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MySQL的练习题及答案
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导出现有数据库数据：
- mysqldump -u用户名 -p密码 数据库名称 >导出文件路径 # 结构+数据
- mysqldump -u用户名 -p密码 -d 数据库名称 >导出文件路径 # 结构
导入现有数据库数据：
- mysqldump -uroot -p密码 数据库名称 < 文件路径
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2、查询“生物”课程比“物理”课程成绩高的所有学生的学号；
思路：
获取所有有生物课程的人（学号，成绩） - 临时表
获取所有有物理课程的人（学号，成绩） - 临时表
根据【学号】连接两个临时表：
学号 物理成绩 生物成绩
然后再进行筛选
select A.student_id,sw,ty from
(select student_id,num as sw from score left join course on score.course_id = course.cid where course.cname = ‘生物‘) as A
left join
(select student_id,num as ty from score left join course on score.course_id = course.cid where course.cname = ‘体育‘) as B
on A.student_id = B.student_id where sw > if(isnull(ty),0,ty);
3、查询平均成绩大于60分的同学的学号和平均成绩；
思路：
根据学生分组，使用avg获取平均值，通过having对avg进行筛选
select student_id,avg(num) from score group by student_id having avg(num) > 60
4、查询所有同学的学号、姓名、选课数、总成绩；
select score.student_id,sum(score.num),count(score.student_id),student.sname
from
score left join student on score.student_id = student.sid
group by score.student_id
5、查询姓“李”的老师的个数；
select count(tid) from teacher where tname like ‘李%‘
select count(1) from (select tid from teacher where tname like ‘李%‘) as B
6、查询没学过“叶平”老师课的同学的学号、姓名；
思路：
先查到“李平老师”老师教的所有课ID
获取选过课的所有学生ID
学生表中筛选
select * from student where sid not in (
select DISTINCT student_id from score where score.course_id in (
select cid from course left join teacher on course.teacher_id = teacher.tid where tname = ‘李平老师‘
)
)
7、查询学过“001”并且也学过编号“002”课程的同学的学号、姓名；
思路：
先查到既选择001又选择002课程的所有同学
根据学生进行分组，如果学生数量等于2表示，两门均已选择
select student_id,sname from
(select student_id,course_id from score where course_id = 1 or course_id = 2) as B
left join student on B.student_id = student.sid group by student_id HAVING count(student_id) > 1
8、查询学过“叶平”老师所教的所有课的同学的学号、姓名；
同上，只不过将001和002变成 in (叶平老师的所有课)
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；
同第1题
10、查询有课程成绩小于60分的同学的学号、姓名；
select sid,sname from student where sid in (
select distinct student_id from score where num < 60
)
11、查询没有学全所有课的同学的学号、姓名；
思路：
在分数表中根据学生进行分组，获取每一个学生选课数量
如果数量 == 总课程数量，表示已经选择了所有课程
select student_id,sname
from score left join student on score.student_id = student.sid
group by student_id HAVING count(course_id) = (select count(1) from course)
12、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；
思路：
获取 001 同学选择的所有课程
获取课程在其中的所有人以及所有课程
根据学生筛选，获取所有学生信息
再与学生表连接，获取姓名
select student_id,sname, count(course_id)
from score left join student on score.student_id = student.sid
where student_id != 1 and course_id in (select course_id from score where student_id = 1) group by student_id
13、查询至少学过学号为“001”同学所有课的其他同学学号和姓名；
先找到和001的学过的所有人
然后个数 ＝ 001所有学科 ＝＝》 其他人可能选择的更多
select student_id,sname, count(course_id)
from score left join student on score.student_id = student.sid
where student_id != 1 and course_id in (select course_id from score where student_id = 1) group by student_id having count(course_id) ＝ (select count(course_id) from score where student_id = 1)
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；
个数相同
002学过的也学过
select student_id,sname from score left join student on score.student_id = student.sid where student_id in (
select student_id from score where student_id != 1 group by student_id HAVING count(course_id) = (select count(1) from score where student_id = 1)
) and course_id in (select course_id from score where student_id = 1) group by student_id HAVING count(course_id) = (select count(1) from score where student_id = 1)
15、删除学习“叶平”老师课的score表记录；
delete from score where course_id in (
select cid from course left join teacher on course.teacher_id = teacher.tid where teacher.name = ‘叶平‘
)
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；
思路：
由于insert 支持
inset into tb1(xx,xx) select x1,x2 from tb2;
所有，获取所有没上过002课的所有人，获取002的平均成绩
insert into score(student_id, course_id, num) select sid,2,(select avg(num) from score where course_id = 2)
from student where sid not in (
select student_id from score where course_id = 2
)
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；
select sc.student_id,
(select num from score left join course on score.course_id = course.cid where course.cname = "生物" and score.student_id=sc.student_id) as sy,
(select num from score left join course on score.course_id = course.cid where course.cname = "物理" and score.student_id=sc.student_id) as wl,
(select num from score left join course on score.course_id = course.cid where course.cname = "体育" and score.student_id=sc.student_id) as ty,
count(sc.course_id),
avg(sc.num)
from score as sc
group by student_id desc
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；
select course_id, max(num) as max_num, min(num) as min_num from score group by course_id;
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；
思路：case when .. then
select course_id, avg(num) as avgnum,sum(case when score.num > 60 then 1 else 0 END)/count(1)*100 as percent from score group by course_id order by avgnum asc,percent desc;
20、课程平均分从高到低显示（现实任课老师）；
select avg(if(isnull(score.num),0,score.num)),teacher.tname from course
left join score on course.cid = score.course_id
left join teacher on course.teacher_id = teacher.tid
group by score.course_id
21、查询各科成绩前三名的记录:(不考虑成绩并列情况)
select score.sid,score.course_id,score.num,T.first_num,T.second_num from score left join
(
select
sid,
(select num from score as s2 where s2.course_id = s1.course_id order by num desc limit 0,1) as first_num,
(select num from score as s2 where s2.course_id = s1.course_id order by num desc limit 3,1) as second_num
from
score as s1
) as T
on score.sid =T.sid
where score.num <= T.first_num and score.num >= T.second_num
22、查询每门课程被选修的学生数；
select course_id, count(1) from score group by course_id;
23、查询出只选修了一门课程的全部学生的学号和姓名；
select student.sid, student.sname, count(1) from score
left join student on score.student_id = student.sid
group by course_id having count(1) = 1
24、查询男生、女生的人数；
select * from
(select count(1) as man from student where gender=‘男‘) as A ,
(select count(1) as feman from student where gender=‘女‘) as B
25、查询姓“张”的学生名单；
select sname from student where sname like ‘张%‘;
26、查询同名同姓学生名单，并统计同名人数；
select sname,count(1) as count from student group by sname;
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 course_id,avg(if(isnull(num), 0 ,num)) as avg from score group by course_id order by avg asc,course_id desc;
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 student_id,sname, avg(if(isnull(num), 0 ,num)) from score left join student on score.student_id = student.sid group by student_id;
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 student.sname,score.num from score
left join course on score.course_id = course.cid
left join student on score.student_id = student.sid
where score.num < 60 and course.cname = ‘生物‘
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 * from score where score.student_id = 3 and score.num > 80
31、求选了课程的学生人数
select count(distinct student_id) from score
select count(c) from (
select count(student_id) as c from score group by student_id) as A
32、查询选修“杨艳”老师所授课程的学生中，成绩最高的学生姓名及其成绩；
select sname,num from score
left join student on score.student_id = student.sid
where score.course_id in (select course.cid from course left join teacher on course.teacher_id = teacher.tid where tname=‘张磊老师‘) order by num desc limit 1;
33、查询各个课程及相应的选修人数；
select course.cname,count(1) from score
left join course on score.course_id = course.cid
group by course_id;
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；
select DISTINCT s1.course_id,s2.course_id,s1.num,s2.num from score as s1, score as s2 where s1.num = s2.num and s1.course_id != s2.course_id;
35、查询每门课程成绩最好的前两名；
select score.sid,score.course_id,score.num,T.first_num,T.second_num from score left join
(
select
sid,
(select num from score as s2 where s2.course_id = s1.course_id order by num desc limit 0,1) as first_num,
(select num from score as s2 where s2.course_id = s1.course_id order by num desc limit 1,1) as second_num
from
score as s1
) as T
on score.sid =T.sid
where score.num <= T.first_num and score.num >= T.second_num
36、检索至少选修两门课程的学生学号；
select student_id from score group by student_id having count(student_id) > 1
37、查询全部学生都选修的课程的课程号和课程名；
select course_id,count(1) from score group by course_id having count(1) = (select count(1) from student);
38、查询没学过“叶平”老师讲授的任一门课程的学生姓名；
select student_id,student.sname from score
left join student on score.student_id = student.sid
where score.course_id not in (
select cid from course left join teacher on course.teacher_id = teacher.tid where tname = ‘张磊老师‘
)
group by student_id
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 student_id,count(1) from score where num < 60 group by student_id having count(1) > 2
40、检索“004”课程分数小于60，按分数降序排列的同学学号；
select student_id from score where num< 60 and course_id = 4 order by num desc;
41、删除“002”同学的“001”课程的成绩；
delete from score where course_id = 1 and student_id = 2
|
MySQL的练习题及答案
标签：均值 x64 int 数据 物理 col having end number
本文系统来源：https://www.cnblogs.com/xied/p/12500107.html
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