50道sql练习题和答案

Posted 2020-08-18

最近两年的工作没有写过多少SQL，感觉水平下降十分严重，网上找了50道练习题学习和复习

原文地址：50道SQL练习题及答案与详细分析

1.0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 分数

2.0 数据表创建

学生表Student

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‘ , ‘孙风‘ , ‘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);

3.0 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
• 1.1. 查询同时存在" 01 "课程和" 02 "课程的情况
• 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 )
• 1.3 查询不存在" 01 "课程但存在" 02 "课程的情况
• 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
• 3.查询在 SC 表存在成绩的学生信息
• 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 )
• 4.1 查有成绩的学生信息
• 5.查询「李」姓老师的数量
• 6.查询学过「张三」老师授课的同学的信息
• 7.查询没有学全所有课程的同学的信息
• 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
• 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
• 10.查询没学过"张三"老师讲授的任一门课程的学生姓名
• 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
• 12.检索" 01 "课程分数小于 60，按分数降序排列的学生信息
• 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
• 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，若人数相同，按课程号升序排列

• 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
• 15.1 按各科成绩进行排序，并显示排名， Score 重复时合并名次
• 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
• 16.1 查询学生的总成绩，并进行排名，总分重复时不保留名次空缺
• 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
• 18.查询各科成绩前三名的记录
• 19.查询每门课程被选修的学生数
• 20.查询出只选修两门课程的学生学号和姓名
• 21.查询男生、女生人数
• 22.查询名字中含有「风」字的学生信息
• 23.查询同名同性学生名单，并统计同名人数
• 24.查询 1990 年出生的学生名单
• 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
• 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
• 27.查询课程名称为「数学」，且分数低于 60 的学生姓名和分数
• 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）
• 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
• 30.查询不及格的课程
• 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
• 32.求每门课程的学生人数
• 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
• 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
• 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
• 36.查询每门功成绩最好的前两名
• 37.统计每门课程的学生选修人数（超过 5 人的课程才统计）。
• 38.检索至少选修两门课程的学生学号
• 39.查询选修了全部课程的学生信息
• 40.查询各学生的年龄，只按年份来算
• 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
• 42.查询本周过生日的学生
• 43.查询下周过生日的学生
• 44.查询本月过生日的学生
• 45.查询下月过生日的学生

4.0答案和分析(如有更好解法，欢迎留言交流)

use DataBaseStudy
--1.查询" 01 "课程比" 02 "课程成绩高的学生的信息及课程分数
--因为需要全部的学生信息，则需要在sc表中得到符合条件的SId后与student表进行join，可以left join 也可以 right join 

select * from (
select   t1.sid as sid, t1.score as class1,t2.score as class2  from 
(select *from sc where sc.cid=‘01‘) as t1 ,
(select * from sc where sc.cid=‘02‘) as t2
where  t1.sid=t2.sid and  t1.score>t2.score
 ) as r

left join student
on r.sid=student.sid


select * from Student RIGHT JOIN (
    select t1.SId, class1, class2 from
          (select SId, score as class1 from sc where sc.CId = ‘01‘)as t1, 
          (select SId, score as class2 from sc where sc.CId = ‘02‘)as t2
    where t1.SId = t2.SId AND t1.class1 > t2.class2
)r 
on Student.SId = r.SId;

--join (inner join)属于等值连接，只返回两个表中联结字段相等的行，select * from t1,t2 where...这种属于隐式表连接，逐渐被抛弃
select * from 
  (select SId, score as class1 from sc where sc.CId = ‘01‘)as t1
join
  (select SId, score as class2 from sc where sc.CId = ‘02‘)as t2

  on t1.SId=t2.SId


--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 * from 
(select * from sc where sc.CId = ‘01‘) as t1
left join 
(select * from sc where sc.CId = ‘02‘) as t2
on t1.SId = t2.SId

--1.3 查询不存在" 01 "课程但存在" 02 "课程的情况
--考察子查询

select * from sc where sc.CId = ‘02‘

and sc.[SId] not in

(select sid from sc where sc.cid=‘01‘)


--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
--考察group by 分组和常用聚合函数

select * from 
(select sid,AVG(score) as AVGsource from sc
where score>60
group by sid) as t1

left join  student

on t1.SId=Student.SId

--3 查询在 SC 表存在成绩的学生信息

select * from 

(select SId from sc 
group  by SId) as t1
left join Student
on t1.SId=Student.SId


select distinct Student.* from Student
right join sc
on Student.SId=sc.SId

--下面这种写法相当于innerjoin，属于隐式的表连接，已经被逐渐抛弃
select DISTINCT student.*
from student,sc
where student.SId=sc.SId

--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 * from 
(select SId,COUNT(CId) as CCount,SUM(score)as SumScore from  SC
group by SId ) as t1
right join student
on t1.SId=Student.SId

--1 查有成绩的学生信息
select * from Student right join
(select  Sid from sc
group by Sid) as t1
on t1.SId=Student.SId

--5 查询「李」姓老师的数量
select count(*) from Teacher
where Tname like ‘李%‘

--6,查询学过「张三」老师授课的同学的信息

select count(*) from sc,Student
select count (*) from  sc
inner join Student
on sc.SId=Student.SId

--7查询没有学全所有课程的同学的信息 
select * from Student where SId not in 
(
select SId from sc
group by 
SId
having COUNT(CId)= (select count(CId) from Course)
)
--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 * from Student right join 

(select sid from SC 
where CId in (select cid from sc where SId=‘01‘) and SId !=‘01‘
group by sid) as t1

on Student.SId=t1.SId

--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

--不会
--10查询没学过"张三"老师讲授的任一门课程的学生姓名

select * from sc where sid=6
select * from Course where tid=‘01‘

select * from Student where sid not in

(select sid from sc where cid in (select cid from Course where TId=‘01‘))


select * from student
where student.sid not in(
    select sc.sid from sc,course,teacher 
    where
        sc.cid = course.cid
        and course.tid = teacher.tid
        and teacher.tname= ‘张三‘
)

--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 * from sc  where score<=60

select  Student.Sname,t1.SId,t1.avgScore,t1.num from Student
right join
(
select sid,count(score)as num,AVG(score)as avgScore from sc
where score<60
group by SId 
having count(score)>=2) as t1
on t1.SId=Student.SId

--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 right join
(
select * from sc where cid=‘01‘ and score<=60

)as t1
on Student.sid=t1.sid
order by score desc

select student.*, sc.score from student, sc
where student.sid = sc.sid
and sc.score < 60
and cid = ‘01‘
ORDER BY sc.score DESC;

--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 *  from sc 
left join (
    select sid,avg(score) as avscore from sc 
    group by sid
    )r 
on sc.sid = r.sid
order by avscore desc;

select sid,Count(*) from sc
group by sid
select * from Course

--下面是带行转列的优化版本,pivot函数用法稍微有些复杂。
select Student.Sname,t3.* from Student right join
(select sid, [01]as‘语文‘,[02] as ‘数学‘,[03] as ‘英语‘,avgScore
from 
(
select sc.*,t1.avgScore from sc right join 
(
select sid,avg(score) as avgScore from sc
group by sid ) as t1
on sc.SId=t1.SId
)as t2
pivot
(
sum(score) for cid in([01],[02],[03])
)tbl) as t3
on Student.SId=t3.SId
order by avgScore desc

--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，若人数相同，按课程号升序排列

--sqlserver百分比除法，全是整型结果也是整形，需要进行类型转换

select cid
,max(score) as 最高分
,min(score) as 最低分
,avg(score) as 平均分
,count(*) as 选修人数
,sum(case when sc.score>=60 then 1 else 0 end)/cast(count(*) as float )as 及格率
from SC
group by CId
order by  选修人数 desc,CId asc



select 
sc.CId ,
max(sc.score)as 最高分,
min(sc.score)as 最低分,
AVG(sc.score)as 平均分,
count(*)as 选修人数,
sum(case when sc.score>=60 then 1 else 0 end )/count(*)as 及格率,
sum(case when sc.score>=70 and sc.score<80 then 1 else 0 end )/count(*)as 中等率,
sum(case when sc.score>=80 and sc.score<90 then 1 else 0 end )/count(*)as 优良率,
sum(case when sc.score>=90 then 1 else 0 end )/count(*)as 优秀率 
from sc
GROUP BY sc.CId
ORDER BY count(*)DESC, sc.CId ASC

--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
-- left join ,right join,如果副表有多条数据对应主表，那查询出来就是多条数据
select a.cid, a.sid, a.score, count(b.score)+1 as rank
from sc as a 
left join sc as b 
on a.score<b.score and a.cid = b.cid
group by a.cid, a.sid,a.score
order by a.cid, rank ASC;

select sc.CId,sc.SId,sc.score,count(b.score)+1 from sc
left join sc as b
on sc.score<b.score and sc.CId=b.CId
group by sc.cid,sc.sid,sc.score
order by sc.CId,score desc

--未聚合前的结果集，
select * from sc
left join sc as b
on sc.score<b.score and sc.CId=b.CId
order by sc.CId
--使用排序函数会更加简洁
select sid,cid,score,rank() over(partition by cid order by score desc) as ranking from sc;
--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
--考察sql四大排名函数的用法,row_number,rank,dense_rank,ntile,

with t1 as (
select sid,sum(score) as sumScore from sc
group by sid)

select dense_rank() over(order by sumScore desc)as num,*  from t1
--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 Course.Cname,t1.* from course right join

(select CId,count(score) as 课程人数
,sum(case  when score>=85 then 1 else 0 end)*100/CAST( count(score) as float) as [100-85]
,sum(case  when score>=70 and score<85 then 1 else 0 end)*100/CAST( count(score) as float) as [85-70]
,sum(case  when score>=60 and score<70 then 1 else 0 end)*100/CAST( count(score) as float) as [70-60]
,sum(case  when score<60 then 1 else 0 end)*100/CAST( count(score) as float) as [60-0]
 from sc
 group by CId) as t1
 on Course.CId=t1.CId


--18 查询各科成绩前三名的记录
select * from (
select sid,cid,score, RANK() over(partition by cid order by score desc)as rankNum from SC) as t1
where t1.rankNum<=3

--19查询每门课程被选修的学生数
select cid,count(SId)as snum from sc
group by cid
--20查询出只选修两门课程的学生学号和姓名
select Student.Sname,t1.* from Student right join
(
select sid from sc
group by sid
having count(cid)=2) as t1
on Student.SId=t1.SId
--21查询男生、女生人数
select Ssex,count(sid) as num from Student group by Ssex
--22查询名字中含有「风」字的学生信息
select * from Student where Sname like ‘%风%‘
--23 查询同名同性学生名单，并统计同名人数
select  Sname,count(*) as 人数 from Student
group by Sname
having count(*)>1
--24 查询 1990 年出生的学生名单
--考察时间函数的熟悉程度
select * from Student where Sage>=‘1990-01-01‘ and Sage<=‘1990-12-31‘

--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)as avgScore from sc
group by cid
order by avgScore desc,CId asc

--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 Student.Sname,t1.* from Student
right join
(
select sid,avg(score)as avgScore from sc
group by sid
having avg(score)>=85) as t1
on Student.SId=t1.SId
order by avgScore desc

--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,t1.score from Student right join
(
select * from sc where cid=(select cid from Course where cname=‘数学‘) and score<60) as t1
on  Student.sid=t1.sid
--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 * from Student
left join sc
on Student.SId=sc.SId
--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,Course.Cname,sc.score from Student,sc,Course
where  Student.SId=sc.SId and sc.CId=Course.CId and    sc.score>70

select Student.Sname,Course.Cname,sc.score  from Student
join sc on Student.SId=sc.SId
join Course on sc.CId=Course.CId
where sc.score>70
--30查询不及格的课程
select * from sc where score<60

--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,Student.Sname,sc.score from Student join sc
on Student.SId=sc.SId
where sc.CId=‘01‘ and score>80

--32求每门课程的学生人数
select cid,count(*)as num from sc
group by CId

--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 top(1)* from sc where sc.CId=
(
select cid from Course join Teacher
on Course.TId=Teacher.TId
where Teacher.Tname=‘李四‘)
order by score desc
--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 * from (
select *,rank() over(partition by cid order by score desc) as 名次 from sc 
where cid in (

select CId from Course 
join Teacher
on Course.TId=Teacher.TId
where Teacher.Tname=‘张三‘
)) as t1
where t1.名次=1
--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
--思路-》查询的结果集中，不同的结果中不同字段的对比，可以用join连接，去重用group by去掉类似[1,3]和[3,1]这样的重复数据
select t1.SId,t1.CId,t1.score,t2.score from sc as t1
join sc as t2
on t1.SId=t2.SId
and t1.CId!=t2.CId
and t1.score=t2.score
group by t1.SId,t1.CId,t1.score,t2.score
order by t1.SId

--36查询每门功成绩最好的前两名
select cid,sid,名次 from
(select sid,cid,rank() over(partition by cid order by score desc)as 名次 from sc) as t1
where t1.名次<3
--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(*) as num from sc
group by cid
having count(*)>5

--38检索至少选修两门课程的学生学号
select sid,count(cid) as num from sc
group by sid
having count(cid)>=2

--39查询选修了全部课程的学生信息
select Student.* from Student
right join
(
select sid,count(cid) as num from sc
group by sid
having count(cid)=3) as t1
on Student.SId=t1.SId

--40查询各学生的年龄，只按年份来算 
日期函数不太熟悉，后五题还没做