MySQL数据库学习【第八篇】多表查询

 答案

1、查询所有的课程的名称以及对应的任课老师姓名
select cname 课程名称,tname 老师姓名 from course  left join teacher on course.teacher_id = teacher.tid;

2、查询学生表中男女生各有多少人
select gender ,count(gender) from student group by gender;

3、查询物理成绩等于100的学生的姓名
select sname from student inner join score on student.sid = score.student_id
join course on score.course_id = course.cid
where cname = ‘物理‘ and num = 100;

4、查询平均成绩大于八十分的同学的姓名和平均成绩
(1):
select sname 姓名,avg(num) 平均成绩 from student inner join score on student.sid = score.student_id 
group by student_id having avg(num)>80;
(2):
select sname,avg(num) from student inner join score on student.sid = score.student_id 
group by student_id having avg(num)>80

5、查询所有学生的学号，姓名，选课数，总成绩
select student.sid 学号,sname 姓名,count(course_id) 选课数,sum(num) 总成绩 from student left join score on student.sid = score.student_id
group by student.sid ;
6、 查询姓李老师的个数
select count(tid) from teacher where tname like ‘李%‘;

7、 查询没有报李平老师课的学生姓名
select sname 姓名 from student where sid not in (
select student_id from score where course_id in(
select cid from course where teacher_id =(
select tid from teacher where tname = ‘李平老师‘
)));

8、 查询物理课程比生物课程高的学生的学号
select t1.student_id  学号 from 
(select student_id ,num from score inner join course on score.course_id=course.cid  where cname=‘物理‘ )as t1
inner join 
(select student_id , num from score inner join course on score.course_id=course.cid  where cname = ‘生物‘) as t2
on t1.student_id = t2.student_id
where t1.num>t2.num;



9、 查询没有同时选修物理课程和体育课程的学生姓名
(1):
select sname from student inner join score on student.sid = score.student_id
join course on course.cid=score.course_id and cname in (‘物理‘,‘体育‘) 
group by student_id having count(course_id)!=2;

(2):
select sname from student join score  on student.sid = score.student_id
join course on course.cid=score.course_id where cname =‘物理‘ or cname= ‘体育‘
group by student_id having count(course_id) !=2;

10、查询挂科超过两门(包括两门)的学生姓名和班级
select sname 姓名,caption 班级 from student inner join score on student.sid = score.course_id
join class on class.cid = score.course_id
where num<60 group by student_id having count(course_id)>=2;

11、查询选修了所有课程的学生姓名
(1)
select sname 姓名 ,所有的课程数 from student inner join 
(select student_id,count(course_id) 所有的课程数 from score group by student_id having count(course_id) = (
select count(cid) from course)) as t1
on t1.student_id = student.sid;

(2)
select sname,count(course_id) from student inner join score on student.sid = score.student_id
group by student_id having count(course_id)=(select count(cid) from course);

12、查询李平老师教的课程的所有成绩记录
(1):
select num from score inner join course on course.cid=score.course_id
join teacher on teacher.tid=course.teacher_id
where tname = ‘李平老师‘;

(2):
select num from score where course_id in(select cid from course where teacher_id=
(select tid from teacher where tname=‘李平老师‘));

13、查询全部学生都选修了的课程号和课程名
select cid 课程号,cname 课程名 from course
select ;
14、查询每门课程被选修的次数 
(1)
select course.cname,count(student_id) 选课人数 from score inner join course on score.course_id=course.cid
group by course_id;

(2):也可以按照name分组
select course.cname,count(student_id) 选课人数 from score inner join course on score.course_id=course.cid
group by cname;

15、查询之选修了一门课程的学生姓名和学号
select sname 姓名,student_id 学号 from student inner join score on student.sid = score.student_id
group by student_id having count(course_id)=1;

16、查询所有学生考出的成绩并按从高到低排序（成绩去重）
select distinct num from score order by num desc;

17、查询平均成绩大于85的学生姓名和平均成绩
(1):
select sname 姓名,avg(num) 平均成绩 from student inner join score on student.sid = score.student_id
group by student_id having avg(num)>85;

(2):
select student.sname,avg_num from student inner join
(select student_id,avg(num) as avg_num from score group by student_id having avg(num) > 85
) t1
on student.sid=t1.student_id;


18、查询生物成绩不及格的学生姓名和对应生物分数
(1):
select student.sname ,num 生物成绩 from student inner join score on student.sid = score.student_id
join course on course.cid=score.course_id
where cname=‘生物‘ and num<60;

(2):
select student.sname,t1.num from student inner join
(
select student_id,num from score
where course_id=(select cid from course where cname=‘生物‘) and num < 60
) t1
on t1.student_id=student.sid
;

19、查询在所有选修了李平老师课程的学生中，这些课程(李平老师的课程，不是所有课程)平均成绩最高的学生姓名
select sname from student where sid=(
select student_id from score where course_id in (
select cid from course where teacher_id=(select tid from teacher where tname=‘李平老师‘)
) group by student_id order by avg(num) desc limit 1
)

MySQL数据库学习【第八篇】多表查询

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