50道SQL练习题及答案与详细分析

川长思鸟来 2024-04-08 10:09 10阅读 0赞

## 一、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

## 二、插入数据 ##

#### 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);

## 三、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

## 四、题目答案 ##

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 student s right join (
    select s1.SId, s1.score as score1, s2.score as score2 from
                (select SId, score FROM sc where CId = 01) s1,
                (select SId, score from sc where CId = 02) s2
    where s1.score > s2.score and s1.SId = s2.SId) ss
    on s.SId = ss.SId;

select * from (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
    left join student s
    on r.SId = s.SId;

1.1 查询同时存在" 01 “课程和” 02 "课程的情况

select * from 
        (select * from sc where sc.CId = '01') as t1, 
        (select * from sc where sc.CId = '02') as t2
    where 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;

select * from 
    (select * from sc where sc.CId = '02') as t2
    right join 
    (select * from sc where sc.CId = '01') as t1
    on t1.SId = t2.SId;

1.3 查询不存在" 01 “课程但存在” 02 "课程的情况

select * from sc
    where sc.SId not in (
        select SId from sc 
        where sc.CId = '01'
    ) 
    AND sc.CId= '02';

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。根据学生ID把成绩分组，对分组中的score求平均值，最后在选取结果中AVG大于60的即可

select student.SId,sname,ss from student,(
        select SId, AVG(score) as ss from sc  
        GROUP BY SId 
        HAVING AVG(score)> 60
        )r
    where student.sid = r.sid;

select Student.SId, Student.Sname, r.ss from Student right join(
          select SId, AVG(score) AS ss from sc
          GROUP BY SId
          HAVING AVG(score)> 60
    )r on Student.SId = r.SId;

select s.SId,ss,Sname from(
    select SId, AVG(score) as ss from sc  
    GROUP BY SId 
    HAVING AVG(score)> 60
    )r left join 
    (select Student.SId, Student.Sname from
    Student)s on s.SId = r.SId;

3. 查询在 SC 表存在成绩的学生信息

select * from student s where s.SId in
    (select distinct SId from sc);

4. 查询所有同学的学生编号、学生姓名、选课总数、所有课程的成绩总和。联合查询不显示没选课的学生

select s2.SId, s2.Sname, s3.counts, s3.sums from student s2 right join
    (select s1.sid, count(CId) counts, sum(s1.score) sums from sc s1 group by s1.SId)s3
    on s2.SId = s3.SId;

4.1 查有成绩的学生信息

select * from student where SId in
    (select SId from sc);

select * from student 
    where exists (select sc.sid from sc where student.sid = sc.sid);

5. 查询「李」姓老师的数量

select count(TId) as nums from teacher where Tname like "李%";

6. 查询学过「张三」老师授课的同学的信息

select * from student where SId in (
    select SId from sc where CId = (
    select CId from course where TId = (
    select TId from teacher where Tname = "张三")));

select student.* from teacher, course, student, sc where teacher.Tname="张三"
    and course.TId=teacher.TId
    and course.CId=sc.CId
    and 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 in (
    select s1.SId from sc s1 inner join (select CId from sc where SId = 1) s2
    on s1.CId = s2.CId and s1.SId != 1 group by SId);

select * from student where SId in (
    select sid from sc group by SId having count(*) < (
        select count(*) 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 s1.* from student s1,
    (select s1.SId from sc s1 inner join (select CId from sc where SId = 1) s2
                  on s1.CId = s2.CId and s1.SId != 1 group by SId) s3
    where s1.SId=s3.SId;

select * from student where SId in (
    select s1.SId from sc s1 inner join (select CId from sc where SId = 1) s2
    on s1.CId = s2.CId and s1.SId != 1 group by 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

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 student where student.SId not in(
    select sc.SId from sc where CId in (
    select course.cid from course where TId in (
    select teacher.TId from teacher where Tname = '张三')));

select * from student where student.SId not in(
    select s2.SId from teacher t1, course c1, sc s2 
    where t1.Tname = '张三' and t1.TId = c1.TId and c1.CId = s2.CId);

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

思路: **这里可以转换成先把不及格的课程得到, 然后统计两门及其以上的课程数量**

select s1.SId, s1.Sname, avg(s2.score) 
    from student as s1, sc as s2 
    where s1.SId = s2.SId and s2.score<60 
    group by s1.SId 
    having count(s2.CId) >1;

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 s2 
    where s2.SId in (
    select s1.SId from sc s1 
    where s1.CId = 01 and s1.score < 60 order by s1.score desc);

select * from student s1, sc s2 
    where s1.SId = s2.SId and s2.CId = 01 and s2.score < 60 
    order by s2.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 s2.*, s3.* from sc s2,
    (select s1.SId, avg(s1.score) as avgs from sc s1 group by s1.SId) s3
    where s2.SId = s3.SId order by s3.avgs desc;

select * from sc s1 left join (
        select s2.SId, avg(s2.score) avgs from sc s2 group by s2.SId) s3
    on s1.SId = s3.SId order by avgs desc;

原文链接: [50道SQL练习题及答案与详细分析\_弥川的博客-CSDN博客\_sql选择题及答案][50_SQL_-CSDN_sql]

[50_SQL_-CSDN_sql]: https://blog.csdn.net/qq_41577017/article/details/124131848