1. 首页
  2. 自学中心
  3. 软件
  4. SQL

SQL经典50题&答案

作者 胖熊酱

数据表介绍

1.学生表

Student(SId,Sname,Sage,Ssex)

SId 学生编号,Sname 学生姓名,Sage 出生年年月,Ssex 学生性别

2.课程表

Course(CId,Cname,TId)

CId 课程编号,Cname 课程名称,TId 教师编号

3.教师表

Teacher(TId,Tname)

TId 教师编号,Tname 教师姓名

4.成绩表

SC(SId,CId,score)

SId 学生编号,CId 课程编号,score

create table Student(SId varchar(10),Sname varchar(10),Sage datetime,Ssex varchar(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' , '孙⻛' , '1990-12-20' , '男');
insert into Student values('04' , '李云' , '1990-12-06' , '男');
insert into Student values('05' , '周梅' , '1991-12-01' , '⼥');
insert into Student values('06' , '吴兰' , '1992-01-01' , '⼥');
insert into Student values('07' , '郑⽵' , '1989-01-01' , '⼥');
insert into Student values('09' , '张三' , '2017-12-20' , '⼥');
insert into Student values('10' , '李四' , '2017-12-25' , '⼥');
insert into Student values('11' , '李四' , '2012-06-06' , '⼥');
insert into Student values('12' , '赵六' , '2013-06-13' , '⼥');
insert into Student values('13' , '孙七' , '2014-06-01' , '⼥');

科目表 Course

create table Course(CId varchar(10),Cname nvarchar(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');

教师表 Teacher

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');

成绩表 SC

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);

练习题目

1.查询” 01 “课程比” 02 “课程成绩高的学生的信息及课程分数

解法1:(因为题目没有说明全部课程还是仅有这些课程,故列出了全部课程)

select s.*,c.CId,c.score from student s  
left join sc c on s.SId=c.SId 
where s.SId in(select a.SId FROM (
select *from sc where CId=01)a  INNER JOIN  (
select *from sc where CId=02
)b ON a.SId=b.SId where a.score>b.score group by a.SId)
SQL经典50题&答案

解法2:(仅列出01,02课)

SELECT Student.*,01_score.CId,1_score,02_score.CId,2_score
FROM Student JOIN
(SELECT SId,CId,score AS 1_score FROM SC WHERE CId ='01') AS 01_score
ON Student.SId=01_score.SId 
JOIN (SELECT SId,CId,score AS 2_score FROM SC WHERE CId ='02') AS 02_score 
ON 01_score.SId = 02_score.SId
WHERE 1_score>2_score
SQL经典50题&答案

1.1 查询同时存在”01″课程和”02″课程的情况

select student.*,a.CId,a.score,b.CId,b.score 
from student join  (select * from sc where CId=01
)a
on a.SId=student.SId
join (select * from sc where CId=02
)b 
on a.SId=b.SId
SQL经典50题&答案

1.2 查询存在”01″课程但可能不存在”02″课程的情况(不存在时显示为 null )

select student.*,a.CId,a.score,b.CId,
b.score from student join(
select * from sc where CId=01
)a  on student.SId=a.SId
left join (
select * from sc where CId=02
)b 
on a.SId=b.SId
SQL经典50题&答案

1.3 查询不存在“01”课程但存在”02″课程的情况(文字陷阱啊,好几次搞错了。)

select student.*,a.CId,a.score,b.CId,b.score 
from student
join (select * from sc where CId=02)a 
on student.SId=a.SId
left join (select * from sc where CId=01)b 
on a.SId=b.SId
where b.CId is null 
SQL经典50题&答案

2.查询平均成绩大于等于 60 分的同学的学生编号和学生姓名和平均成绩

select a.SId,b.Sname,avgs 
from(select SId,
AVG(score) as avgs from sc 
group by SId having AVG(score)>=60)a
left join student b
on a.SId=b.SId
SQL经典50题&答案

3.查询在 SC 表存在成绩的学生信息(不停精进代码,让代码更优化。)

select distinct student.* from student 
join sc on student.SId=sc.SId
SQL经典50题&答案

4.查询所有同学的学生编号、学生姓名、选课总数、所有课程的总成绩(没成绩的显示

为 null )

select s.SId,s.Sname,CCId,sscorefrom student s 
left join (select SId,count(CId) as CCId,
sum(score) as sscore from sc group by SId)a
on s.SId=a.SId
SQL经典50题&答案

4.1 查有成绩的学生信息(有成绩不等于学生信息在sc表里,如果表中学生成绩为0呢?逻辑要对,避免逻辑漏洞。而且,以后尽量避免用查询嵌套。)

SELECT DISTINCT Student.*
FROM Student LEFT JOIN sc
ON Student.SId = sc.SId
WHERE score IS NOT NULL
SQL经典50题&答案

5.查询「李」姓老师的数量

select count(TId) from teacher 
where Tname like '李%'
SQL经典50题&答案

6.查询学过张三老师授课的同学的信息

select student.*,sc.CId,sc.score,
course.Cname,teacher.* from student 
join sc on student.SId=sc.SId
join course on sc.CId=course.CId
join teacher
on course.TId=teacher.TId
where teacher.Tname='张三'
SQL经典50题&答案

7.查询没有学全所有课程的同学的信息

解法1

SELECT Student.*,COUNT(CId)
FROM Student LEFT JOIN sc
ON Student.SId = sc.SId
GROUP BY student.SId
HAVING COUNT(CId)<(SELECT COUNT(CId) FROM Course)
SQL经典50题&答案

解法2(不是很熟练这种用法。多练习。)

SELECT * FROM Student 
WHERE EXISTS(SELECT * FROM Course WHERE
NOT EXISTS (SELECT * FROM SC 
WHERE cid = Course.CId AND SId = Student.SId))
SQL经典50题&答案

8.查询至少有一门课与学号为” 01 “的同学所学相同的同学的信息

SELECT DISTINCT Student.*
FROM Student JOIN SC
ON Student.SId = SC.SId
WHERE CId IN (SELECT CId FROM SC WHERE SId = 01) and Student.SId !=01
SQL经典50题&答案

9.查询和” 01 “号的同学学习的课程完全相同的其他同学的信息

select * from student where SId in
(select SId from (
select * from sc a where CId in (
select CId from sc where SId=01))b
group by SId having count(CId) =(
select count(CId) from sc c where 
SId=01))
and SId !=01
SQL经典50题&答案

10.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 SId from sc 
left join course on sc.CId=course.CId
left join teacher on course.TId=teacher.TId 
where Tname='张三' )
SQL经典50题&答案

11.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 sc.SId,
student.Sname,
avg(score) from sc 
left join student on sc.SId=student.SId 
where sc.SId in (select SId from sc  
where score<60 group by SId
having count(CId) >=2) group by SId
SQL经典50题&答案

12.检索” 01 “课程分数小于 60,按分数降序排列的学生信息

SELECT * FROM Student JOIN SC ON Student.SId=SC.SId
WHERE CId = 01 AND score < 60
ORDER BY score DESC
SQL经典50题&答案

13.按平均成绩从高到低显示所有学生的所有课程的成绩以及平均成绩

select a.*,avgs from sc a 
left join (
select SId,avg(score) as avgs 
from sc group by SId
)b on a.SId=b.SId 
order by avgs desc
SQL经典50题&答案

14.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,中等率,

优良率,优秀率

及格为>=60,中等为:70-80,优良为:80-90,优秀为:>=90

要求输出课程号和选修人数,查询结果按人数降序排列列,若人数相同,按课程号升序

排列

select CId,
count(SId),
max(score),
min(score),
avg(score),
sum(及格)/count(SId) as 及格率, 
sum(中等)/count(SId) as 中等率,
sum(优良)/count(SId) as 优良率,
sum(优秀)/count(SId) as 优秀率
from (select *,
case when score>=60  then 1
else 0  end as 及格,
case when  score>=70 and score<80 then 1
else 0  end as 中等 ,
case when  score>=80 and score<90 then 1
else 0  end as 优良 ,
case when  score>=90 then 1
else 0  end as 优秀 
from sc)a group by CId
order by count(SId) desc,CId
SQL经典50题&答案

15.按各科成绩进行排序,并显示排名, Score 重复时保留名次空缺

select a.*,count(a.score) rank from sc a 
left join sc b on a.cid = b.cid and a.score < b.score
group by a.cid,a.sid
order by a.cid,rank
SQL经典50题&答案

15.1 按各科成绩进行行排序,并显示排名, Score 重复时合并名次

select a.*,count(b.score)+1 rank from sc a 
left join sc b on a.cid = b.cid and a.score < b.score
group by a.cid,a.sid
order by a.cid,rank
SQL经典50题&答案

16.查询学生的总成绩,并进行排名,总分重复时保留名次空缺

select a.*,@rank:= @rank+1 as rank
from (select SId, sum(score)
from sc
GROUP BY SId 
ORDER BY sum(score) desc)a,(select @rank:=0)b
SQL经典50题&答案

16.1 查询学生的总成绩,并进行行排名,总分重复时不保留名次空缺

select a.*, case when @fscore=a.sumscore then @rank  
when @fscore:=a.sumscore then @rank:=@rank+1 end as rank
from (select sc.SId, sum(score) as sumscore
from sc
GROUP BY SId 
ORDER BY sum(score) desc) as a,(
select @rank:=0,@fscore:=null) as t
SQL经典50题&答案

17. 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],[60-0] 及所占百分比

select sc.CId,Cname,
sum(case when score>=0 and score<=60 then 1
else 0  end) as '[60-0]',
sum(case when score>=0 and score<=60 then 1
else 0  end)/count(SId) as '[60-0]百分比',
sum(case when  score>=60 and score<=70 then 1
else 0  end) as '[70-60]',
sum(case when  score>=60 and score<=70 then 1
else 0  end)/count(SId) as '[70-60]百分比',
sum(case when  score>=70 and score<=85 then 1
else 0  end)as '[85-70]',
sum(case when  score>=70 and score<=85 then 1
else 0  end)/count(SId) as '[85-70]百分比',
sum(case when  score>=85 and score<=100 then 1
else 0  end) as '[100-85]',
sum(case when  score>=85 and score<=100 then 1
else 0  end)/count(SId) as '[100-85]百分比'
from sc 
join course on sc.CId=course.CId group by sc.CId,Cname
SQL经典50题&答案

18.查询各科成绩前三名的记录

SELECT a.*,COUNT(B.score) +1 as ranking
FROM SC a LEFT JOIN SC b 
ON a.CId = b.CId AND a.score<b.score
GROUP BY a.CId,a.SId,a.score
HAVING ranking <= 3
ORDER BY a.CId,ranking
SQL经典50题&答案

19.查询每门课程被选修的学生数

select CId,count(SId)  from sc group by CId
SQL经典50题&答案

20.查询出只选修两门课程的学生学号和姓名

select student.SId,student.Sname,count(CId) from student 
join sc on student.SId=sc.SId 
group by student.SId having count(CId)=2
SQL经典50题&答案

21. 查询男生、女生人数

select Ssex,count(Ssex)  
from student group by Ssex
SQL经典50题&答案

22. 查询名字中含有「风」字的学生信息

select * from student where 
Sname LIKE '%风%'
SQL经典50题&答案

23查询同名同性学生名单,并统计同名人数

select Sname,
count(SId) from student 
group by Sname having count(Sname) >1
SQL经典50题&答案

24.查询 1990 年年出生的学生名单

select * from student where Sage like '1990%'
SQL经典50题&答案

25.查询每门课程的平均成绩,结果按平均成绩降序排列,平均成绩相同时,按课程编

号升序排列

select CId,avg(score) from sc 
group by CId order by avg(score) desc,CId
SQL经典50题&答案

26.查询平均成绩大于等于 85 的所有学生的学号、姓名和平均成绩

select sc.SId,
student.Sname,
avg(score) from student 
join sc on student.SId=sc.SId
group by sc.SId having avg(score)>=85
SQL经典50题&答案

27.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,course.Cname  
from student 
join sc on student.SId=sc.SId
join course on sc.CId=course.CId
where course.Cname='数学' and score<60
SQL经典50题&答案

28. 查询所有学生的课程及分数情况(存在学生没成绩,没选课的情况)

select student.Sname,
sc.CId,
sc.score from student left join sc 
on student.SId=sc.SId
SQL经典50题&答案

29.查询任何一门课程成绩在 70 分以上的姓名、课程名称和分数

select student.Sname,Cname,score from student 
join sc on student.SId=sc.SId
join course on sc.CId=course.CId
where score>70
SQL经典50题&答案

30.查询不及格的课程

select distinct Cname from course 
join sc on course.CId=sc.CId
where score<60
SQL经典50题&答案

31.查询课程编号为 01 且课程成绩在 80 分以上的学生的学号和姓名

select student.SId,Sname 
from student join sc on student.SId=sc.SId
where CId=01 and score>=80
SQL经典50题&答案

32.求每门课程的学生人数

select CId,count(SId)  
from sc group by sc.CId
SQL经典50题&答案

33.成绩不重复,查询选修「张三」老师所授课程的学生中,成绩最高的学生信息及其成绩

SELECT Student.*,SC.CId,score
FROM Student JOIN SC ON Student.SId = SC.SId
JOIN Course ON SC.CId = Course.CId
JOIN Teacher ON Course.TId = Teacher.TId
WHERE Tname = '张三' 
ORDER BY score DESC LIMIT 1
SQL经典50题&答案

34.成绩有重复的情况下,查询选修「张三」老师所授课程的学生中,成绩最高的学生

信息及其成绩

SELECT Student.*,SC.CId,score
FROM Student JOIN SC ON Student.SId = SC.SId
JOIN Course ON SC.CId = Course.CId
JOIN Teacher ON Course.TId = Teacher.TId
WHERE Tname = '张三' AND score IN
(SELECT MAX(score) FROM
SC JOIN Course ON SC.CId = Course.CId
JOIN Teacher ON Course.TId = Teacher.TId
WHERE Tname = '张三')
SQL经典50题&答案

35.查询不同课程成绩相同的学生的学生编号、课程编号、学生成绩

select distinct a.* from sc a join sc b on a.SId=b.SId
where a.score=b.score and a.CId!=b.CId
SQL经典50题&答案

36. 查询每门成绩最好的前两名

SELECT A.CId,A.SId,A.score,COUNT(B.score) +1 as ranking
FROM SC AS A LEFT JOIN SC AS B 
ON A.CId = B.CId AND A.score<B.score
GROUP BY A.CId,A.SId,A.score
HAVING ranking <= 2 
ORDER BY A.CId,ranking
SQL经典50题&答案

37. 统计每门课程的学生选修人数(超过 5 人的课程才统计)。

select CId,count(SId) from sc group by CId 
having count(SId)>5
SQL经典50题&答案

38.检索至少选修两门课程的学生学号

select SId from sc group by SId having count(CId)>=2 
SQL经典50题&答案

39.查询选修了全部课程的学生信息

select student.* from student 
join sc on student.SId=sc.SId
group by student.SId 
having count(CId)=(select count(CId) from course)
SQL经典50题&答案

40.查询各学生的年龄,只按年份来算

select Sname,Sage,
year(now())-year(Sage) as 年龄 
from student
SQL经典50题&答案

41. 按照出生日期来算,当前月日 < 出生年月的月日则,年龄减一

解法1:

SELECT Sname,Sage,
case when (DATE_FORMAT(NOW(),'%m-%d')- DATE_FORMAT(Sage,'%m-%d')) <0
then year(now())-year(Sage)+1
else year(now())-year(Sage) 
end as 年龄
FROM student
SQL经典50题&答案

解法2:

SELECT SId,Sname,TIMESTAMPDIFF(YEAR,Sage,NOW()) AS age
FROM Student
SQL经典50题&答案

42.查询本周过生日的学生

SELECT * FROM student where 
week(Sage)=week(now())
SQL经典50题&答案

43. 查询下周过生日的学生

SELECT * FROM student where 
week(Sage)=week(now())+1
SQL经典50题&答案

44.查询本月过生日的学生

SELECT * FROM student
where month(Sage)=month(now())
SQL经典50题&答案

45.查询下月过生日的学生

SELECT * FROM student
where month(Sage)=month(now())+1
SQL经典50题&答案

本文来自知乎,观点不代表一起大数据-技术文章心得立场,如若转载,请注明出处:https://zhuanlan.zhihu.com/p/67645448

联系我们

在线咨询:点击这里给我发消息

邮件:23683716@qq.com

跳至工具栏