互联网经典SQL面试题及答案解析
练习前的准备
SQL环境搭建
建表语句
create table Student(sid varchar(10),sname varchar(10),sage datetime,ssex nvarchar(10));
insert into Student values('01' , '赵雷' , '1990-01-01' , '男');
insert into Student values('02' , '钱电' , '1990-12-21' , '男');
insert into Student values('03' , '孙风' , '1990-05-20' , '男');
insert into Student values('04' , '李云' , '1990-08-06' , '男');
insert into Student values('05' , '周梅' , '1991-12-01' , '女');
insert into Student values('06' , '吴兰' , '1992-03-01' , '女');
insert into Student values('07' , '郑竹' , '1989-07-01' , '女');
insert into Student values('08' , '王菊' , '1990-01-20' , '女');
create table Course(cid varchar(10),cname varchar(10),tid varchar(10));
insert into Course values('01' , '语文' , '02');
insert into Course values('02' , '数学' , '01');
insert into Course values('03' , '英语' , '03');
create table Teacher(tid varchar(10),tname varchar(10));
insert into Teacher values('01' , '张三');
insert into Teacher values('02' , '李四');
insert into Teacher values('03' , '王五');
create table SC(sid varchar(10),cid varchar(10),score decimal(18,1));
insert into SC values('01' , '01' , 80);
insert into SC values('01' , '02' , 90);
insert into SC values('01' , '03' , 99);
insert into SC values('02' , '01' , 70);
insert into SC values('02' , '02' , 60);
insert into SC values('02' , '03' , 80);
insert into SC values('03' , '01' , 80);
insert into SC values('03' , '02' , 80);
insert into SC values('03' , '03' , 80);
insert into SC values('04' , '01' , 50);
insert into SC values('04' , '02' , 30);
insert into SC values('04' , '03' , 20);
insert into SC values('05' , '01' , 76);
insert into SC values('05' , '02' , 87);
insert into SC values('06' , '01' , 31);
insert into SC values('06' , '03' , 34);
insert into SC values('07' , '02' , 89);
insert into SC values('07' , '03' , 98);表结构预览
--学生表
Student(SId,Sname,Sage,Ssex)
--SId 学生编号,Sname 学生姓名,Sage 出生年月,Ssex 学生性别
--课程表
Course(CId,Cname,TId)
--CId 课程编号,Cname 课程名称,TId 教师编号
--教师表
Teacher(TId,Tname)
--TId 教师编号,Tname 教师姓名
--成绩表
SC(SId,CId,score)
--SId 学生编号,CId 课程编号,score 分数
题目
1、查询“01”课程比“02”课程成绩高的所有学生的学号；
2、查询平均成绩大于60分的同学的学号和平均成绩；
3、查询所有同学的学号、姓名、选课数、总成绩
4、查询姓“李”的老师的个数；
5、查询没学过“张三”老师课的同学的学号、姓名；
6、查询学过编号“01”并且也学过编号“02”课程的同学的学号、姓名；
7、查询学过“张三”老师所教的课的同学的学号、姓名；
8、查询课程编号“01”的成绩比课程编号“02”课程低的所有同学的学号、姓名；
9、查询所有课程成绩小于60分的同学的学号、姓名；
10、查询没有学全所有课的同学的学号、姓名；
11、查询至少有一门课与学号为“01”的同学所学相同的同学的学号和姓名；
12、查询和"01"号的同学学习的课程完全相同的其他同学的学号和姓名
13、把“SC”表中“张三”老师教的课的成绩都更改为此课程的平均成绩；
14、查询没学过"张三"老师讲授的任一门课程的学生姓名
15、查询两门及其以上不及格课程的同学的学号，姓名及其平均成绩
16、检索"01"课程分数小于60，按分数降序排列的学生信息
17、按平均成绩从高到低显示所有学生的平均成绩
18、查询各科成绩最高分、最低分和平均分：以如下形式显示：课程ID，课程name，最高分，最低分，平均分，及格率
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、查询学生的总成绩并进行排名
21、查询不同老师所教不同课程平均分从高到低显示
22、查询所有课程的成绩第2名到第3名的学生信息及该课程成绩
23、统计各科成绩各分数段人数：课程编号,课程名称,100-85,85-70,70-60,0-60及所占百分比
24、查询学生平均成绩及其名次
25、查询各科成绩前三名的记录
26、查询每门课程被选修的学生数
27、查询出只选修了一门课程的全部学生的学号和姓名
28、查询男生、女生人数
29、查询名字中含有"风"字的学生信息
30、查询同名同性学生名单，并统计同名人数
31、查询1990年出生的学生名单(注：Student表中Sage列的类型是datetime)
32、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
37、查询不及格的课程，并按课程号从大到小排列
38、查询课程编号为"01"且课程成绩在60分以上的学生的学号和姓名；
40、查询选修“张三”老师所授课程的学生中，成绩最高的学生姓名及其成绩
42、查询每门功课成绩最好的前两名
43、统计每门课程的学生选修人数（超过5人的课程才统计）。要求输出课程号和选修人数，查询结果按人数降序排列，若人数相同，按课程号升序排列
44、检索至少选修两门课程的学生学号
45、查询选修了全部课程的学生信息
46、查询各学生的年龄
47、查询本周过生日的学生
48、查询下周过生日的学生
49、查询本月过生日的学生
50、查询下月过生日的学生
答案
1、查询“01”课程比“02”课程成绩高的所有学生的学号；
select distinct t1.sid as sid
from
(select * from sc where cid='01')t1
left join
(select * from sc where cid='02')t2
on t1.sid=t2.sid
where t1.score>t2.score2、查询平均成绩大于60分的同学的学号和平均成绩；
select
sid
,avg(score)
from sc
group by sid
having avg(score>60)3、查询所有同学的学号、姓名、选课数、总成绩
select
student.sid as sid
,sname
,count(distinct cid) course_cnt
,sum(score) as total_score
from student
left join sc
on student.sid=sc.sid
group by sid,sname4、查询姓“李”的老师的个数；
select
count(distinct tid) as teacher_cnt
from teacher
where tname like '李%'5、查询没学过“张三”老师课的同学的学号、姓名；
select
sid,sname
from student
where sid not in
(
select
sc.sid
from teacher
left join course
on teacher.tid=course.tid
left join sc
on course.cid=sc.cid
where teacher.tname='张三'
)6、查询学过“01”并且也学过编号“02”课程的同学的学号、姓名；
select
t.sid as sid
,sname
from
(
select
sid
,count(if(cid='01',score,null)) as count1
,count(if(cid='02',score,null)) as count2
from sc
group by sid
having count(if(cid='01',score,null))>0 and count(if(cid='02',score,null))>0
)t
left join student
on t.sid=student.sid7、查询学过“张三”老师所教的课的同学的学号、姓名；
select
student.sid
,sname
from
(
select
distinct cid
from course
left join teacher
on course.tid=teacher.tid
where teacher.tname='张三'
)course
left join sc
on course.cid=sc.cid
left join student
on sc.sid=student.sid
group by student.sid,sname8、查询课程编号“01”的成绩比课程编号“02”课程低的所有同学的学号、姓名；
select
t1.sid,sname
from
(
select distinct t1.sid as sid
from
(select * from sc where cid='01')t1
left join
(select * from sc where cid='02')t2
on t1.sid=t2.sid
where t1.score>t2.score
)t1
left join student
on t1.sid=student.sid9、查询所有课程成绩小于60分的同学的学号、姓名；
select
t1.sid,sname
from
(
select
sid,max(score)
from sc
group by sid
having max(score<60)
)t1
left join student
on t1.sid=student.sid10、查询没有学全所有课的同学的学号、姓名；
select
t1.sid,sname
from
(
select
count(cid),sid
from sc
group by sid
having count(cid) < (select count(distinct cid) from course)
)t1
left join student
on t1.sid=student.sid11、查询至少有一门课与学号为“01”的同学所学相同的同学的学号和姓名；
select
distinct sc.sid
from
(
select
cid
from sc
where sid='01'
)t1
left join sc
on t1.cid=sc.cid12、查询和"01"号的同学学习的课程完全相同的其他同学的学号和姓名
#注意是和'01'号同学课程完全相同但非学习课程数相同的,这里我用左连接解决这个问题
select
t1.sid,sname
from
(
select
sc.sid
,count(distinct sc.cid)
from
(
select
cid
from sc
where sid='01'
)t1 #选出01的同学所学的课程
left join sc
on t1.cid=sc.cid
group by sc.sid
having count(distinct sc.cid)= (select count(distinct cid) from sc where sid = '01')
)t1
left join student
on t1.sid=student.sid
where t1.sid!='01'13、把“SC”表中“张三”老师教的课的成绩都更改为此课程的平均成绩；
#暂跳过update题目14、查询没学过"张三"老师讲授的任一门课程的学生姓名
select
sname
from student
where sid not in
(
select
distinct sid
from sc
left join course
on sc.cid=course.cid
left join teacher
on course.tid=teacher.tid
where tname='张三'
)15、查询两门及其以上不及格课程的同学的学号，姓名及其平均成绩
select
t1.sid,sname,avg_score
from
(
select
sid,count(if(score<60,cid,null)),avg(score) as avg_score
from sc
group by sid
having count(if(score<60,cid,null)) >=2
)t1
left join student
on t1.sid=student.sid16、检索"01"课程分数小于60，按分数降序排列的学生信息
select
sid,if(cid='01',score,100)
from sc
where if(cid='01',score,100)<60
order by if(cid='01',score,100) desc17、按平均成绩从高到低显示所有学生的平均成绩
select sid,avg(score)
from sc
group by sid
order by avg(score) desc18、查询各科成绩最高分、最低分和平均分：以如下形式显示：课程ID，课程name，最高分，最低分，平均分，及格率
select
sc.cid
,cname
,max(score) as max_score
,min(score) as min_score
,avg(score) as avg_score
,count(if(score>=60,sid,null))/count(sid) as pass_rate
from sc
left join course
on sc.cid=course.cid
group by sc.cid19、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
#这里先按照平均成绩排序，再按照及格百分数排序，题目有点奇怪
select
cid
,avg(score) as avg_score
,count(if(score>=60,sid,null))/count(sid) as pass_rate
from sc
group by cid
order by avg_score,pass_rate desc20、查询学生的总成绩并进行排名
select
sid
,sum(score) as sum_score
from sc
group by sid
order by sum_score desc21、查询不同老师所教不同课程平均分从高到低显示
select
tid
,avg(score) as avg_score
from course
left join sc
on course.cid=sc.cid
group by tid
order by avg_score desc22、查询所有课程的成绩第2名到第3名的学生信息及该课程成绩
select
sid,rank_num,score,cid
from
(
select
rank() over(partition by cid order by score desc) as rank_num
,sid
,score
,cid
from sc
)t
where rank_num in (2,3)23、统计各科成绩各分数段人数：课程编号,课程名称,100-85,85-70,70-60,0-60及所占百分比
select
sc.cid
,cname
,count(if(score between 85 and 100,sid,null))/count(sid)
,count(if(score between 70 and 85,sid,null))/count(sid)
,count(if(score between 60 and 70,sid,null))/count(sid)
,count(if(score between 0 and 60,sid,null))/count(sid)
from sc
left join course
on sc.cid=course.cid
group by sc.cid,cname24、查询学生平均成绩及其名次
select
sid
,avg_score
,rank() over (order by avg_score desc)
from
(
select
sid
,avg(score) as avg_score
from sc
group by sid
)t25、查询各科成绩前三名的记录
select
sid,cid,rank1
from
(
select
cid
,sid
,rank() over(partition by cid order by score desc) as rank1
from sc
)t
where rank1<=326、查询每门课程被选修的学生数
select
count(sid)
,cid
from sc
group by cid27、查询出只选修了一门课程的全部学生的学号和姓名
#只查出来sid即可，后面懒得交student表
select
sid
from sc
group by sid
having count(cid) =128、查询男生、女生人数
select
ssex
,count(distinct sid)
from student
group by ssex29、查询名字中含有"风"字的学生信息
select
sid,sname
from student
where sname like '%风%'30、查询同名同性学生名单，并统计同名人数
#题目有歧义，这套题的质量感觉有点差
select
ssex
,sname
,count(sid)
from student
group by ssex,sname
having count(sid)>=231、查询1990年出生的学生名单(注：Student表中Sage列的类型是datetime)
select
sid,sname,sage
from student
where year(sage)=199032、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
cid,avg(score) as avg_score
from sc
group by cid
order by avg_score,cid desc37、查询不及格的课程，并按课程号从大到小排列
#有问题的题目
select
cid,sid,score
from sc
where score<60
order by cid desc,sid38、查询课程编号为"01"且课程成绩在60分以上的学生的学号和姓名；
select
sid,cid,score
from sc
where cid='01' and score>6040、查询选修“张三”老师所授课程的学生中，成绩最高的学生姓名及其成绩
select
sc.sid,sname,cname,score
from sc
left join course
on sc.cid=course.cid
left join teacher
on course.tid=teacher.tid
left join student
on sc.sid=student.sid
where tname='张三'
order by score desc
limit 1;42、查询每门功课成绩最好的前两名
##感觉题目重复了
select
cid,sid,rank1
from
(
select
cid
,sid
,rank() over(partition by cid order by score desc) as rank1
from sc
)t
where rank1 <=243、统计每门课程的学生选修人数（超过5人的课程才统计）。要求输出课程号和选修人数，查询结果按人数降序排列，若人数相同，按课程号升序排列
select
cid
,count(sid) as cnt
from sc
group by cid
having cnt>=5
order by count(sid) desc,cid44、检索至少选修两门课程的学生学号
select
sid
,count(cid)
from sc
group by sid
having count(cid)>=245、查询选修了全部课程的学生信息
#不太严谨，但实务中应该没问题，如需严谨见12题思路
select
sid
,count(cid)
from sc
group by sid
having count(cid)=(select count(distinct cid) from sc)46、查询各学生的年龄
select
sid,sname,year(curdate())-year(sage) as sage
from student47、查询本周过生日的学生
select
sid,sname,sage
from student
where weekofyear(sage)=weekofyear(curdate())48、查询下周过生日的学生
select
sid,sname,sage
from student
where weekofyear(sage) = weekofyear(date_add(curdate(),interval 1 week))49、查询本月过生日的学生
select
sid,sname,sage
from student
where month(sage) = month(curdate())50、查询下月过生日的学生
select
sid,sname,sage
from student
where month(date_sub(sage,interval 1 month)) = month(curdate())