MySQL 练习题附答案，未完

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标签：mic class ble 课程 core 没有 ber -418 相同

综合练习题

表结构

整合一下方便查看

teacher

技术图片

student

course

技术图片

scors

技术图片

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;

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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;

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5、查询姓“长”的老师的个数；

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

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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

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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

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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‘)

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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

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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

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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)

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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‘

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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

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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‘;

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25、查询姓“张”的学生名单；

select * from student where sname like ‘萌%‘

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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 sname

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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

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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

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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

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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

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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‘

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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

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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)

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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

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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

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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

MySQL 练习题附答案，未完

标签：mic class ble 课程 core 没有 ber -418 相同

原文地址：https://www.cnblogs.com/shijieli/p/10408638.html

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周排行

MySQL 练习题 附答案，未完

综合练习题

表结构

整合一下方便查看

练习题

MySQL 练习题附答案，未完