SQL数据库面试题以及答案

Student(S# ,Sname,Sage,Ssex)学生表
S#：学号
Sname：学生姓名
Sage：学生年龄
Ssex：学生性别
 
Course(C#,Cname,T#)课程表
C#：课程编号
Cname：课程名称
T#：教师编号
 
SC(S#,C#,score)成绩表
S#：学号
C#：课程编号
score：成绩
 
Teacher(T#,Tname)教师表
T#：教师编号：
Tname：教师名字
个人评价：这50套题很经典，都是一些常用的查询语句。有利于锻炼思维，理解关系数据库
问题：

1、查询“001”课程比“002”课程成绩高的所有学生的学号
select a.S# from (select S#,score from SC where C#='001')a, 
(select s#,score from SC where c#='002')b Where a.score>b.score and a.s# = b.s#;
2、查询平均成绩大于60分的同学的学号和平均成绩
select S#, avg(score) from sc group by S# having avg(score)>60
3、查询所有同学的学号、姓名、选课数、总成绩
select student.S_id, student.Sname, count(sc.C_id), sum(score) from student 
left join SC on student.S_id = SC.S_id group by Student.S_id, Sname
4、查询姓‘李’的老师的个数：
select count(distinct(Tname)) 
from teacher 
where tname like '李%';
5、查询没有学过“叶平”老师可的同学的学号、姓名：
select student.S#, student.Sname 
from Student 
where S# not in (select distinct(SC.S#) from SC,Course,Teacher 
where sc.c#=course.c# AND teacher.T#=course.T# AND Teahcer.Tname ='叶平');
6、查询学过“叶平”老师所教的所有课的同学的学号、姓名：
select S#,Sname   from Student    
where S# in (select S# from SC ,Course ,Teacher 
where SC.C#=Course.C# and Teacher.T#=Course.T# 
and Teacher.Tname='叶平' group by S# 
having count(SC.C#)=(select count(C#) from Course,Teacher  
where Teacher.T#=Course.T# and Tname='叶平'));
7、查询学过“011”并且也学过编号“002”课程的同学的学号、姓名：
select Student.S#,Student.Sname 
from Student,SC where Student.S#=SC.S# 
and SC.C#='001'and 
exists( Select * from SC as SC_2 where SC_2.S#=SC.S# and SC_2.C#='002');
8、查询课程编号“002”的成绩比课程编号“001”课程低的所有同学的学号、姓名：
Select S#,Sname 
from (select Student.S#,Student.Sname,score ,
(select score from SC SC_2 where SC_2.S#=Student.S# and SC_2.C#='002') score2    
from Student,SC 
where Student.S#=SC.S# and C#='001') S_2 
where score2 < score; 
9、查询所有课程成绩小于60的同学的学号、姓名：
select S#, sname 
from student 
where s# not in 
(select student.s# from student, sc where s.s# = sc.s# and score>60);
10、查询没有学全所有课的同学的学号、姓名：
select student.s#, student.sname 
from student, sc
where student.s#=sc.s# 
group by student.s#, student.sname 
having count(c#)<(select count(c#) from course);
11、查询至少有一门课与学号为“1001”同学所学相同的同学的学号和姓名：
select s#, Sname 
from Student, SC 
where student.s# = sc.s# 
and c# in (select c# from SC where s#='1001');
12、查询至少学过学号为“001”同学所有一门课的其他同学学号和姓名；
select distinct sc.s# , sname 
from student, sc 
where student.s#=sc.s# 
and c# in (select C# from sc where s#='001');
13、把“SC”表中“叶平”老师教的课的成绩都更改为此课程的平均成绩：
Update Sc Set Score=(Select Avg(s2_Score) From sc s2 Where s2.c#=sc.c#)  
Where c# IN
(Select c# From sc cs INNER JOIN Teacher tc ON cs.t#=tc.t# WHERE tname ='叶平')
14、查询和“1002”号的同学学习的课程完全相同的其他同学学号和姓名：
select s# from sc where c#  in 
(select c# from sc where s#='1002') 
group by s# having count(*)=
(select count(*) from sc where s#='1002');
 这个写法是有问题的，如果“1002”同学的学习课程是其它同学的子集，那么也会筛选出来；
正确写法如下：子集且记录数相同
select t1.s_id from 
(select s_id,count(distinct c_id) as cnt1 from sc where c_id in
(select c_id from sc where s_id=2) and s_id <> 2 group by s_id 
having count(distinct c_id)=(select count(distinct c_id) from sc where s_id=2)) t1,
(select s_id,count(distinct c_id) as cnt2 from sc  group by s_id) t2
where t1.s_id=t2.s_id and t1.cnt1=t2.cnt2 
15、删除学习“叶平”老师课的SC表记录：
delect sc 
from course, Teacher 
where course.c#=sc.c# 
and course.t#=teacher.t# 
and tname='叶平';
16、向SC表中插入一些记录，这些记录要求符合以下条件：没有上过编号“003”课程的同学学号、002号课的平均成绩：
Insert SC select S#,'002',
(Select avg(score) from SC where C#='002') 
from Student where S# not in (Select S# from SC where C#='002');
17、按平均成绩从高到低显示所有学生的“数据库”、“企业管理”、“英语”三门的课程成绩，按如下形式显示：学生ID，数据库，企业管理，英语，有效课程数，有效平均分：
select s_id as 学生ID,
(select score from sc where sc.s_id=t.s_id and c_id=1) as 数据库,
(select score from sc where sc.s_id=t.s_id and c_id=2) as 企业管理,
(select score from sc where sc.s_id=t.s_id and c_id=3) as 英语,
count(*) as 有效课程数, avg(t.score) as 平均成绩,rank() over(order by avg(t.score) desc) as 名次
from sc  t
group by s_id 
order by avg(t.score) asc
此题很经典，没做出来。rank over() 里面是是计算名次的排序；order by是记录展示的排序
18、查询各科成绩最高和最低的分： 以如下的形式显示：课程ID，最高分，最低分
select L.c# as 课程ID, L.score as 最高分,
R.score as 最低分
from sc L, sc R 
where L.c# = R.c# 
and L.score = (select max(IL.score) 
        from sc IL, student as IM 
        where L.c#=IL.c# and IM.s#=IL.s#
        group by IL.c#)
and R.score = (select min(IR.score)
        from sc as IR
        where R.c#=IR.c#
        group by IR.c#);
select c_id,max(score),min(score) from sc group by c_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：
SELECT t.C# AS 课程号,
max(course.Cname)AS 课程名,
isnull(AVG(score),0) AS 平均成绩,
100 * SUM(CASE WHEN  isnull(score,0)>=60 THEN 1 ELSE 0 END)/COUNT(*) AS 及格百分数     
FROM SC T,Course     
where t.C#=course.C#     
GROUP BY t.C#      
ORDER BY 100 * SUM(CASE WHEN  isnull(score,0)>=60 THEN 1 ELSE 0 END)/COUNT(*) DESC 
select c_id 课程编号,round(avg(score),2) 平均分,
round(sum(case when score>59 then 1 else 0 end)/count(1)*100.00,2)||'%' 及格率 from sc
group by c_id order by 平均分 asc,及格率 desc;
20、查询如下课程平均成绩和及格率的百分数(用”1行”显示): 企业管理（001），马克思（002），OO&UML （003），数据库（004）：
21、查询不同老师所教不同课程平均分从高到低显示：
 SELECT max(Z.T#) AS 教师ID,
 MAX(Z.Tname) AS 教师姓名,
 C.C# AS 课程ID,
 AVG(Score) AS 平均成绩     
 FROM SC AS T,Course AS C ,Teacher AS Z    
 where T.C#=C.C# and C.T#=Z.T#   
 GROUP BY C.C#    
 ORDER BY AVG(Score) DESC
22、查询如下课程成绩第3名到第6名的学生成绩单：企业管理(001)，马克思(002)，UML(003)，数据库(004)：
23、统计下列各科成绩，各分数段人数：课程ID，课程名称，[100-85],[85-70],[70-60],[ 小于60] ：
SELECT SC.C# as 课程ID, Cname as 课程名称,
SUM(CASE WHEN score BETWEEN 85 AND 100 THEN 1 ELSE 0 END) AS [100 - 85]  ,
SUM(CASE WHEN score BETWEEN 70 AND 85 THEN 1 ELSE 0 END) AS [85 - 70],
SUM(CASE WHEN score BETWEEN 60 AND 70 THEN 1 ELSE 0 END) AS [70 - 60],
SUM(CASE WHEN score < 60 THEN 1 ELSE 0 END) AS [60 -]     
FROM SC,Course     
where SC.C#=Course.C#     
GROUP BY SC.C#,Cname;
select c_id 课程ID,
       (select cname from course t1 where t1.c_id=t2.c_id) 课程名称,
       count(distinct case when score between 85 and 100 then s_id end ) "[85-100分]人数",
       count(distinct case when score between 70 and 85 then s_id end ) "[75-85分]人数",
       count(distinct case when score between 60 and 70 then s_id end ) "[60-70]人数",
       count(distinct case when score <60 then s_id end ) "小于60分人数"
       from sc t2 group by c_id ;
注：这种写法更好一点，可以排除，同人同科目错误录入了两条记录的情况；
24、查询学生平均成绩及其名次：
SELECT 1+(SELECT COUNT( distinct 平均成绩)                
FROM (SELECT S#,AVG(score) AS 平均成绩                       
FROM SC                   
GROUP BY S#  ) AS T1  WHERE 平均成绩 > T2.平均成绩) as 名次,       
S# as 学生学号,平均成绩      
FROM (SELECT S#,AVG(score) 平均成绩             
FROM SC         
GROUP BY S# ) AS T2      
ORDER BY 平均成绩 desc; 
select s_id 学号,avg(score) 平均成绩,rank() over(order by avg(score) desc) 名次 from sc group by s_id;
25、查询各科成绩前三名的记录（不考虑成绩并列情况）：
SELECT t1.S# as 学生ID,t1.C# as 课程ID,Score as 分数       
FROM SC t1        
WHERE score IN 
(SELECT TOP 3 score               
FROM SC               
WHERE t1.C#= C#             
ORDER BY score DESC)   
select s_id,c_id,dense_rank() over(partition by c_id order by score desc) rank from sc;  
注：oracle没有top N的写法；  
 
26、查询每门课程被选修的学生数：
select c#, count(s#) 
from sc 
group by c#;
27、查询出只选修一门课程的全部学生的学号和姓名：
select t2.s_Id,t2.sname from student t2 where t2.s_id 
in(select s_Id from sc group by sc.s_id having count(distinct sc.c_id)=1);
 
select t1.s_id,t2.sname from sc t1,student t2 where t1.s_id=t2.s_id 
group by t1.s_id,t2.sname having count(distinct t1.c_id)=1
 注：写法2 可加深对group by的理解，实际根据t1.s_id已经能够唯一定位，加上t2.sname完全是语法需要
28、查询男生、女生人数：
select count(Ssex) as 男生人数 
from student 
group by Ssex 
having Ssex='男'；
select count(Ssex) as 女生人数 
from student 
group by Ssex 
having Ssex='女';
select (case when ssex='M' then '男' else '女' end) 性别,count(1) 人数 from student group by ssex;
注：注意单引号，数据库里只识别单引号。
29、查询姓“张”的学生名单：
select sname 
from student 
where sname like '张%';
30、查询同名同姓的学生名单，并统计同名人数：
select sanme,count(*) 
from student 
group by sname 
havang count(*)>1;
31、1981年出生的学生名单（注：student表中sage列的类型是datetime）:
select sname, convert(char(11),DATEPART(year,sage)) as age
from student 
where convert(char(11),DATEPART(year,Sage))='1981';
32、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 Sname,SC.S# ,avg(score)     
from Student,SC      
where Student.S#=SC.S# 
group by SC.S#,Sname 
having    avg(score)>85;
33、查询每门课程的平均成绩，结果按平均成绩升序排序，平均成绩相同时，按课程号降序排列：
select C#, avg(score) 
from sc 
group by c# 
order by avg(score), c# desc;
34、查询课程名称为“数据库”，且分数低于60的学生名字和分数：
select sname, isnull(score,0) 
from student, sc ,course 
where sc.s#=student.s#  and sc.c#=course.c# and course.cname='数据库' and score<60;
35、查询所有学生的选课情况：
select sc.s#,sc.c#,sname,cname 
from sc,student course 
where sc.s#=student.s# and sc.c#=course.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 distinct student.s#,student.sname,sc.c#,sc.score 
from student,sc 
where sc.score>=70 and sc.s#=student.s#;
37、查询课程和不及格人数，并按课程号从大到小的排列：
select c# 
from sc 
where score<60 
order by c#;
select c_id , count(1) from sc where score < 60 group by c_id order by c_id;
注：很好的考察了对group by 用法的理解
38、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 sc.s#,student.sname 
from sc,student 
where sc.s#=student.s# and score>80 and c#='003';
39、求选了课程的学生人数：
select count(*) from sc;
select count(distinct c_id) from sc where score is not null;
注：感觉这样更严谨
40、查询选修“叶平”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,score 
from student,sc,course c, teacher 
where student.s#=sc.S# and sc.c#=c.c#
and c.T#=teacher.T#
and teacher.tname='叶平' 
and sc.score=(select max(score) from sc where c#=c.c#);
select t4.sname 姓名,t3.score 成绩 from sc t3,student t4 where t3.s_id=t4.s_id and  t3.score=
(select max(score) from sc 
where c_id in(select t1.c_id from course t1,teacher t2 where t1.t_id=t2.t_id and t2.tname='叶平'));
注：连接4张表
41、查询各个课程及相应的选修人数：
select count(*) from sc group by c#;
42、查询不同课程成绩相同的学生和学号、课程号、学生成绩：
select distinct a.s#,b.score 
from sc a ,sc b 
where a.score=b.score 
and a.c#<>b.c#;
select t1.* from 
sc t1,sc t2 where 
t1.score=t2.score and t1.s_id<>t2.s_id and t1.c_id<>t2.c_id order by t1.score;
注：使用自连接的例子
43、查询每门课程成绩最好的前两名：
select t1.s# as 学生ID,t1.c#  课程ID, Score as 分数
from sc t1 
where score in (select top 2 score from sc 
        where t1.c#=c#
        order by score desc)
order by t1.c#;
select (select cname from course where c_id=t1.c_id) 课程名,t2.sname 学生姓名,t1.rank 排名 from 
(select c_id,s_id,rank() over(partition by c_id order by score desc) rank from sc) t1,student t2 
where t1.s_id=t2.s_id and t1.rank <3 
注：partition by 以前没用过，注意下用法；
44、统计每门课程的学生选修人数(超过10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序，若人数相同，按课程号升序排序：
select c_id,count(distinct s_id) from sc group by c_id 
having count(distinct s_id)>10 order by count(distinct s_id) desc,c_id asc;
45、检索至少选修两门课程的学生学号：
select s# 
from sc 
group by s# 
having count(*)>=2;
46、查询全部学生选修的课程和课程号和课程名：
select c# ,cname
from course 
where c# in (select c# from sc group by c#);
47、查询没学过”叶平”老师讲授的任一门课程的学生姓名：
select sname 
from student 
where s# not in (select s# from course,teacher,sc where course.t#=teacher.t# and sc.c#=course.c# 
and tname='叶平');
select distinct  t3.s_id,t3.sname from student t3,sc t4 where t3.s_id=t4.s_id and t4.c_id 
not in(select t1.c_id from course t1,teacher t2 where t1.t_id=t2.t_id and t2.tname='叶平');
注：此题集合的思想，1、学过叶平的课的学生；2、叶平教过的课    取相反数
48、查询两门以上不及格课程的同学的学号以及其平均成绩：
select s#,avg(isnull(score,0)) 
from sc 
where s# in (select s# from sc where score<60 group by s# having count(*)>2)
group by s#;
select s_id,avg(score) from sc group by s_id 
having count(distinct case when score < 60 then c_id end)>2
49、检索“004”课程分数小于60，按分数降序排列的同学学号：
select s# 
from sc 
where c#='004' 
and score<60 
order by score desc;
50、删除“002”同学的“001”课程的成绩：
delect from sc 
where s#='002' 
and c#='001';