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  • 创建表的设计

    数据库的设计:

    一、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');


    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;

    二、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('p0001','王军',20,1,'c101');
    INSERT INTO student VALUES('p0002','张宇',21,1,'c101');
    INSERT INTO student VALUES('p0003','刘飞',22,1,'c102');
    INSERT INTO student VALUES('p0004','赵燕',18,0,'c103');
    INSERT INTO student VALUES('p0005','曾婷',19,0,'c103');
    INSERT INTO student VALUES('p0006','周慧',21,0,'c104');
    INSERT INTO student VALUES('p0007','小红',23,0,'c104');
    INSERT INTO student VALUES('p0008','杨晓',18,0,'c104');
    INSERT INTO student VALUES('p0009','李杰',20,1,'c105');
    INSERT INTO student VALUES('p0010','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');


    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);
    1
    2
    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 WHER
    ---------------------
    转载

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  • 原文地址:https://www.cnblogs.com/lingtiaoti/p/9874709.html
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