SQL 面试必备50题

Dear 丶 2022-05-15 04:29 294阅读 0赞

# SQL 先思考，试着写写，然后作答 #

问题及描述：  
–1.学生表  
Student(S\#,Sname,Sage,Ssex) –S\# 学生编号,Sname 学生姓名,Sage 出生年月,Ssex 学生性别  
–2.课程表  
Course(C\#,Cname,T\#) –C\# –课程编号,Cname 课程名称,T\# 教师编号  
–3.教师表  
Teacher(T\#,Tname) –T\# 教师编号,Tname 教师姓名  
–4.成绩表  
SC(S\#,C\#,score) –S\# 学生编号,C\# 课程编号,score 分数  
\*/  
–创建测试数据  
create table Student(S\# varchar(10),Sname nvarchar(10),Sage datetime,Ssex nvarchar(10))  
insert into Student values(‘01’ , N’赵雷’ , ‘1990-01-01’ , N’男’)  
insert into Student values(‘02’ , N’钱电’ , ‘1990-12-21’ , N’男’)  
insert into Student values(‘03’ , N’孙风’ , ‘1990-05-20’ , N’男’)  
insert into Student values(‘04’ , N’李云’ , ‘1990-08-06’ , N’男’)  
insert into Student values(‘05’ , N’周梅’ , ‘1991-12-01’ , N’女’)  
insert into Student values(‘06’ , N’吴兰’ , ‘1992-03-01’ , N’女’)  
insert into Student values(‘07’ , N’郑竹’ , ‘1989-07-01’ , N’女’)  
insert into Student values(‘08’ , N’王菊’ , ‘1990-01-20’ , N’女’)  
create table Course(C\# varchar(10),Cname nvarchar(10),T\# varchar(10))  
insert into Course values(‘01’ , N’语文’ , ‘02’)  
insert into Course values(‘02’ , N’数学’ , ‘01’)  
insert into Course values(‘03’ , N’英语’ , ‘03’)  
create table Teacher(T\# varchar(10),Tname nvarchar(10))  
insert into Teacher values(‘01’ , N’张三’)  
insert into Teacher values(‘02’ , N’李四’)  
insert into Teacher values(‘03’ , N’王五’)  
create table SC(S\# varchar(10),C\# 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)  
go

–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  
select a.\* , b.score \[课程’01’的分数\],c.score \[课程’02’的分数\] from Student a , SC b , SC c  
where a.S\# = b.S\# and a.S\# = c.S\# and b.C\# = ‘01’ and c.C\# = ‘02’ and b.score > c.score  
–1.2、查询同时存在”01”课程和”02”课程的情况和存在”01”课程但可能不存在”02”课程的情况(不存在时显示为null)(以下存在相同内容时不再解释)  
select a.\* , b.score \[课程”01”的分数\],c.score \[课程”02”的分数\] from Student a  
left join SC b on a.S\# = b.S\# and b.C\# = ‘01’  
left join SC c on a.S\# = c.S\# and c.C\# = ‘02’  
where b.score > isnull(c.score,0)

–2、查询”01”课程比”02”课程成绩低的学生的信息及课程分数  
–2.1、查询同时存在”01”课程和”02”课程的情况  
select a.\* , b.score \[课程’01’的分数\],c.score \[课程’02’的分数\] from Student a , SC b , SC c  
where a.S\# = b.S\# and a.S\# = c.S\# and b.C\# = ‘01’ and c.C\# = ‘02’ and b.score < c.score  
–2.2、查询同时存在”01”课程和”02”课程的情况和不存在”01”课程但存在”02”课程的情况  
select a.\* , b.score \[课程”01”的分数\],c.score \[课程”02”的分数\] from Student a  
left join SC b on a.S\# = b.S\# and b.C\# = ‘01’  
left join SC c on a.S\# = c.S\# and c.C\# = ‘02’  
where isnull(b.score,0) < c.score

–3、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.S\# , a.Sname , cast(avg(b.score) as decimal(18,2)) avg\_score  
from Student a , sc b  
where a.S\# = b.S\#  
group by a.S\# , a.Sname  
having cast(avg(b.score) as decimal(18,2)) >= 60  
order by a.S\#

–4、查询平均成绩小于60分的同学的学生编号和学生姓名和平均成绩  
–4.1、查询在sc表存在成绩的学生信息的SQL语句。  
select a.S\# , a.Sname , cast(avg(b.score) as decimal(18,2)) avg\_score  
from Student a , sc b  
where a.S\# = b.S\#  
group by a.S\# , a.Sname  
having cast(avg(b.score) as decimal(18,2)) < 60  
order by a.S\#  
–4.2、查询在sc表中不存在成绩的学生信息的SQL语句。  
select a.S\# , a.Sname , isnull(cast(avg(b.score) as decimal(18,2)),0) avg\_score  
from Student a left join sc b  
on a.S\# = b.S\#  
group by a.S\# , a.Sname  
having isnull(cast(avg(b.score) as decimal(18,2)),0) < 60  
order by a.S\#

–5、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  
–5.1、查询所有有成绩的SQL。  
select a.S\# \[学生编号\], a.Sname \[学生姓名\], count(b.C\#) 选课总数, sum(score) \[所有课程的总成绩\]  
from Student a , SC b  
where a.S\# = b.S\#  
group by a.S\#,a.Sname  
order by a.S\#  
–5.2、查询所有(包括有成绩和无成绩)的SQL。  
select a.S\# \[学生编号\], a.Sname \[学生姓名\], count(b.C\#) 选课总数, sum(score) \[所有课程的总成绩\]  
from Student a left join SC b  
on a.S\# = b.S\#  
group by a.S\#,a.Sname  
order by a.S\#

–6、查询”李”姓老师的数量  
–方法1  
select count(Tname) \[“李”姓老师的数量\] from Teacher where Tname like N’李%’  
–方法2  
select count(Tname) \[“李”姓老师的数量\] from Teacher where left(Tname,1) = N’李’

–7、查询学过”张三”老师授课的同学的信息  
select distinct Student.\* from Student , SC , Course , Teacher  
where Student.S\# = SC.S\# and SC.C\# = Course.C\# and Course.T\# = Teacher.T\# and Teacher.Tname = N’张三’  
order by Student.S\#

–8、查询没学过”张三”老师授课的同学的信息  
select m.\* from Student m where S\# not in (select distinct SC.S\# from SC , Course , Teacher where SC.C\# = Course.C\# and Course.T\# = Teacher.T\# and Teacher.Tname = N’张三’) order by m.S\#

–9、查询学过编号为”01”并且也学过编号为”02”的课程的同学的信息  
–方法1  
select Student.\* from Student , SC where Student.S\# = SC.S\# and SC.C\# = ‘01’ and exists (Select 1 from SC SC\_2 where SC\_2.S\# = SC.S\# and SC\_2.C\# = ‘02’) order by Student.S\#  
–方法2  
select Student.\* from Student , SC where Student.S\# = SC.S\# and SC.C\# = ‘02’ and exists (Select 1 from SC SC\_2 where SC\_2.S\# = SC.S\# and SC\_2.C\# = ‘01’) order by Student.S\#  
–方法3  
select m.\* from Student m where S\# in  
(  
select S\# from  
(  
select distinct S\# from SC where C\# = ‘01’  
union all  
select distinct S\# from SC where C\# = ‘02’  
) t group by S\# having count(1) = 2  
)  
order by m.S\#

–10、查询学过编号为”01”但是没有学过编号为”02”的课程的同学的信息  
–方法1  
select Student.\* from Student , SC where Student.S\# = SC.S\# and SC.C\# = ‘01’ and not exists (Select 1 from SC SC\_2 where SC\_2.S\# = SC.S\# and SC\_2.C\# = ‘02’) order by Student.S\#  
–方法2  
select Student.\* from Student , SC where Student.S\# = SC.S\# and SC.C\# = ‘01’ and Student.S\# not in (Select SC\_2.S\# from SC SC\_2 where SC\_2.S\# = SC.S\# and SC\_2.C\# = ‘02’) order by Student.S\#

–11、查询没有学全所有课程的同学的信息  
–11.1、  
select Student.\*  
from Student , SC  
where Student.S\# = SC.S\#  
group by Student.S\# , Student.Sname , Student.Sage , Student.Ssex having count(C\#) < (select count(C\#) from Course)  
–11.2  
select Student.\*  
from Student left join SC  
on Student.S\# = SC.S\#  
group by Student.S\# , Student.Sname , Student.Sage , Student.Ssex having count(C\#) < (select count(C\#) from Course)

–12、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 , SC where Student.S\# = SC.S\# and SC.C\# in (select C\# from SC where S\# = ‘01’) and Student.S\# <> ‘01’

–13、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 Student.\* from Student where S\# in  
(select distinct SC.S\# from SC where S\# <> ‘01’ and SC.C\# in (select distinct C\# from SC where S\# = ‘01’)  
group by SC.S\# having count(1) = (select count(1) from SC where S\#=’01’))

–14、查询没学过”张三”老师讲授的任一门课程的学生姓名  
select student.\* from student where student.S\# not in  
(select distinct sc.S\# from sc , course , teacher where sc.C\# = course.C\# and course.T\# = teacher.T\# and teacher.tname = N’张三’)  
order by student.S\#

–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 student.S\# , student.sname , cast(avg(score) as decimal(18,2)) avg\_score from student , sc  
where student.S\# = SC.S\# and student.S\# in (select S\# from SC where score < 60 group by S\# having count(1) >= 2)  
group by student.S\# , student.sname

–16、检索”01”课程分数小于60，按分数降序排列的学生信息  
select student.\* , sc.C\# , sc.score from student , sc  
where student.S\# = SC.S\# and sc.score < 60 and sc.C\# = ‘01’  
order by sc.score desc

–17、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  
–17.1 SQL 2000 静态  
select a.S\# 学生编号 , a.Sname 学生姓名 ,  
max(case c.Cname when N’语文’ then b.score else null end) \[语文\],  
max(case c.Cname when N’数学’ then b.score else null end) \[数学\],  
max(case c.Cname when N’英语’ then b.score else null end) \[英语\],  
cast(avg(b.score) as decimal(18,2)) 平均分  
from Student a  
left join SC b on a.S\# = b.S\#  
left join Course c on b.C\# = c.C\#  
group by a.S\# , a.Sname  
order by 平均分 desc  
–17.2 SQL 2000 动态  
declare @sql nvarchar(4000)  
set @sql = ‘select a.S\# ’ + N’学生编号’ + ’ , a.Sname ’ + N’学生姓名’  
select @sql = @sql + ‘,max(case c.Cname when N”’+Cname+”’ then b.score else null end) \[‘+Cname+’\]’  
from (select distinct Cname from Course) as t  
set @sql = @sql + ’ , cast(avg(b.score) as decimal(18,2)) ’ + N’平均分’ + ’ from Student a left join SC b on a.S\# = b.S\# left join Course c on b.C\# = c.C\#  
group by a.S\# , a.Sname order by ’ + N’平均分’ + ’ desc’  
exec(@sql)

–24、查询学生平均成绩及其名次  
–24.1 查询学生的平均成绩并进行排名，sql 2000用子查询完成，分平均成绩重复时保留名次空缺和不保留名次空缺两种。  
select t1.\* , px = (select count(1) from  
(  
select m.S\# \[学生编号\] ,  
m.Sname \[学生姓名\] ,  
isnull(cast(avg(score) as decimal(18,2)),0) \[平均成绩\]  
from Student m left join SC n on m.S\# = n.S\#  
group by m.S\# , m.Sname  
) t2 where 平均成绩 > t1.平均成绩) + 1 from  
(  
select m.S\# \[学生编号\] ,  
m.Sname \[学生姓名\] ,  
isnull(cast(avg(score) as decimal(18,2)),0) \[平均成绩\]  
from Student m left join SC n on m.S\# = n.S\#  
group by m.S\# , m.Sname  
) t1  
order by px

select t1.\* , px = (select count(distinct 平均成绩) from  
(  
select m.S\# \[学生编号\] ,  
m.Sname \[学生姓名\] ,  
isnull(cast(avg(score) as decimal(18,2)),0) \[平均成绩\]  
from Student m left join SC n on m.S\# = n.S\#  
group by m.S\# , m.Sname  
) t2 where 平均成绩 >= t1.平均成绩) from  
(  
select m.S\# \[学生编号\] ,  
m.Sname \[学生姓名\] ,  
isnull(cast(avg(score) as decimal(18,2)),0) \[平均成绩\]  
from Student m left join SC n on m.S\# = n.S\#  
group by m.S\# , m.Sname  
) t1  
order by px  
–24.2 查询学生的平均成绩并进行排名，sql 2005用rank,DENSE\_RANK完成，分平均成绩重复时保留名次空缺和不保留名次空缺两种。  
select t.\* , px = rank() over(order by \[平均成绩\] desc) from  
(  
select m.S\# \[学生编号\] ,  
m.Sname \[学生姓名\] ,  
isnull(cast(avg(score) as decimal(18,2)),0) \[平均成绩\]  
from Student m left join SC n on m.S\# = n.S\#  
group by m.S\# , m.Sname  
) t  
order by px

select t.\* , px = DENSE\_RANK() over(order by \[平均成绩\] desc) from  
(  
select m.S\# \[学生编号\] ,  
m.Sname \[学生姓名\] ,  
isnull(cast(avg(score) as decimal(18,2)),0) \[平均成绩\]  
from Student m left join SC n on m.S\# = n.S\#  
group by m.S\# , m.Sname  
) t  
order by px  
  
–25、查询各科成绩前三名的记录  
–25.1 分数重复时保留名次空缺  
select m.\* , n.C\# , n.score from Student m, SC n where m.S\# = n.S\# and n.score in  
(select top 3 score from sc where C\# = n.C\# order by score desc) order by n.C\# , n.score desc  
–25.2 分数重复时不保留名次空缺，合并名次  
–sql 2000用子查询实现  
select \* from (select t.\* , px = (select count(distinct score) from SC where C\# = t.C\# and score >= t.score) from sc t) m where px between 1 and 3 order by m.c\# , m.px  
–sql 2005用DENSE\_RANK实现  
select \* from (select t.\* , px = DENSE\_RANK() over(partition by c\# order by score desc) from sc t) m where px between 1 and 3 order by m.C\# , m.px

–26、查询每门课程被选修的学生数  
select c\# , count(S\#)\[学生数\] from sc group by C\#

–27、查询出只有两门课程的全部学生的学号和姓名  
select Student.S\# , Student.Sname  
from Student , SC  
where Student.S\# = SC.S\#  
group by Student.S\# , Student.Sname  
having count(SC.C\#) = 2  
order by Student.S\#

–28、查询男生、女生人数  
select count(Ssex) as 男生人数 from Student where Ssex = N’男’  
select count(Ssex) as 女生人数 from Student where Ssex = N’女’  
select sum(case when Ssex = N’男’ then 1 else 0 end) \[男生人数\],sum(case when Ssex = N’女’ then 1 else 0 end) \[女生人数\] from student  
select case when Ssex = N’男’ then N’男生人数’ else N’女生人数’ end \[男女情况\] , count(1) \[人数\] from student group by case when Ssex = N’男’ then N’男生人数’ else N’女生人数’ end

–29、查询名字中含有”风”字的学生信息  
select \* from student where sname like N’%风%’  
select \* from student where charindex(N’风’ , sname) > 0

–30、查询同名同性学生名单，并统计同名人数  
select Sname \[学生姓名\], count(\*) \[人数\] from Student group by Sname having count(\*) > 1

–31、查询1990年出生的学生名单(注：Student表中Sage列的类型是datetime)  
select \* from Student where year(sage) = 1990  
select \* from Student where datediff(yy,sage,’1990-01-01’) = 0  
select \* from Student where datepart(yy,sage) = 1990  
select \* from Student where convert(varchar(4),sage,120) = ‘1990’

–32、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 m.C\# , m.Cname , cast(avg(n.score) as decimal(18,2)) avg\_score  
from Course m, SC n  
where m.C\# = n.C\#  
group by m.C\# , m.Cname  
order by avg\_score desc, m.C\# asc

–33、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 a.S\# , a.Sname , cast(avg(b.score) as decimal(18,2)) avg\_score  
from Student a , sc b  
where a.S\# = b.S\#  
group by a.S\# , a.Sname  
having cast(avg(b.score) as decimal(18,2)) >= 85  
order by a.S\#

–34、查询课程名称为”数学”，且分数低于60的学生姓名和分数  
select sname , score  
from Student , SC , Course  
where SC.S\# = Student.S\# and SC.C\# = Course.C\# and Course.Cname = N’数学’ and score < 60

–35、查询所有学生的课程及分数情况；  
select Student.\* , Course.Cname , SC.C\# , SC.score  
from Student, SC , Course  
where Student.S\# = SC.S\# and SC.C\# = Course.C\#  
order by Student.S\# , SC.C\#

–36、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.\* , Course.Cname , SC.C\# , SC.score  
from Student, SC , Course  
where Student.S\# = SC.S\# and SC.C\# = Course.C\# and SC.score >= 70  
order by Student.S\# , SC.C\#

–37、查询不及格的课程  
select Student.\* , Course.Cname , SC.C\# , SC.score  
from Student, SC , Course  
where Student.S\# = SC.S\# and SC.C\# = Course.C\# and SC.score < 60  
order by Student.S\# , SC.C\#

–38、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.\* , Course.Cname , SC.C\# , SC.score  
from Student, SC , Course  
where Student.S\# = SC.S\# and SC.C\# = Course.C\# and SC.C\# = ‘01’ and SC.score >= 80  
order by Student.S\# , SC.C\#

–39、求每门课程的学生人数  
select Course.C\# , Course.Cname , count(\*) \[学生人数\]  
from Course , SC  
where Course.C\# = SC.C\#  
group by Course.C\# , Course.Cname  
order by Course.C\# , Course.Cname

–40、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  
–40.1 当最高分只有一个时  
select top 1 Student.\* , Course.Cname , SC.C\# , SC.score  
from Student, SC , Course , Teacher  
where Student.S\# = SC.S\# and SC.C\# = Course.C\# and Course.T\# = Teacher.T\# and Teacher.Tname = N’张三’  
order by SC.score desc  
–40.2 当最高分出现多个时  
select Student.\* , Course.Cname , SC.C\# , SC.score  
from Student, SC , Course , Teacher  
where Student.S\# = SC.S\# and SC.C\# = Course.C\# and Course.T\# = Teacher.T\# and Teacher.Tname = N’张三’ and  
SC.score = (select max(SC.score) from SC , Course , Teacher where SC.C\# = Course.C\# and Course.T\# = Teacher.T\# and Teacher.Tname = N’张三’)

–41、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  
–方法1  
select m.\* from SC m ,(select C\# , score from SC group by C\# , score having count(1) > 1) n  
where m.C\#= n.C\# and m.score = n.score order by m.C\# , m.score , m.S\#  
–方法2  
select m.\* from SC m where exists (select 1 from (select C\# , score from SC group by C\# , score having count(1) > 1) n  
where m.C\#= n.C\# and m.score = n.score) order by m.C\# , m.score , m.S\#

–42、查询每门功成绩最好的前两名  
select t.\* from sc t where score in (select top 2 score from sc where C\# = T.C\# order by score desc) order by t.C\# , t.score desc

–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  
select Course.C\# , Course.Cname , count(\*) \[学生人数\]  
from Course , SC  
where Course.C\# = SC.C\#  
group by Course.C\# , Course.Cname  
having count(\*) >= 5  
order by \[学生人数\] desc , Course.C\#

–44、检索至少选修两门课程的学生学号  
select student.S\# , student.Sname  
from student , SC  
where student.S\# = SC.S\#  
group by student.S\# , student.Sname  
having count(1) >= 2  
order by student.S\#

–45、查询选修了全部课程的学生信息  
–方法1 根据数量来完成  
select student.\* from student where S\# in  
(select S\# from sc group by S\# having count(1) = (select count(1) from course))  
–方法2 使用双重否定来完成  
select t.\* from student t where t.S\# not in  
(  
select distinct m.S\# from  
(  
select S\# , C\# from student , course  
) m where not exists (select 1 from sc n where n.S\# = m.S\# and n.C\# = m.C\#)  
)  
–方法3 使用双重否定来完成  
select t.\* from student t where not exists(select 1 from  
(  
select distinct m.S\# from  
(  
select S\# , C\# from student , course  
) m where not exists (select 1 from sc n where n.S\# = m.S\# and n.C\# = m.C\#)  
) k where k.S\# = t.S\#  
)

–46、查询各学生的年龄  
–46.1 只按照年份来算  
select \* , datediff(yy , sage , getdate()) \[年龄\] from student  
–46.2 按照出生日期来算，当前月日 < 出生年月的月日则，年龄减一  
select \* , case when right(convert(varchar(10),getdate(),120),5) < right(convert(varchar(10),sage,120),5) then datediff(yy , sage , getdate()) - 1 else datediff(yy , sage , getdate()) end \[年龄\] from student

–47、查询本周过生日的学生  
select \* from student where datediff(week,datename(yy,getdate()) + right(convert(varchar(10),sage,120),6),getdate()) = 0

–48、查询下周过生日的学生  
select \* from student where datediff(week,datename(yy,getdate()) + right(convert(varchar(10),sage,120),6),getdate()) = -1

–49、查询本月过生日的学生  
select \* from student where datediff(mm,datename(yy,getdate()) + right(convert(varchar(10),sage,120),6),getdate()) = 0

–50、查询下月过生日的学生  
select \* from student where datediff(mm,datename(yy,getdate()) + right(convert(varchar(10),sage,120),6),getdate()) = -1

drop table Student,Course,Teacher,SC

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