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hibernate入门教程3-----多对多的例子

2007-03-30 10:47 555 查看

Hibernate入门－－多对多的例子

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经典的多对多例子，一个学生可以选n门课，一门课可以被n个人选。
所有代码使用上一篇中的。
在Person类中加入：


public Set getCourses() {
        return courses;
    }
    private void setCourses(Set courses) {
        this.courses = courses;
    }

在Person.hbm.xml文件<property name="sex"/>后加入：



<set name="courses" table="PERSON_COURSE">
            <key column="PERSON_ID"/>
            <many-to-many column="COURSE_ID" class="examples.Course"/>
        </set>

这个set对应Person类中的courses，我们知道多对多要从两个表变成三个表，talb
e指明中间关系表的名称，key指明中间表中Person的标识字段，many-to-many表示
是一个多对多关系
新建一个Course类：

package examples;
import java.util.HashSet;
import java.util.Set;
public class Course {
    private Long id;
    private String name;
    private Set persons = new HashSet();
    public Set getPersons() {
        return persons;
    }
    private void setPersons(Set persons) {
        this.persons = persons;
    }
    public Long getId() {
        return id;
    }
    private void setId(Long id) {
        this.id = id;
    }
    public String getName() {
        return name;
    }
    public void setName(String name) {
        this.name = name;
    }
}

新建相应的Course.hbm.xml文件：

<?xml version="1.0"?>
<!DOCTYPE hibernate-mapping PUBLIC
    "-//Hibernate/Hibernate Mapping DTD 3.0//EN"
    "http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd">
<hibernate-mapping>
    <class name="examples.Course">
        <id name="id">
            <generator class="increment"/>
        </id>
        <property name="name"/>
        <set name="persons" inverse="true" table="PERSON_COURSE">
            <key column="COURSE_ID"/>
            <many-to-many column="PERSON_ID" class="examples.Person"/>
        </set>
    </class>
</hibernate-mapping>

需要说明的是inverse=true的属性，多对多有一方是主控方，这里应该是Person类
，所以Course类加上inverse=true
在hibernate.cfg.xml中加入

<mapping resource="examples/Course.hbm.xml"/>

在Test.java中加入：


Course c1 = new Course();
        Course c2 = new Course();
        Course c3 = new Course();
        c1.setName("c1");
        c2.setName("c2");
        c3.setName("c3");

        p1.getCourses().add(c2);
        p2.getCourses().add(c3);
        p3.getCourses().add(c1);
        session.save(c1);
        session.save(c2);
        session.save(c3);

修改Test.java的fetchData方法中的打印语句为：

System.out.println("" + p.getId() +" :" + p.getName() + " :" + p.getSex(
) + " :" + ((Course)p.getCourses().iterator().next()).getName());

运行，应该有结果：
1 :p1 :M :c2
3 :p3 :M :c1
相应的sql语句为：
11:21:34,921 DEBUG SchemaUpdate:142 - create table Course (id bigint not
null, name varchar(255), primary key (id))
11:21:34,921 DEBUG SchemaUpdate:142 - create table PERSON_COURSE (PERSON
_ID bigint not null, COURSE_ID bigint not null, primary key (PERSON_ID,
COURSE_ID))
11:21:34,921 DEBUG SchemaUpdate:142 - create table Person (id bigint not
null, name varchar(255), sex char(1), primary key (id))
11:21:34,937 DEBUG SchemaUpdate:142 - alter table PERSON_COURSE add cons
traint FKE448459882F6FF foreign key (PERSON_ID) references Person
11:21:34,937 DEBUG SchemaUpdate:142 - alter table PERSON_COURSE add cons
traint FKE44845A766793F foreign key (COURSE_ID) references Course
11:21:34,937 INFO SchemaUpdate:153 - schema update complete
11:21:34,937 INFO SessionFactoryImpl:379 - Checking 0 named queries
Hibernate: insert into Person (name, sex, id) values (?, ?, ?)
Hibernate: insert into Person (name, sex, id) values (?, ?, ?)
Hibernate: insert into Person (name, sex, id) values (?, ?, ?)
Hibernate: insert into Course (name, id) values (?, ?)
Hibernate: insert into Course (name, id) values (?, ?)
Hibernate: insert into Course (name, id) values (?, ?)
Hibernate: insert into PERSON_COURSE (PERSON_ID, COURSE_ID) values (?, ?
)
Hibernate: insert into PERSON_COURSE (PERSON_ID, COURSE_ID) values (?, ?
)
Hibernate: insert into PERSON_COURSE (PERSON_ID, COURSE_ID) values (?, ?
)
Hibernate: select person0_.id as col_0_0_ from Person person0_ where per
son0_.sex=?
Hibernate: select person0_.id as id0_, person0_.name as name0_0_, person
0_.sex as sex0_0_ from Person person0_ where person0_.id=?
Hibernate: select person0_.id as id0_, person0_.name as name0_0_, person
0_.sex as sex0_0_ from Person person0_ where person0_.id=?
Hibernate: select courses0_.PERSON_ID as PERSON1_1_, courses0_.COURSE_ID
as COURSE2_1_, course1_.id as id0_, course1_.name as name2_0_ from PERS
ON_COURSE courses0_ inner join Course course1_ on courses0_.COURSE_ID=co
urse1_.id where courses0_.PERSON_ID=?
Hibernate: select courses0_.PERSON_ID as PERSON1_1_, courses0_.COURSE_ID
as COURSE2_1_, course1_.id as id0_, course1_.name as name2_0_ from PERS
ON_COURSE courses0_ inner join Course course1_ on courses0_.COURSE_ID=co
urse1_.id where courses0_.PERSON_ID=?

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