50道SQL笔试编程题及答案整理

blackcatbiubiubiu 
写这篇整理文，一是给自己多多练习SQL的机会（SQL还是比较难的，
一下子写出很溜的语句还是要花时间的），二也是方便各位同路的小
伙伴，每道题附上难度值，从简单到难做完之后你肯定会有新的体会。



--建表
--学生表
CREATE TABLE `Student`(
`s_id` VARCHAR(20),
`s_name` VARCHAR(20) NOT NULL DEFAULT \'\',
`s_birth` VARCHAR(20) NOT NULL DEFAULT \'\',
`s_sex` VARCHAR(10) NOT NULL DEFAULT \'\',
PRIMARY KEY(`s_id`)
);
--课程表
CREATE TABLE `Course`(
`c_id` VARCHAR(20),
`c_name` VARCHAR(20) NOT NULL DEFAULT \'\',
`t_id` VARCHAR(20) NOT NULL,
PRIMARY KEY(`c_id`)
);
--教师表
CREATE TABLE `Teacher`(
`t_id` VARCHAR(20),
`t_name` VARCHAR(20) NOT NULL DEFAULT \'\',
PRIMARY KEY(`t_id`)
);
--成绩表
CREATE TABLE `Score`(
`s_id` VARCHAR(20),
`c_id` VARCHAR(20),
`s_score` INT(3),
PRIMARY KEY(`s_id`,`c_id`)
);
--插入学生表测试数据
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\');
--课程表测试数据
insert into Course values(\'01\' , \'语文\' , \'02\');
insert into Course values(\'02\' , \'数学\' , \'01\');
insert into Course values(\'03\' , \'英语\' , \'03\');

--教师表测试数据
insert into Teacher values(\'01\' , \'张三\');
insert into Teacher values(\'02\' , \'李四\');
insert into Teacher values(\'03\' , \'王五\');

--成绩表测试数据
insert into Score values(\'01\' , \'01\' , 80);
insert into Score values(\'01\' , \'02\' , 90);
insert into Score values(\'01\' , \'03\' , 99);
insert into Score values(\'02\' , \'01\' , 70);
insert into Score values(\'02\' , \'02\' , 60);
insert into Score values(\'02\' , \'03\' , 80);
insert into Score values(\'03\' , \'01\' , 80);
insert into Score values(\'03\' , \'02\' , 80);
insert into Score values(\'03\' , \'03\' , 80);
insert into Score values(\'04\' , \'01\' , 50);
insert into Score values(\'04\' , \'02\' , 30);
insert into Score values(\'04\' , \'03\' , 20);
insert into Score values(\'05\' , \'01\' , 76);
insert into Score values(\'05\' , \'02\' , 87);
insert into Score values(\'06\' , \'01\' , 31);
insert into Score values(\'06\' , \'03\' , 34);
insert into Score values(\'07\' , \'02\' , 89);
insert into Score values(\'07\' , \'03\' , 98);

笔试题

1、查询姓“张”的老师的个数

难度：**

select count(t_id)
from Teacher
where t_name like \'张%\';

2、查询课程编号为“02”的总成绩

要点：聚合

难度：**

select sum(s_score)
from Score
where c_id = \'02\';

3、查询每门课程被选修的学生数

难度：**

select c_id, count(s_id)
from Score
group by c_id;

4、查询选了课程的学生人数

难度：**

select count(DISTINCT s_id)  as 学生人数 from Score

5、查询名字中含有“风”字的学生信息

难度：**

select * 
from Student
where s_name like \'%风%\';

6、查询各个课程及相应的选修人数

难度：**

select a.c_id as 课程号,c_name as 课程名称, count(s_id)as 选修人数
from Score a join Course b on a.c_id=b.c_id
group by a.c_id;

7、删除学生编号为“02”的课程编号为“01”的成绩

难度：**

delete from Score
where s_id=\'02\'
and c_id=\'01\';

8、检索课程编号为“04”且分数小于60的学生学号，结果按按分数降序排列

难度：**

select s_id
from Score
where c_id=\'04\'
and s_score <60
order by s_score DESC;

 

 

9、查询至少选修两门课程的学生学号

难度：**

select s_id
from Score
group by s_id
having count(c_id)>=2;

10、查询出只选修了两门课程的全部学生的学号和姓名

难度：***

写法一：
select Student.s_id,s_name, count(c_id)
from Student join Score on Student.s_id= Score.s_id
group by s_id,s_name
having count(c_id)=2;
写法二：
select s_id,s_name
from Student
WHERE s_id IN (SELECT s_id FROM Score GROUP BY s_id having count(c_id)=2)

11、查询男生、女生人数

难度：***

select s_sex, count(*) from Student group by s_sex;

 

12、查询同名同姓学生名单并统计同名人数

难度：***

写法一：
select a.s_id, a.s_name,a.s_birth,a.s_sex
from Student AS a JOIN Student AS b
ON a.s_name=b.s_name
WHERE a.s_id<>b.s_id
写法二：
select s_name,count(*)
from Student
group by s_name
having count(*)>1

13、1990年出生的学生名单（注：Student表中s_birth列的类型是datetime）

难度：***

-- 方法一
select s_name
from Student
where s_birth like \'1990%\';

-- 方法二
select s_name
from Student
where year(s_birth)=1990;

14、查询平均成绩大于85的所有学生的学号、姓名和平均成绩

难度：***

select a.s_id as 学号, s_name as 姓名, avg(b.s_score) as 平均分
from Student a join Score b  on a.s_id=b.s_id
group by a.s_id,s_name
having avg(b.s_score) >85;

15、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

难度：***

select c_id, avg(s_score) 
from Score
group by c_id
order by avg(s_score),c_id DESC;

16、查询课程名称为“数学”且分数低于60的学生姓名和分数

难度：***

select s_name as 学生姓名, s_score as 分数
from Student a join Score b on a.s_id=b.s_id
join Course c on c.c_id=b.c_id
where c_name=\'数学\'
and s_score <60;

17、查询所有学生的选课情况

难度:***

select a.s_id as 学号, s_name as 姓名 , c.c_id as 课程号,c_name as 课程名称
from Student a join Score b on a.s_id=b.s_id
join Course c on b.c_id=c.c_id

 

 

 

18、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

难度：***

select s_name as 姓名,c_name as 课程名称,s_score as 分数
from Student a join Score b on at.s_id=b.s_id
join Course c on b.c_id=c.c_id
where s_score >70

 

 

 

19、查询不及格的课程并按课程号从大到小排列

难度：***

select c_id as 课程号 ,s_score as 分数
from Score 
where s_score<60
order by c_id;

20、查询平均成绩大于60分的学生的学号和平均成绩

难度：***

select s_id, avg(s_score)
from Score
group by s_id
having avg(s_score)>60;

21、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

难度：***

 

select c_id as 课程ID,
max(s_score) as 最高分,
min(s_score) as 最低分
from Score
group by c_id;

22、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序，若人数相同，按课程号升序排序

要点：分组，排序

难度:***

select c_id as \'课程号\', count(s_id) as \'选修人数\'
from Score
group by c_id
having count(s_id)>5
order by count(s_id) DESC,c_id;

 

 

 

23、查询所有学生的学号、姓名、选课数、总成绩

要点：内联结

难度：****

select Score.s_id AS \'学号\', s_name AS \'姓名\',count(c_id)AS \'选课数\',sum(s_score)AS \'总成绩\'
from Score join Student on Score.s_id = Student.s_id
group by Score.s_id,s_name;

24、查询没有学全所有课的学生的学号、姓名

难度：****

写法一：
select Student.s_id, Student.s_name
from Student join Score on Score.s_id = Student.s_id
group by s_id, s_name
having count(Score.c_id) < (select count(c_id) from Course);
写法二：
select s_id, s_name
from Student 
where s_id IN (SELECT s_id FROM Score group by s_id 
having count(c_id)<(select count(c_id) from Course))
小于还可以是不等于

25、查询不同老师所教不同课程平均分从高到低显示

要点：联结

难度：****

select Teacher.t_id,Teacher.t_name,Course.c_name,avg(Score.s_score)
from Teacher join Course on Teacher.t_id=Course.t_id
join Score on Course.c_id=Score.c_id
group by t_id,t_name,c_name
order by avg(Score.s_score) DESC;

26、查询课程编号为03且课程成绩在80分以上的学生的学号和姓名

难度：****

写法一：
select a.s_id as 学号 ,s_name as 姓名
from Student a join Score b on a.s_id=b.s_id
where c_id=\'03\'
and s_score>80;
写法二：
select s_id as 学号 ,s_name as 姓名
from Student 
where s_id IN(SELECT s_id FROM Score WHERE  c_id=\'03\'
and s_score>80);

 

27、查询选修了全部课程的学生信息

难度：****

写法一：
select a.s_id, s_name,s_birth, s_sex
from Student a join Score b on a.s_id=b.s_id
group by a.s_id
having count(Score.c_id)=(select count(distinct c_id) from Score);

写法二：
select * from Student 
where s_id in(select s_id from Score GROUP BY s_id 
having count(c_id)=(select count(distinct c_id) from Score))

28、查询没学过“张三”老师讲授的任一门课程的学生姓名

要点：三表联结

难度：****
select s_name from Student
where s_id not in 
(select s_id from Score join Course on Score.c_id=Course.c_id
join Teacher on Teacher.t_id=Course.t_id
where t_name = \'张三\');

29、查询两门以上不及格课程的同学的学号及其平均成绩

要点：执行顺序

难度：****

select s_id as 学号, avg(s_score)as 平均成绩
from Score
where s_score <60
group by s_id
having count(c_id)>=2;

 

 

 

 

 

30、查询课

 

 

 

 

程编号为“01”的课程比“02”的课程成绩高的所有学生的学号。

要点：联结

难度：*****

-- 写法一：
select a.s_id
from Score a join Score b on a.s_id = b.s_id and a.s_score >b.s_score
where a.c_id = \'01\' and b.c_id= \'02\';

-- 写法二：
select a.s_id
from (select * from Score where c_id = \'01\') as a 
join (select * from Score where c_id=\'02\') as b
on a.s_id = b.s_id
where a.s_score > b.s_score;

 

 

 

31.查询没学过“张三”老师课的学生的学号、姓名(重点)

要点: 三表联结

难度：*****

select s_id, s_name
from Student
where s_id not in 
(select s_id from Score join Course on Score.c_id = Course.c_id
join Teacher on Course.t_id = Teacher.t_id
where t_name = \'张三\');

32、查询学过“张三”老师所教的所有课的同学的学号、姓名

要点: 同上第5题

难度：*****

写法一：
select s_id, s_name
from Student
where s_id in
(select s_id from Score join Course on Score.c_id = Course.c_id
join Teacher on Course.t_id = Teacher.t_id
where t_name = \'张三\');
写法二：
select Student.s_id, s_name
from Student JOIN Score on Student.s_id = Score.s_id
JOIN Course on Score.c_id = Course.c_id
JOIN Teacher on Teacher.t_id = Course.t_id
where t_name=\'张三\';

33、查询学过编号为“01”的课程并且也学过编号为“02”的课程的学生的学号、姓名

难度：*****

--写法一：
select s_id, s_name
from Student
where s_id in
(select a.s_id from 
(select s_id from Score where c_id = \'01\') as a
join (select s_id from Score where c_id =\'02\') as b
on a.s_id= b.s_id);
--写法二：
select s_id,s_name
from Student 
where s_id in
(select s_id from Score where c_id = \'01\') 
AND s_id in
(select s_id from Score where c_id = \'02\')
--写法三：
select a.s_id,a.s_name
from Student a JOIN Score b ON a.s_id=b.s_id
JOIN Score c ON a.s_id=c.s_id
where b.c_id=\'01\' and c.c_id=\'02\'
and b.s_id=c.s_id ;

 

34、查询所有课程成绩小于60分的学生的学号、姓名

难度：*****

select s_id, s_name
from Student 
where s_id NOT IN (SELECT s_id FROM Score where s_score >=60)

 

35、查询至少有一门课与学号为“01”的学生所学课程相同的学生的学号和姓名

要点：联结

难度：*****
select distinct a.s_id, a.s_name
from Student a join Score b on a.s_id= b.s_id
where c_id in
(select c_id from Score where s_id = \'01\')
and a.s_id<>\'01\';
陷阱两处你知道吗

36、查询和“01”号同学所学课程完全相同的其他同学的学号

难度：*****

select s_id
from Score
where c_id in
(select c_id from Score where s_id=\'01\')
and s_id <> \'01\'
group by s_id
having count(c_id)=(select count(c_id) from Score where s_id=\'01\');

37、把“SCORE”表中“张三”老师教的课的成绩都更改为此课程的平均成绩

难度：*****

update Score as a join 
(select avg(s_score) as t, Score.c_id from Score 
join Course on Score.c_id= Course.c_id
join Teacher on Teacher.t_id= Course.t_id
where t_name =\'张三\' group by c_id) as b#张三老师教的课与平均分
on a.c_id= b.c_id
set a.s_score= b.t;

 

38、查询和“02”号的同学学习的课程完全相同的其他同学学号和姓名(同12题）

难度：*****

select a.s_id,a.s_name 
from Student a join Score b on a.s_id=b.s_id 
where c_id in (select c_id from Score where s_id=\'02\') 
and a.s_id <> \'02\' 
group by a.s_id 
having count(c_id)=(select count(c_id) from Score where s_id=\'02\');

 

39、删除学习“张三”老师课的SC表记录

难度：*****

delete from Score
where c_id in 
(select c_id from Course join Teacher on Course.t_id=Teacher.t_id
where t_name =\'张三\');

40、按平均成绩从高到低显示所有学生的“数据库”（c_id=\'04\'）、“企业管理”（c_id=\'01\'）、“英语”（c_id=\'06\'）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

难度：*****

select s_id as \'学生ID\',
(case when c_id=\'04\' then s_score else NULL end) as \'数据库\',
(case when c_id=\'01\' then s_score else NULL end) as \'企业管理\',
(case when c_id=\'06\' then s_score else NULL end) as \'英语\',
count(c_id) as 有效课程数,
avg(s_score) as 有效平均分
from Score
group by s_id
order by avg(s_score) DESC;

 

41、按各科平均成绩从低到高和及格率的百分数从高到低排列，以如下形式显示：课程号,课程名,平均成绩,及格百分数

要点：关联查询

难度：*****

select a.c_id as \'课程号\',c_name as \'课程名\',avg(s_score) as \'平均成绩\',
concat((select count(b.s_score) from Score b where b.s_score>=60 and a.c_id=b.c_id)/
(select count(b.s_score) from Score b where a.c_id=b.c_id)*100,\'%\') as \'及格百分数\'
from Score a join Course c
on a.c_id=c.c_id
group by a.c_id,c_name
order by 平均成绩, 及格百分数 DESC;

 

 

43、查询如下课程成绩第3名到第6名的学生成绩单，其中企业管理为001，马克思为002，UML为003，数据库为004，以如下形式显示：

学生ID学生姓名企业管理马克思UML数据库平均成绩

题目有问题，这道题不做

 

44、使用分段[100-85],[85-70],[70-60],[<60]来统计各科成绩，分别统计各分数段人数：课程ID和课程名称

要点：case

难度：*****

select a.c_id AS \'课程ID\',c_name \'课程名称\',
sum(case when s_score between 85 and 100 then 1 else 0 end) as \'[100-85]\',
sum(case when s_score >=70 and s_score<85 then 1 else 0 end) as \'[85-70]\',
sum(case when s_score>=60 and s_score<70  then 1 else 0 end) as \'[70-60]\',
sum(case when s_score<60 then 1 else 0 end) as \'[<60]\'
from Score a join Course b on a.c_id=b.c_id
group by a.c_id,c_name;

 

45、查询学生平均成绩及其名次

要点：算出在所有同学中有几个同学的平均分高于某个ID,然后+1，就是名次

难度：*****

SELECT s_id as \'学号\',平均成绩,
      (SELECT COUNT(*) FROM(SELECT s_id,AVG(s_score)AS \'平均成绩\'
                            FROM Score GROUP BY s_id)AS b 
       WHERE b.平均成绩>a.平均成绩)+1 as RANK
FROM (select s_id,avg(S_score) as 平均成绩 from Score group by s_id)AS a
order by 平均成绩 desc;

46、查询各科成绩前三名的记录（不考虑成绩并列情况）

要点：超过当前ID的人最多2人

难度：*****

这里笔者大部分满足了题意，但是没有满足“不考虑成绩并列情况”，欢迎给出建议！

SELECT c_id,s_id,s_score FROM Score a 
WHERE (SELECT COUNT(*) FROM Score b WHERE a.c_id=b.c_id AND a.s_score<b.s_score)<=2 
ORDER BY c_id ASC,s_score DESC

 

 

 

 

47、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

难度：*****

select s_name as 学生姓名, s_score as 成绩
from Student a join Score b on a.s_id=b.s_id
join Course c on c.c_id=b.c_id
join Teacher d on d.t_id=c.t_id
where t_name=\'张三\'
order by s_score DESC
limit 1;

 

 

48、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

难度：*****

select distinct a.s_id as 学生编号 ,a.c_id as 课程编号,a.s_score as 学生成绩
from Score a join Score b
on a.s_id=b.s_id and a.c_id<> b.c_id
where a.s_score=b.s_score;

49、查询每门课程成绩最好的前两名

难度：*****

SELECT a.c_id as 课程号,c_name as 课程名称,s_name as 姓名,b.s_score as chengji  FROM 
Course a join Score b on a.c_id=b.c_id
join Student c on c.s_id=b.s_id
WHERE (SELECT COUNT(*) FROM Score d WHERE a.c_id=d.c_id AND b.s_score<d.s_score)<=1
ORDER BY a.c_id ASC,b.s_score DESC

 

50、查询如下课程平均成绩和及格率的百分数（用1行显示），其中企业管理为001，马克思为002，UML为003，数据库为004

这道题有问题不用做

 

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