MySQL - 数据库设计(表之间的 3 种关联关系)

表与表之间一般存在三种关系,即一对一,一对多,多对多关系。

下面分别就三种关系讲解数据库相关设计的思路和思考过程。

一、一对一关系
例如,下面的一张表,保存了人的相关信息,有男有女,要求查处所有的夫妻。

sql代码:`CREATE TABLE IF NOT EXISTS person(
id INT PRIMARY KEY AUTO_INCREMENT,
sname VARCHAR(10),
sex CHAR(1),
husband INT,
wife INT
);

INSERT INTO person VALUES(1,'小花','0',3,0);
INSERT INTO person VALUES(2,'小明','1',0,4);
INSERT INTO person VALUES(3,'张三','1',0,1);
INSERT INTO person VALUES(4,'小丽','0',2,0);
INSERT INTO person VALUES(5,'王五','1',0,0);`

sql代码对应的person表:

从表中可以看出,小花和张三是夫妻,小明和小丽是夫妻。通常为了查询方便,需要两个表,但实际项目中为了省空间,通常只建一个表,要实现一对一查询,可以建立两个视图(虚表)

  • 有个小细节提一下,视图只是个逻辑概念,并不独立占用物理内存,它依附于数据表,对视图进行修改时相应的表数据也会盖改动 *
    CREATE VIEW men AS (SELECT * FROM person WHERE sex='1'); CREATE VIEW women AS (SELECT * FROM person WHERE sex='0');
    men视图:

    women视图:

    执行查询:
    方式一(原始方法)
    SELECT men.sname AS husband,women.sname AS wife FROM men,women WHERE men.id = women.husband AND women.id = men.wife;
    方式二(连接方法)
    SELECT men.sname AS husband,women.sname AS wife FROM men INNER JOIN women ON men.id = women.husband AND women.id = men.wife;
    二、一对多关系
    例如:一个人可以拥有多辆汽车,要求查询某个人拥有的所有车辆。

分析:这种情况其实也可以采用 一张表,但因为一个人可以拥有多辆汽车,如果采用一张表,会造成冗余信息过多。好的设计方式是,人和车辆分别单独建表,那么如何将两个表关联呢?有个巧妙的方法,在车辆的表中加个外键字段(人的编号)即可。

  • (思路小结:’建两个表,一’方不动,’多’方添加一个外键字段) *
    sql代码:
    `// 建立人员表
    CREATE TABLE human(
    id VARCHAR(12) PRIMARY KEY,
    sname VARCHAR(12),
    age INT,
    sex CHAR(1)
    );
    INSERT INTO human VALUES('H001','小王',27,'1');
    INSERT INTO human VALUES('H002','小明',24,'1');
    INSERT INTO human VALUES('H003','张慧',28,'0');
    INSERT INTO human VALUES('H004','李小燕',35,'0');
    INSERT INTO human VALUES('H005','王大拿',29,'1');
    INSERT INTO human VALUES('H006','周强',36,'1');

// 建立车辆信息表
CREATE TABLE car(
id VARCHAR(12) PRIMARY KEY,
mark VARCHAR(24),
price NUMERIC(6,2),
hid VARCHAR(12),
CONSTRAINT fk_human FOREIGN KEY(hid) REFERENCES human(id)
);
INSERT INTO car VALUES('C001','BMW',65.99,'H001');
INSERT INTO car VALUES('C002','BenZ',75.99,'H002');
INSERT INTO car VALUES('C003','Skoda',23.99,'H001');
INSERT INTO car VALUES('C004','Peugeot',20.99,'H003');
INSERT INTO car VALUES('C005','Porsche',295.99,'H004');
INSERT INTO car VALUES('C006','Honda',24.99,'H005');
INSERT INTO car VALUES('C007','Toyota',27.99,'H006');
INSERT INTO car VALUES('C008','Kia',18.99,'H002');
INSERT INTO car VALUES('C009','Bentley',309.99,'H005');sql代码对应的人员表: ![](https://img2023.cnblogs.com/blog/1423504/202308/1423504-20230822093937539-736404089.png) sql代码对应的车辆信息表: ![](https://img2023.cnblogs.com/blog/1423504/202308/1423504-20230822093955773-1116458173.png) 执行查询 SELECT human.sname AS 车主,car.mark AS 车辆 FROM human,car WHERE human.id = car.hid; SELECT human.sname AS 车主,car.mark AS 车辆 FROM human INNER JOIN car WHERE human.id = car.hid; 三、多对多关系 例如:学生选课,一个学生可以选修多门课程,每门课程可供多个学生选择。 分析:这种方式可以按照类似一对多方式建表,但冗余信息太多,好的方式是实体和关系分离并单独建表,实体表为学生表和课程表,关系表为选修表,其中关系表采用联合主键的方式(由学生表主键和课程表主键组成)建表。 sql代码:// 建立学生表
CREATE TABLE student(
id VARCHAR(10) PRIMARY KEY,
sname VARCHAR(12),
age INT,
sex CHAR(1),
class VARCHAR(6)
);
INSERT INTO student VALUES('S0001','王军',20,1,'c101');
INSERT INTO student VALUES('S0002','张宇',21,1,'c101');
INSERT INTO student VALUES('S0003','刘飞',22,1,'c102');
INSERT INTO student VALUES('S0004','赵燕',18,0,'c103');
INSERT INTO student VALUES('S0005','曾婷',19,0,'c103');
INSERT INTO student VALUES('S0006','周慧',21,0,'c104');
INSERT INTO student VALUES('S0007','小红',23,0,'c104');
INSERT INTO student VALUES('S0008','杨晓',18,0,'c104');
INSERT INTO student VALUES('S0009','李杰',20,1,'c105');
INSERT INTO student VALUES('S0010','张良',22,1,'c105');

// 建立课程表
CREATE TABLE course(
id VARCHAR(10) PRIMARY KEY,
sname VARCHAR(12),
credit NUMERIC(2,1),
teacher VARCHAR(12)
);
INSERT INTO course VALUES('C001','Java',3.5,'李老师');
INSERT INTO course VALUES('C002','高等数学',5.0,'赵老师');
INSERT INTO course VALUES('C003','JavaScript',3.5,'王老师');
INSERT INTO course VALUES('C004','离散数学',3.5,'卜老师');
INSERT INTO course VALUES('C005','数据库',3.5,'廖老师');
INSERT INTO course VALUES('C006','操作系统',3.5,'张老师');

// 建立选修表
CREATE TABLE sc(
sid VARCHAR(10),
cid VARCHAR(10)
);

ALTER TABLE sc ADD CONSTRAINT pk_sc PRIMARY KEY(sid,cid);
ALTER TABLE sc ADD CONSTRAINT fk_student FOREIGN KEY(sid) REFERENCES student(id);
ALTER TABLE sc ADD CONSTRAINT fk_course FOREIGN KEY(cid) REFERENCES course(id);

INSERT INTO sc VALUES('S0001','C001');
INSERT INTO sc VALUES('S0001','C002');
INSERT INTO sc VALUES('S0001','C003');
INSERT INTO sc VALUES('S0002','C001');
INSERT INTO sc VALUES('S0002','C004');
INSERT INTO sc VALUES('S0003','C002');
INSERT INTO sc VALUES('S0003','C005');
INSERT INTO sc VALUES('S0004','C003');
INSERT INTO sc VALUES('S0005','C001');
INSERT INTO sc VALUES('S0006','C004');
INSERT INTO sc VALUES('S0007','C002');
INSERT INTO sc VALUES('S0008','C003');
INSERT INTO sc VALUES('S0009','C001');
INSERT INTO sc VALUES('S0009','C005');`
sql代码对应的学生表:

sql代码对应的课程表:

sql代码对应的选课表:

执行查询:(没有选课的学生信息和无学生选的课程信息)

普通方式:SELECT * FROM student WHERE student.id NOT IN (SELECT sid FROM sc); SELECT * FROM course WHERE course.id NOT IN (SELECT cid FROM sc);
关联方式:SELECT student.* FROM student LEFT JOIN sc ON student.id=sc.sid LEFT JOIN course ON course.id = sc.cid WHERE course.sname IS NULL; SELECT course.* FROM course LEFT JOIN sc ON course.id=sc.cid LEFT JOIN student ON student.id = sc.sid WHERE student.sname IS NULL;

posted @ 2023-08-22 09:41  kisshappyboy  阅读(565)  评论(0编辑  收藏  举报