数据库经典50(45+5)例题

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于 2025-04-06 21:08:30 发布

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文章标签：数据库 sql mysql

本文链接：https://blog.youkuaiyun.com/qq_73704466/article/details/146819300

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1.数据库结构

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.创建数据表

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.搭建环境

选择新建数据库

插入数据，student表，course科目表，teacher教师表，SC成绩表

4.45例题

1.查询" 01 "课程比" 02 "课程成绩高的学生的信息及课程分数

我们需要将同学生表内每一个学生（课程表）的01课程和02课程先查询出来，再条件判断01课程是否大于02，然后再输出结果

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;

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 )

a:join到01课程和02的ID，就可以判断出两个课程了

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;

1-1. 查询同时存在" 01 "课程和" 02 "课程的情况

a:不使用join那么存在的就是01 02 课程都存在的情况

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;

1-3.查询不存在" 01 "课程但存在" 02 "课程的情况

a: 使用not in 即可

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

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;

3.查询在 SC 表存在成绩的学生信息

mysql> select DISTINCT student.*
    -> from student,sc
    -> where student.SId=sc.SId
    -> ;

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没选课的学生

mysql> select student.sid, student.sname,r.coursenumber,r.scoresum
    -> from student,
    -> (select sc.sid, sum(sc.score) as scoresum, count(sc.cid) as coursenumber from sc
    -> group by sc.sid)r
    -> where student.sid = r.sid;

4.1查有成绩的学生信息

select * from student where student.sid in (select sc.sid from sc);

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 student.* from student,teacher,course,sc
where student.sid = sc.sid and course.cid=sc.cid  and course.tid = teacher.tid and tname = '张三';

7.查询没有学全所有课程的同学的信息

SELECT s.*
FROM student s
WHERE s.sid NOT IN (
  SELECT sc.sid
  FROM sc
  GROUP BY sc.sid
  HAVING COUNT(sc.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 DISTINCT s.*
FROM student s
JOIN sc sc1 ON s.sid = sc1.sid
JOIN sc sc2 ON sc1.cid = sc2.cid
WHERE sc2.sid = '01';

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

SELECT s.*
FROM Student s
WHERE s.SId != '01'
AND NOT EXISTS (
    -- 找出学号为 '01' 的同学所学的课程
    SELECT sc.CId
    FROM SC sc
    WHERE sc.SId = '01'
    EXCEPT
    -- 找出其他同学所学的课程
    SELECT sc2.CId
    FROM SC sc2
    WHERE sc2.SId = s.SId
)
AND NOT EXISTS (
    -- 找出其他同学所学的课程
    SELECT sc2.CId
    FROM SC sc2
    WHERE sc2.SId = s.SId
    EXCEPT
    -- 找出学号为 '01' 的同学所学的课程
    SELECT sc.CId
    FROM SC sc
    WHERE sc.SId = '01'
);

10.查询没学过"张三"老师讲授的任一门课程的学生姓名

SELECT DISTINCT s.Sname
FROM Student s
WHERE s.SId NOT IN (
    SELECT sc.SId
    FROM SC sc
    JOIN Course c ON sc.CId = c.CId
    JOIN Teacher t ON c.TId = t.TId
    WHERE t.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 s.SId, s.Sname, AVG(sc.score) AS average_score
FROM Student s
JOIN SC sc ON s.SId = sc.SId
WHERE sc.score < 60
GROUP BY s.SId, s.Sname
HAVING COUNT(sc.CId) >= 2;

12.检索" 01 "课程分数小于 60，按分数降序排列的学生信息

SELECT s.SId, s.Sname, sc.score
FROM Student s
JOIN SC sc ON s.SId = sc.SId
WHERE sc.CId = '01' AND sc.score < 60
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 s.SId, s.Sname, c.Cname, sc.score, AVG(sc.score) OVER (PARTITION BY s.SId) AS average_score
FROM Student s
JOIN SC sc ON s.SId = sc.SId
JOIN Course c ON sc.CId = c.CId
ORDER BY average_score DESC, s.SId, c.Cname;

14.查询各科成绩最高分、最低分和平均分

SELECT c.CId, c.Cname, MAX(sc.score) AS max_score, MIN(sc.score) AS min_score, AVG(sc.score) AS average_score
FROM Course c
JOIN SC sc ON c.CId = sc.CId
GROUP BY c.CId, c.Cname;

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

16.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

SELECT c.CId, c.Cname,
       SUM(CASE WHEN sc.score BETWEEN 85 AND 100 THEN 1 ELSE 0 END) AS '100-85',
       SUM(CASE WHEN sc.score BETWEEN 70 AND 85 THEN 1 ELSE 0 END) AS '85-70',
       SUM(CASE WHEN sc.score BETWEEN 60 AND 70 THEN 1 ELSE 0 END) AS '70-60',
       SUM(CASE WHEN sc.score < 60 THEN 1 ELSE 0 END) AS '60-0',
       COUNT(sc.SId) AS total_students,
       (SUM(CASE WHEN sc.score BETWEEN 85 AND 100 THEN 1 ELSE 0 END) / COUNT(sc.SId)) * 100 AS '100-85%',
       (SUM(CASE WHEN sc.score BETWEEN 70 AND 85 THEN 1 ELSE 0 END) / COUNT(sc.SId)) * 100 AS '85-70%',
       (SUM(CASE WHEN sc.score BETWEEN 60 AND 70 THEN 1 ELSE 0 END) / COUNT(sc.SId)) * 100 AS '70-60%',
       (SUM(CASE WHEN sc.score < 60 THEN 1 ELSE 0 END) / COUNT(sc.SId)) * 100 AS '60-0%'
FROM Course c
JOIN SC sc ON c.CId = sc.CId
GROUP BY c.CId, c.Cname;

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

select cid, count(sid) 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.sid, student.sname from student
where student.sid in
(select sc.sid from sc
group by sc.sid
having count(sc.cid)=2
);

21.查询男生、女生人数

select ssex, count(*) from student
group by ssex;

22.查询名字中含有「风」字的学生信息

select *
from student 
where student.Sname like '%风%';

23.

查询同名学生名单，并统计同名人数 .找到同名的名字并统计个数

select sname, count(*) from student
group by sname
having count(*)>1;

24.查询 1990 年出生的学生名单

SELECT SId, Sname
FROM Student
WHERE YEAR(Sage) = 1990;

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 c.CId, c.Cname, AVG(sc.score) AS average_score
FROM Course c
JOIN SC sc ON c.CId = sc.CId
GROUP BY c.CId, c.Cname
ORDER BY average_score DESC, c.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.sid, student.sname, AVG(sc.score) AS aver
FROM student
JOIN sc ON student.sid = sc.sid
GROUP BY student.sid, student.sname
HAVING aver > 85;

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, sc.score from student, sc, course
where student.sid = sc.sid
and course.cid = sc.cid
and course.cname = "数学"
and sc.score < 60;

28.查询所有学生的课程及分数情况

select student.sname, cid, score 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,course,sc
where sc.score>70
and student.sid = sc.sid
and sc.cid = course.cid;

30.查询存在不及格的课程

SELECT DISTINCT sc.CId, c.Cname
FROM SC sc
JOIN Course c ON sc.CId = c.CId
WHERE sc.score < 60;

31.查询课程编号为 01 且课程成绩在 80 分及以上的学生的学号和姓名

select student.sid,student.sname 
from student,sc
where cid="01"
and score>=80
and student.sid = sc.sid;

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.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 s.SId, s.Sname, sc.CId, c.Cname, sc.score
FROM Student s
JOIN SC sc ON s.SId = sc.SId
JOIN Course c ON sc.CId = c.CId
JOIN Teacher t ON c.TId = t.TId
WHERE t.Tname = '张三'
ORDER BY sc.score DESC
LIMIT 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

SELECT sc1.SId, sc1.CId, sc1.score
FROM SC sc1
JOIN SC sc2 ON sc1.SId = sc2.SId AND sc1.CId != sc2.CId AND sc1.score = sc2.score;

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）

SELECT c.CId, c.Cname, COUNT(sc.SId) AS num_students
FROM Course c
JOIN SC sc ON c.CId = sc.CId
GROUP BY c.CId, c.Cname
HAVING COUNT(sc.SId) > 5;

38.检索至少选修两门课程的学生学号

SELECT sid, COUNT(cid) AS cc
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 s.*
FROM Student s
JOIN SC sc ON s.SId = sc.SId
GROUP BY s.SId, s.Sname, s.Sage, s.Ssex
HAVING COUNT(DISTINCT sc.CId) = (SELECT COUNT(*) FROM Course);

40.查询各学生的年龄，只按年份来算

SELECT SId, Sname, (YEAR(CURDATE()) - YEAR(Sage)) AS age
FROM Student;

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

5.5例题

学生表：Student (Sno, Sname, Ssex , Sage, Sdept)
学号，姓名，性别，年龄，所在系 Sno为主键
课程表：Course (Cno, Cname,)
课程号，课程名 Cno为主键
学生选课表：SC (Sno, Cno, Score)
学号，课程号，成绩 Sno,Cno为主键

1.用SQL语句创建学生表student，定义主键，姓名不能重名，性别只能输入男或女，所在系的默认值是 “计算机”。

CREATE TABLE Student (
    Sno VARCHAR(10) PRIMARY KEY,
    Sname VARCHAR(50) UNIQUE,
    Ssex ENUM('男', '女') NOT NULL,
    Sage INT,
    Sdept VARCHAR(50) DEFAULT '计算机'
);

2.修改student 表中年龄（age）字段属性，数据类型由int 改变为smallint。

ALTER TABLE Student MODIFY Sage SMALLINT;

3.为SC表建立按学号（sno）和课程号（cno）组合的升序的主键索引，索引名为SC_INDEX 。

CREATE TABLE SC (
    Sno VARCHAR(10),
    Cno VARCHAR(10),
    Score DECIMAL(5, 2),
    PRIMARY KEY (Sno, Cno)
);

-- 创建索引
CREATE INDEX SC_INDEX ON SC (Sno ASC, Cno ASC);

4.创建一视图 stu_info,查询全体学生的姓名，性别，课程名，成绩。

CREATE VIEW stu_info AS
SELECT s.Sname, s.Ssex, c.Cname, sc.Score
FROM Student s
JOIN SC sc ON s.Sno = sc.Sno
JOIN Course c ON sc.Cno = c.Cno;