MySQL练习题

参考：

1：表格结构及参考答案

2：练习题博客

3：B站视频讲解

1、2两个连接很重要，第3个参考是视频练习题讲解。

导入表格数据可以新建一个数据库后用命令直接导入，mysqldump -u root -d db1 < db2.sql -p #db2.sql数据写入db1中（db1提前需要建立好）。或者看着练习题博客中的表格建立后，插入数据，然后对着参考1中实际插入的数据使用update table set 列名='XXX' from XXX;来改一下，数据量大就使用可视化操作来编辑即可，主要是练习要会写。

#2 查询“生物”课程比“物理”课程成绩高的所有学生的学号
# 第一步应当查询上过生物课程的学生分数
#select score.sid,score.student_id,course.cname,score.number from score left join course on score.course_id = course.cid where course.cname='生物';
# 第二步查物理课程成绩分数
#select score.sid,score.student_id,course.cname,score.number from score left join course on score.course_id = course.cid where course.cname='物理';
# 第三步，将一、二步联合起来，进行比较分数,根据学生id连接，这样就相当于上过生物、物理课程的学生
select A.student_id from 
(select score.sid,score.student_id,course.cname,score.number from score left join course on score.course_id = course.cid where course.cname='生物') as A
inner join
(select score.sid,score.student_id,course.cname,score.number from score left join course on score.course_id = course.cid where course.cname='物理') as B
on A.student_id = B.student_id where A.number > B.number; 

#3 查询平均成绩大于60分的同学的学号和平均成绩 首先分组 利用聚合函数avg和having
select student_id,avg(number) from score group by student_id having avg(number) > 60;

#3 问题3扩充，加一个联表，看学生名字，会用一个临时表B和as来表示列名
select B.student_id,student.sname,B.平均成绩 from (select student_id,avg(number) as 平均成绩 from score group by student_id having avg(number) > 60) as B 
left join student on B.student_id = student.sid;

#4 查询所有同学的学号、姓名、选课数、总成绩 两种方法查询 核心一样 第一种查询和问题3扩充类似 第二种查询老师写的简单 都要经过联表、分组、聚合
    #第一种
select student.sid,student.sname,B.选课数,B.总成绩 from  (select student_id,count(student_id) as 选课数,sum(number) as 总成绩 from score group by student_id) as B
left join student on student.sid = B.student_id;

    #第二种
select score.student_id,student.sname,count(student_id),sum(number) from score left join student on score.student_id = student.sid group by score.student_id;

#5 查询姓李老师的个数，模糊查询 这个好理解，%代表任意 _则是一位
select count(tname) as 姓李老师个数 from teacher  where tname like('李%'); 

#6 查询没学过"李平"老师课的同学的学号、姓名
#思路 找学过李平老师的所有学生，进行分组后关联学生表查询
#看视频比看文字好
select student.sid,student.sname from student where sid not in (#利用学生表来反向选择没有上过李平老师课程的
select #这里筛选的是的是上过李平老师课程的
score.student_id
from score 
where course_id  in (
    select course.cid #这里筛选的是李平老师所带课程
    from course 
    left join teacher 
    on course.teacher_id = teacher.tid 
    where teacher.tname = "李平老师"
    )group by student_id
);

#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 A.sid1,A.sname1 from#第三个将查出的两个表整合起来比较筛选
(select student.sid as sid1,student.sname as sname1,score.course_id as course_id1 from student left join score on student.sid = score.student_id where score.course_id = 1) as A
#第一个先查出学过1课程的学生
inner join
#第二个查出学过2课程的学生
(select student.sid as sid2,student.sname as sname2,score.course_id as course_id2 from student left join score on student.sid = score.student_id where score.course_id = 2) as B
on A.sid1 = B.sid2 where A.course_id1 = 1 and B.course_id2 = 2;

# 8查询学过“张磊老师“所教的所有课的同学的学号、姓名；
select student.sid,student.sname from student left join score on student.sid = score.student_id
where score.course_id in(select teacher.tid from course left join teacher on teacher.tid = course.teacher_id where teacher.tname = '张磊老师');

#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
# 这题和第7题思路一样
select A.sid1,A.sname1 from#第三个将查出的两个表整合起来比较筛选
(select student.sid as sid1,student.sname as sname1,score.number as number1 from student left join score on student.sid = score.student_id where score.course_id = 1) as A
#第一个先查出学过1课程的学生
inner join
#第二个查出学过2课程的学生
(select student.sid as sid2,student.sname as sname2,score.number as number2 from student left join score on student.sid = score.student_id where score.course_id = 2) as B
on A.sid1 = B.sid2 where A.number1 > B.number2;

# 10查询有课程成绩小于60分的同学的学号、姓名; 这题我自己要注意分组和where顺序
select student.sid,student.sname from student left join score on score.student_id = student.sid  where number < 60 group by sid;

#24 查询男生、女生人数，首先分组,gender中男女就各自分为一组，然后聚合函数统计个数
select gender,count(gender) from student group by gender;