MySql教程

MySQL多表查询练习题

本文主要是介绍MySQL多表查询练习题,对大家解决编程问题具有一定的参考价值,需要的程序猿们随着小编来一起学习吧!

一:准备数据

#创建表及插入记录

CREATE TABLE class (
  cid int(11) NOT NULL AUTO_INCREMENT,
  caption varchar(32) NOT NULL,
  PRIMARY KEY (cid)
) ENGINE=InnoDB CHARSET=utf8;
INSERT INTO class VALUES
(1, '三年二班'), 
(2, '三年三班'), 
(3, '一年二班'), 
(4, '二年九班');

CREATE TABLE course(
  cid int(11) NOT NULL AUTO_INCREMENT,
  cname varchar(32) NOT NULL,
  teacher_id int(11) NOT NULL,
  PRIMARY KEY (cid),
  KEY fk_course_teacher (teacher_id),
  CONSTRAINT fk_course_teacher FOREIGN KEY (teacher_id) REFERENCES teacher (tid)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;

INSERT INTO course VALUES
(1, '生物', 1), 
(2, '物理', 2), 
(3, '体育', 3), 
(4, '美术', 2);

CREATE TABLE score (
  sid int(11) NOT NULL AUTO_INCREMENT,
  student_id int(11) NOT NULL,
  course_id int(11) NOT NULL,
  num int(11) NOT NULL,
  PRIMARY KEY (sid),
  KEY fk_score_student (student_id),
  KEY fk_score_course (course_id),
  CONSTRAINT fk_score_course FOREIGN KEY (course_id) REFERENCES course (cid),
  CONSTRAINT fk_score_student FOREIGN KEY (student_id) REFERENCES student(sid)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;

INSERT INTO score VALUES
(1, 1, 1, 10),
(2, 1, 2, 9),
(5, 1, 4, 66),
(6, 2, 1, 8),
(8, 2, 3, 68),
(9, 2, 4, 99),
(10, 3, 1, 77),
(11, 3, 2, 66),
(12, 3, 3, 87),
(13, 3, 4, 99),
(14, 4, 1, 79),
(15, 4, 2, 11),
(16, 4, 3, 67),
(17, 4, 4, 100),
(18, 5, 1, 79),
(19, 5, 2, 11),
(20, 5, 3, 67),
(21, 5, 4, 100),
(22, 6, 1, 9),
(23, 6, 2, 100),
(24, 6, 3, 67),
(25, 6, 4, 100),
(26, 7, 1, 9),
(27, 7, 2, 100),
(28, 7, 3, 67),
(29, 7, 4, 88),
(30, 8, 1, 9),
(31, 8, 2, 100),
(32, 8, 3, 67),
(33, 8, 4, 88),
(34, 9, 1, 91),
(35, 9, 2, 88),
(36, 9, 3, 67),
(37, 9, 4, 22),
(38, 10, 1, 90),
(39, 10, 2, 77),
(40, 10, 3, 43),
(41, 10, 4, 87),
(42, 11, 1, 90),
(43, 11, 2, 77),
(44, 11, 3, 43),
(45, 11, 4, 87),
(46, 12, 1, 90),
(47, 12, 2, 77),
(48, 12, 3, 43),
(49, 12, 4, 87),
(52, 13, 3, 87);


CREATE TABLE student(
  sid int(11) NOT NULL AUTO_INCREMENT,
  gender char(1) NOT NULL,
  class_id int(11) NOT NULL,
  sname varchar(32) NOT NULL,
  PRIMARY KEY (sid),
  KEY fk_class (class_id),
  CONSTRAINT fk_class FOREIGN KEY (class_id) REFERENCES class (cid)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;

INSERT INTO student VALUES
(1, '男', 1, '理解'), 
(2, '女', 1, '钢蛋'), 
(3, '男', 1, '张三'), 
(4, '男', 1, '张一'), 
(5, '女', 1, '张二'), 
(6, '男', 1, '张四'), 
(7, '女', 2, '铁锤'), 
(8, '男', 2, '李三'), 
(9, '男', 2, '李一'), 
(10, '女', 2, '李二'), 
(11, '男', 2, '李四'), 
(12, '女', 3, '如花'), 
(13, '男', 3, '刘三'), 
(14, '男', 3, '刘一'), 
(15, '女', 3, '刘二'), 
(16, '男', 3, '刘四');

CREATE TABLE teacher(
  tid int(11) NOT NULL AUTO_INCREMENT,
  tname varchar(32) NOT NULL,
  PRIMARY KEY (tid)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;

INSERT INTO teacher VALUES
(1, '张磊老师'), 
(2, '李平老师'), 
(3, '刘海燕老师'), 
(4, '朱云海老师'), 
(5, '李杰老师');

二、题目一

1、查询所有的课程的名称以及对应的任课老师姓名

2、查询学生表中男女生各有多少人

3、查询物理成绩等于100的学生的姓名

4、查询平均成绩大于八十分的同学的姓名和平均成绩

5、查询所有学生的学号,姓名,选课数,总成绩

6、 查询姓李老师的个数

7、 查询没有报李平老师课的学生姓名

8、 查询物理课程比生物课程高的学生的学号

9、 查询没有同时选修物理课程和体育课程的学生姓名

10、查询挂科超过两门(包括两门)的学生姓名和班级
、查询选修了所有课程的学生姓名

12、查询李平老师教的课程的所有成绩记录

13、查询全部学生都选修了的课程号和课程名

14、查询每门课程被选修的次数

15、查询之选修了一门课程的学生姓名和学号

16、查询所有学生考出的成绩并按从高到低排序(成绩去重)

17、查询平均成绩大于85的学生姓名和平均成绩

18、查询生物成绩不及格的学生姓名和对应生物分数

19、查询在所有选修了李平老师课程的学生中,这些课程(李平老师的课程,不是所有课程)平均成绩最高的学生姓名

20、查询每门课程成绩最好的前两名学生姓名

21、查询不同课程但成绩相同的学号,课程号,成绩

22、查询没学过“叶平”老师课程的学生姓名以及选修的课程名称;

23、查询所有选修了学号为1的同学选修过的一门或者多门课程的同学学号和姓名;

24、任课最多的老师中学生单科成绩最高的学生姓名

三、答案

#1、查询所有的课程的名称以及对应的任课老师姓名

SELECT
    course.cname,
    teacher.tname
FROM
    course
INNER JOIN teacher ON course.teacher_id = teacher.tid;

#2、查询学生表中男女生各有多少人

SELECT
    gender 性别,
    count(1) 人数
FROM
    student
GROUP BY
    gender;

#3、查询物理成绩等于100的学生的姓名

SELECT
    student.sname
FROM
    student
WHERE
    sid IN (
        SELECT
            student_id
        FROM
            score
        INNER JOIN course ON score.course_id = course.cid
        WHERE
            course.cname = '物理'
        AND score.num = 100
    );

#4、查询平均成绩大于八十分的同学的姓名和平均成绩

SELECT
    student.sname,
    t1.avg_num
FROM
    student
INNER JOIN (
    SELECT
        student_id,
        avg(num) AS avg_num
    FROM
        score
    GROUP BY
        student_id
    HAVING
        avg(num) > 80
) AS t1 ON student.sid = t1.student_id;

#5、查询所有学生的学号,姓名,选课数,总成绩(注意:对于那些没有选修任何课程的学生也算在内)

SELECT
    student.sid,
    student.sname,
    t1.course_num,
    t1.total_num
FROM
    student
LEFT JOIN (
    SELECT
        student_id,
        COUNT(course_id) course_num,
        sum(num) total_num
    FROM
        score
    GROUP BY
        student_id
) AS t1 ON student.sid = t1.student_id;

#6、 查询姓李老师的个数

SELECT
    count(tid)
FROM
    teacher
WHERE
    tname LIKE '李%';

#7、 查询没有报李平老师课的学生姓名(找出报名李平老师课程的学生,然后取反就可以)

SELECT
    student.sname
FROM
    student
WHERE
    sid NOT IN (
        SELECT DISTINCT
            student_id
        FROM
            score
        WHERE
            course_id IN (
                SELECT
                    course.cid
                FROM
                    course
                INNER JOIN teacher ON course.teacher_id = teacher.tid
                WHERE
                    teacher.tname = '李平老师'
            )
    );

#8、 查询物理课程比生物课程高的学生的学号(分别得到物理成绩表与生物成绩表,然后连表即可)

SELECT
    t1.student_id
FROM
    (
        SELECT
            student_id,
            num
        FROM
            score
        WHERE
            course_id = (
                SELECT
                    cid
                FROM
                    course
                WHERE
                    cname = '物理'
            )
    ) AS t1
INNER JOIN (
    SELECT
        student_id,
        num
    FROM
        score
    WHERE
        course_id = (
            SELECT
                cid
            FROM
                course
            WHERE
                cname = '生物'
        )
) AS t2 ON t1.student_id = t2.student_id
WHERE
    t1.num > t2.num;

#9、 查询没有同时选修物理课程和体育课程的学生姓名(没有同时选修指的是选修了一门的,思路是得到物理+体育课程的学生信息表,然后基于学生分组,统计count(课程)=1)

SELECT
    student.sname
FROM
    student
WHERE
    sid IN (
        SELECT
            student_id
        FROM
            score
        WHERE
            course_id IN (
                SELECT
                    cid
                FROM
                    course
                WHERE
                    cname = '物理'
                OR cname = '体育'
            )
        GROUP BY
            student_id
        HAVING
            COUNT(course_id) = 1
    );

#10、查询挂科超过两门(包括两门)的学生姓名和班级(求出<60的表,然后对学生进行分组,统计课程数目>=2)

SELECT
    student.sname,
    class.caption
FROM
    student
INNER JOIN (
    SELECT
        student_id
    FROM
        score
    WHERE
        num < 60
    GROUP BY
        student_id
    HAVING
        count(course_id) >= 2
) AS t1
INNER JOIN class ON student.sid = t1.student_id
AND student.class_id = class.cid;

#11、查询选修了所有课程的学生姓名(先从course表统计课程的总数,然后基于score表按照student_id分组,统计课程数据等于课程总数即可)

SELECT
    student.sname
FROM
    student
WHERE
    sid IN (
        SELECT
            student_id
        FROM
            score
        GROUP BY
            student_id
        HAVING
            COUNT(course_id) = (SELECT count(cid) FROM course)
    );

#12、查询李平老师教的课程的所有成绩记录

SELECT
    *
FROM
    score
WHERE
    course_id IN (
        SELECT
            cid
        FROM
            course
        INNER JOIN teacher ON course.teacher_id = teacher.tid
        WHERE
            teacher.tname = '李平老师'
    );

#13、查询全部学生都选修了的课程号和课程名(取所有学生数,然后基于score表的课程分组,找出count(student_id)等于学生数即可)

SELECT
    cid,
    cname
FROM
    course
WHERE
    cid IN (
        SELECT
            course_id
        FROM
            score
        GROUP BY
            course_id
        HAVING
            COUNT(student_id) = (
                SELECT
                    COUNT(sid)
                FROM
                    student
            )
    );

#14、查询每门课程被选修的次数

SELECT
    course_id,
    COUNT(student_id)
FROM
    score
GROUP BY
    course_id;

#15、查询之选修了一门课程的学生姓名和学号

SELECT
    sid,
    sname
FROM
    student
WHERE
    sid IN (
        SELECT
            student_id
        FROM
            score
        GROUP BY
            student_id
        HAVING
            COUNT(course_id) = 1
    );

#16、查询所有学生考出的成绩并按从高到低排序(成绩去重)

SELECT DISTINCT
    num
FROM
    score
ORDER BY
    num DESC;

#17、查询平均成绩大于85的学生姓名和平均成绩

SELECT
    sname,
    t1.avg_num
FROM
    student
INNER JOIN (
    SELECT
        student_id,
        avg(num) avg_num
    FROM
        score
    GROUP BY
        student_id
    HAVING
        AVG(num) > 85
) t1 ON student.sid = t1.student_id;

#18、查询生物成绩不及格的学生姓名和对应生物分数

SELECT
    sname 姓名,
    num 生物成绩
FROM
    score
LEFT JOIN course ON score.course_id = course.cid
LEFT JOIN student ON score.student_id = student.sid
WHERE
    course.cname = '生物'
AND score.num < 60;

#19、查询在所有选修了李平老师课程的学生中,这些课程(李平老师的课程,不是所有课程)平均成绩最高的学生姓名

SELECT
    sname
FROM
    student
WHERE
    sid = (
        SELECT
            student_id
        FROM
            score
        WHERE
            course_id IN (
                SELECT
                    course.cid
                FROM
                    course
                INNER JOIN teacher ON course.teacher_id = teacher.tid
                WHERE
                    teacher.tname = '李平老师'
            )
        GROUP BY
            student_id
        ORDER BY
            AVG(num) DESC
        LIMIT 1
    );

#20、查询每门课程成绩最好的前两名学生姓名
#查看每门课程按照分数排序的信息,为下列查找正确与否提供依据

SELECT
    *
FROM
    score
ORDER BY
    course_id,
    num DESC;

#表1:求出每门课程的课程course_id,与最高分数first_num

SELECT
    course_id,
    max(num) first_num
FROM
    score
GROUP BY
    course_id;

#表2:去掉最高分,再按照课程分组,取得的最高分,就是第二高的分数

second_num
SELECT
    score.course_id,
    max(num) second_num
FROM
    score
INNER JOIN (
    SELECT
        course_id,
        max(num) first_num
    FROM
        score
    GROUP BY
        course_id
) AS t ON score.course_id = t.course_id
WHERE
    score.num < t.first_num
GROUP BY
    course_id;

#将表1和表2联合到一起,得到一张表t3,包含课程course_id与该们课程的

first_num与second_num
SELECT
    t1.course_id,
    t1.first_num,
    t2.second_num
FROM
    (
        SELECT
            course_id,
            max(num) first_num
        FROM
            score
        GROUP BY
            course_id
    ) AS t1
INNER JOIN (
    SELECT
        score.course_id,
        max(num) second_num
    FROM
        score
    INNER JOIN (
        SELECT
            course_id,
            max(num) first_num
        FROM
            score
        GROUP BY
            course_id
    ) AS t ON score.course_id = t.course_id
    WHERE
        score.num < t.first_num
    GROUP BY
        course_id
) AS t2 ON t1.course_id = t2.course_id;

#查询前两名的学生(有可能出现并列第一或者并列第二的情况)

SELECT
    score.student_id,
    t3.course_id,
    t3.first_num,
    t3.second_num
FROM
    score
INNER JOIN (
    SELECT
        t1.course_id,
        t1.first_num,
        t2.second_num
    FROM
        (
            SELECT
                course_id,
                max(num) first_num
            FROM
                score
            GROUP BY
                course_id
        ) AS t1
    INNER JOIN (
        SELECT
            score.course_id,
            max(num) second_num
        FROM
            score
        INNER JOIN (
            SELECT
                course_id,
                max(num) first_num
            FROM
                score
            GROUP BY
                course_id
        ) AS t ON score.course_id = t.course_id
        WHERE
            score.num < t.first_num
        GROUP BY
            course_id
    ) AS t2 ON t1.course_id = t2.course_id
) AS t3 ON score.course_id = t3.course_id
WHERE
    score.num >= t3.second_num
AND score.num <= t3.first_num;

#排序后可以看的明显点

SELECT
    score.student_id,
    t3.course_id,
    t3.first_num,
    t3.second_num
FROM
    score
INNER JOIN (
    SELECT
        t1.course_id,
        t1.first_num,
        t2.second_num
    FROM
        (
            SELECT
                course_id,
                max(num) first_num
            FROM
                score
            GROUP BY
                course_id
        ) AS t1
    INNER JOIN (
        SELECT
            score.course_id,
            max(num) second_num
        FROM
            score
        INNER JOIN (
            SELECT
                course_id,
                max(num) first_num
            FROM
                score
            GROUP BY
                course_id
        ) AS t ON score.course_id = t.course_id
        WHERE
            score.num < t.first_num
        GROUP BY
            course_id
    ) AS t2 ON t1.course_id = t2.course_id
) AS t3 ON score.course_id = t3.course_id
WHERE
    score.num >= t3.second_num
AND score.num <= t3.first_num
ORDER BY
    course_id;

#可以用以下命令验证上述查询的正确性

SELECT
    *
FROM
    score
ORDER BY
    course_id,
    num DESC;

– 21、查询不同课程但成绩相同的学号,课程号,成绩
– 22、查询没学过“叶平”老师课程的学生姓名以及选修的课程名称;
– 23、查询所有选修了学号为1的同学选修过的一门或者多门课程的同学学号和姓名;
– 24、任课最多的老师中学生单科成绩最高的学生姓名

转载:https://www.cnblogs.com/12345huangchun/p/10123881.html

子查询总结

  • 子查询如果查询出的是一个字段(单列),那就再where后面作为条件使用
  • 子查询如果查询出的是一个表(多列),就当一张表去操作(起别名)

三范式

  • 第一规范 1NF

    • 原子性,做到列不可在拆分
    • 第一范式也是最基本的范式,数据库中表子段都是单一属性,不可拆分
  • 第二范式2NF

    • 再第一范式基础上更进一步,目标是确保表中的每列和主键相关
    • 一张表只能够描述一件事情
  • 第三范式3NF

    • 消除传递依赖
      • 表的信息,如果可以被推导出来,就不应该再单独的去设计一个字段来存放

数据库反三范式

  • 反三范式只等通过增加冗余和重复的数据来提高数据的读性能
  • 浪费存储空间,节省了查询时间(以空间换时间)
  • 冗余字段
    • 设计数据库的时候,某一个字段属于一张表,当他同时出现在另一个表或多个表中,并且完全等同于他在原来所属的意义一样,那么这个字段就被称为一个冗余字段

总结:

  • 创建一个关系型数据库设计,我们有两种选择
    1. 尽量遵循范式理论的规则,尽可能的减少冗余字段,让数据库设计看起来精致、优雅,让人心醉
    2. 合理的加入冗余字段这个润滑剂,较少join,让数据库执行性能更高更快
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