MySQL 40题练习题和答案
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, a, b
from
(select score.student_id, number as a from score left join course on course.cid = score.corse_id where cname='生物') as A
left join
(select score.student_id, number as b from score left join course on course.cid = score.corse_id where cname='物理') as B
on
A.student_id = B.student_id where a > if(isnull(b),0,b);
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(number) from score group by student_id having avg(number)>60;
4、查询所有同学的学号、姓名、选课数、总成绩;
select score.student_id, student.sname, count(score.corse_id), sum(score.number)
from
score
left join
student on student.sid = student_id
group by
student_id;
select count(tid) from teacher where tname like '波%';
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.corse_id in (
select cid from course left join teacher on course.teacher_id = teacher.tid where tname='波多')
);
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
法一:
select student.sid, student.sname from
(select student_id, corse_id from score where corse_id=1 or corse_id=2) as B
left join student on B.student_id = student.sid group by student_id having count(student_id)>1;
法二:
select student.sid, student.sname from score left join student on student.sid= student_id
where corse_id= 1 or corse_id=2 group by student_id having count(corse_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
思路:
同上,只不过将1和2变成in(老师课程)
select student_id, student.sname from score left join student on student.sid=student_id where corse_id in
(select course.cid from course left join teacher on course.teacher_id=teacher.tid where tname='波多')
group by student_id having count(corse_id) = (select count(course.cid) from course left join teacher on
course.teacher_id=teacher.tid where tname='波多');
思路:
先找出分别找出课程1 和2 的成绩
在连表 (id ,A.成绩,B.成绩) 比较成绩大小
找出学生id 然后用 in
select student.sid, student.sname from student where sid in (
select A.student_id as c from (select student_id,number from score where corse_id=1) as A
left join (select student_id, number from score where corse_id=2) as B on A.student_id=B.student_id
where A.number> if(isnull(B.number),0,B.number)
);
思路:
先在score表找到成绩小于60的学生学号
再用 sid in
select student.sid, student.sname from student where sid in (
select student_id from score where number<60
);
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.sid from student where sid not in (
select student.sid from score
left join student on student_id=student.sid group by student_id having
count(corse_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:
法一:
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
student.sid, student.sname from score left join student on score.student_id=student.sid
where student_id !=1 and corse_id in (select corse_id from score where student_id=1);
法二:
select student.sid, student.sname from
(select * from score where corse_id in (select corse_id from score where student_id=1)) as A
left join student on student.sid = A.student_id where student_id !=1;
思路:
法一:
同12
先找到学过001的所有人
然后个数=001所有的学科个数
select student.sid, student.sname from
(select * from score where corse_id in (select corse_id from score where student_id=1)) as A
left join student on student.sid= A.student_id where A.student_id !=1 group by student_id having
count(corse_id)=(select count(corse_id) from score where student_id=1) ;
法二:
select student_id,sname, count(course_id)
from score left join student on score.student_id=student.sid
where stduent_id !=1 and score_id in (select corse_id from score where student_id=1) group by student_id having
count(corse_id)=(select count(corse_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;
思路:
先找到002学过的课程号
在用 in 找到学过所有相同的课程号的人
在通过连表,分组,计数 把相同的人找出来
select student.sid,student.sname from (select * from score where corse_id in (select corse_id from score where student_id=1)) as A
left join student on student.sid = A.student_id where A.student_id!=2 group by A.student_id having
count(A.student_id)= (select count(corse_id) from score where student_id=2);
思路:
先连表查姓名为叶平的老师 然后在删除
delete from score where corse_id in (
select cid from course left join teacher on course.teacher_id=teacher.tid where teacher.tname='饭岛');
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, number) select sid,2,(select avg(number) from score where score_id=2)
from student where sid not in (select stduent_id from score where score_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;
思路:
连表查学生id,有效课程数,平均分
通过已查询到的学生id为约束条件,再分别连表查询课程号们
select student.sid,
(select avg(number) from score left join course on course.cid = score.corse_id where course.cname="语文" and student_id = student.sid) as 语文,
(select avg(number) from score left join course on course.cid = score.corse_id where course.cname="数学" and student_id = student.sid) as 数学,
(select avg(number) from score left join course on course.cid = score.corse_id where course.cname="英语" and student_id = student.sid) as 英语,
count(student_id)as 有效课程数,
avg(number)as 平均成绩
from score
left join student on student.sid = student_id
group by student_id order by avg(number) desc;
思路:
法一:
通过联表把课程表和成绩表关联
再通过corse_id分组 找出最大最小成绩
select course.cid,max(number),min(number) from score left join course on course.cid = corse_id
group by corse_id;
法二:
select corse_id,max(number),min(number) from score group by corse_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
corse_id,
avg(number) as avgnum,
sum(case when score.number > 60 then 1 else 0 END)/count(1)*100 as percent,
from score group by corse_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);
思路:
联表 分组
select
avg(if (isnull(score.number),0,score.number)),
teacher.tname
from course
left join score on course.cid = score.corse_id
left join teacher on course.teacher_id = teacher.tid
group by score.corse_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.corse_id,score.number,T.first_num,T.second_num from score left join
(
SELECT
sid,
(select number from score as s2 where s2.corse_id = s1.corse_id order by number desc limit 0,1)as first_num,
(select number from score as s2 where s2.corse_id = s1.corse_id order by number desc limit 3,1)as second_num
from
score as s1
)as T
on score.sid = T.sid
where score.number <= T.first_num and score.number >= T.second_num;
22、查询每门课程被选修的学生数;
select count(student_id) from score group by corse_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 from score left join student on student.sid = score.student_id
group by corse_id having count(corse_id) = 1;
24、查询男生、女生的人数;
法一:
select gender,count(gender) from student group by gender;
法二:
select * from (select count(1) as man from student where gender="男") as A,(select count(1) as women from student where gender="女")as B;
25、查询姓“张”的学生名单;
select * from student where sname like '%蛋%';
26、查询同名同姓学生名单,并统计同名人数;
select sname, count(sname) as '数量' 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 corse_id, avg(if(isnull(number), 0, number)) as avg from score group by corse_id order by avg asc, corse_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.sid, student.sname, avg(number) from score left join student on student.sid = score.student_id
group by student_id having avg(number)>80;
29、查询课程名称为“生物”,且分数低于60的学生姓名和分数;
select * from course;
select student.sname, number from score
left join course on course.cid = corse_id
left join student on student.sid = student_id
where course.cname = "生物" and score.number < 60;
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 student.sid, student.sname from score left join student on student.sid = student_id where corse_id = 3 and number >80;
31、求选了课程的学生人数
法一:
select count(c) from(select count(student_id) as c from score group by student_id)as c;
法二:
select count(distinct student_id) from score
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 student.sname, number from score left join student on student.sid = score.student_id where score.corse_id in
select course.cid from course left join teacher on course.teacher_id = teacher.tid where tname="波多")
order by number desc limit 1;
33、查询各个课程及相应的选修人数;
select course.cname,count(student_id) from score left join course on course.cid = corse_id
group by corse_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.corse_id, s2.corse_id, s1.number from score as s1, score as s2 WHERE
s1.number = s2.number and s1.corse_id != s2.corse_id;
35、查询每门课程成绩最好的前两名;
select score.sid,score.corse_id,score.number,T.first_num,T.second_num from score left join
(
select
sid,
(select number from score as s2 where s2.corse_id = s1.corse_id order by number desc limit 0,1) as first_num,
(select number from score as s2 where s2.corse_id = s1.corse_id order by number desc limit 1,1) as second_num
from
score as s1
) as T
on score.sid =T.sid
where score.number <= T.first_num and score.number >= 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(corse_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.cid, course.cname from score left join course on course.cid = corse_id group by corse_id;
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.sname from score left join student on student.sid = student_id where score.corse_id not in
(select tid from teacher where tname="波多") 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, avg(number) from score where number<60 group by student_id having count(student_id)>2;
40、检索“002”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 corse_id = 2 and number < 60 order by number desc;
41、删除“002”同学的“001”课程的成绩;
delete from score where corse_id = 2 and student_id = 2;
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