid
where c.cname = '数学') b
on a.student_id = b.student_id and a.number < b.number;
View Code
3、查询平均成绩大于60分的同学的学号和平均成绩；
select student_id, AVG(number)AS avg from score group by student_id having avg > 60;View Code
4、查询所有同学的学号、姓名、选课数、总成绩；
select student.sid, sname, count(*),sum(number)
from student
left join score s on student.sid = s.student_id
group by student.sid;
View Code
5、查询姓“长”的老师的个数；
select count(*) from teacher where tname like "长%" ;View Code
6、查询没学过“教授”老师课的同学的学号、姓名；
select s.sid,s.sname
from score
inner join student s on score.student_id = s.sid
inner join course c on score.corse_id = c.cid
inner join teacher t on c.teacher_id = t.tid
where t.tname <> '教授'
group by s.sid
View Code
7、查询学过“1”并且也学过编号“2”课程的同学的学号、姓名；
select s.sid,s.sname
from score
inner join student s on score.student_id = s.sid
inner join course c on score.corse_id = c.cid
where c.cid in (1,2)
group by s.sid
View Code
8、查询学过“叶平”老师所教的所有课的同学的学号、姓名；
select distinct s.sname
from score
left join student s on score.student_id = s.sid
where corse_id not in (select cid from course
left join teacher t on course.teacher_id = t.tid where tname = '教授')
View Code
9、查询课程编号“002”的成绩比课程编号“001”课程低的所有同学的学号、姓名；
思路 1 拿到成绩表中，课程为1 的记录 2 成绩表链表学生表 作为一个记录 3 重复上面步骤拿到课程为2 的链表记录 两个记录再基于相同的学生id链表。条件设置为 成绩比较
select distinct a.sid,a.sname
from
(select distinct sc.number, st.sname, st.sid
from score as sc
inner join student st on sc.student_id = st.sid
where corse_id = 2) a
inner join
(select distinct sc.number, st.sname, st.sid
from score as sc
inner join student st on sc.student_id = st.sid
where corse_id = 1) b
on a.sid = b.sid
where a.number < b.number
View Code
10、查询有课程成绩小于60分的同学的学号、姓名；
select distinct st.sid, st.sname
from student as st
inner join score s on st.sid = s.student_id
where s.number < 60
View Code
11、查询没有学全所有课的同学的学号、姓名；
思路 1 查询出来所有的课程然后计数 2 成绩表按照学生分组后统计记录个数 3 小于总课程个数即为没学全 4 相反 等于 则为学全
select st.sid, st.sname
from student as st
inner join score sc on st.sid = sc.student_id
group by st.sid
having count(*) < (select count(cid) from course)
View Code
12、查询至少学过学号为“001”同学所选课程中任意一门课的其他同学学号和姓名；
思路
1 先拿到 1 同学所学的所有课程的id
1.1 要链表 student 和 course 以及 score
2 将1同学的课程列表当做条件来判断是否在里面即可
2.1 因为要查其他学生，排除掉 1 同学
from student as st
inner join score sc on st.sid = sc.student_id
inner join course as co on sc.corse_id = co.cid
where st.sid <> 1
and sc.corse_id in
(select co.cid
from student as st
inner join score sc on st.sid = sc.student_id
inner join course as co on sc.corse_id = co.cid
where st.sid = 1)
View Code
13、查询至少有一门课与学号为“001”的同学所学相同的同学的学号和姓名；
同上，只是不需要再排除 1 同学
14、查询和“002”号的同学学习的课程完全相同的其他同学学号和姓名；
同上上，再条件判断的时候改为完全等于
15、删除学习“长老”老师课的成绩表记录；
delete sc
from score as sc
inner join course as co on sc.sid = co.cid
inner join teacher te on co.teacher_id = te.tid
where tname = '长老'
View Code
16、向SC表中插入一些记录，这些记录要求符合以下条件：①没有上过编号“002”课程的同学学号；②插入“002”号课程的平均成绩；
17、按平均成绩从低到高显示所有学生的“语文”、“数学”、“英语”三门的课程成绩，按如下形式显示： 学生ID,语文,数学,英语,有效课程数,有效平均分；
18、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；
19、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；
20、课程平均分从高到低显示（显示任课老师）；
select co.cname, te.tname, avg(sc.number) av
from score as sc,
course as co,
teacher as te
where sc.corse_id = co.cid
and co.teacher_id = te.tid
group by sc.corse_id
order by av desc
View Code
21、查询各科成绩前三名的记录:(不考虑成绩并列情况)
22、查询每门课程被选修的学生数；
23、查询出只选修了一门课程的全部学生的学号和姓名；
24、查询男生、女生的人数；
select count(*) from student where gender ='男'; select count(*) from student where gender ='女';View Code
25、查询姓“张”的学生名单；
select * from student where sname like '萌%'View Code
26、查询同名同姓学生名单，并统计同名人数；
select sname,count(sname) from student group by snameView Code
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 co.cname, avg(sc.number) as av
from score as sc,
course as co
where sc.corse_id = co.cid
group by sc.corse_id
order by av asc, sc.corse_id desc
View Code
28、查询平均成绩大于85的所有学生的学号、姓名和平均成绩；
select st.sname, avg(sc.number)
from student as st,
score as sc
where st.sid = sc.student_id
group by st.sid
View Code
29、查询课程名称为“数学”，且分数低于60的学生姓名和分数；
select st.sname, avg(sc.number) as av
from student as st
inner join score as sc on st.sid = sc.student_id
inner join course as co on sc.corse_id = co.cid
where co.cname = '数学'
group by st.sid
having av < 60
View Code
30、查询课程编号为003且课程成绩在80分以上的学生的学号和姓名；
select st.sid, st.sname
from student as st
inner join score sc on st.sid = sc.student_id
where sc.corse_id = 3
and sc.number > 80
View Code
31、求选了编号 ‘1 ’课程的学生人数
select count(*)
from student as st
inner join score sc on st.sid = sc.student_id
where sc.corse_id = 1
View Code
32、查询选修“长老”老师所授课程的学生中，成绩最高的学生姓名及其成绩；
select st.sid, st.sname, max(sc.number) as max
from student as st,
score as sc,
course as co,
teacher as te
where sc.corse_id = co.cid
and st.sid = sc.student_id
and co.teacher_id = te.tid
and te.tname = '长老'
View Code
33、查询各个课程及相应的选修人数；
select co.cname,count(*)
from score as sc
inner join course as co on sc.corse_id = co.cid
group by sc.corse_id
View Code
34、查询不同课程但成绩相同的学生的学号、课程号、学生成绩；
35、查询每门课程成绩最好的前两名；
36、检索至少选修两门课程的学生学号；
37、查询全部学生都选修的课程的课程号和课程名；
select co.cid, co.cname
from student as st,
score as sc,
course as co
where sc.corse_id = co.cid
and st.sid = sc.student_id
group by sc.corse_id
having count(*) = (select count(*) from student)
View Code
38、查询没学过“长老”老师讲授的任一门课程的学生姓名；
39、查询两门以上不及格课程的同学的学号及其平均成绩；
select st.sid, st.sname, avg(sc.number) as av
from student as st,
score as sc,
course as co
where sc.corse_id = co.cid
and st.sid = sc.student_id
and sc.number < 60
having count(*) > 2
View Code
40、检索“003”课程分数小于60，按分数降序排列的同学学号；
select st.sid
from score as sc
inner join course as co on sc.corse_id = co.cid
inner join student as st on sc.student_id = st.sid
where sc.number < 60
order by sc.number desc
View Code
41、删除“002”同学的“001”课程的成绩；
delete sc
from score as sc,
student as st
where sc.student_id = st.sid
and st.sid = 2
and sc.corse_id = 1
View Code
内容总结
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