Hibernate对联合子类的实现

有四个实体类Person,Programmer,Player,Player1 

他们的关系是，Person是Programmer和Player的父类，Player是Player1的父类

其中Programmer，Player，Player1在数据库中有对应的表，Person为抽象类，没有对应的数据表

CREATE TABLE `test` (
  `name` varchar(10) default NULL,
  `sex` varchar(20) default NULL
) ENGINE=InnoDB DEFAULT CHARSET=gb2312;

CREATE TABLE `player1` (
  `id` varchar(50) NOT NULL,
  `platform` varchar(50) default NULL,
  `name` varchar(50) default NULL,
  `style` varchar(11) default NULL,
  PRIMARY KEY  (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=gb2312;

CREATE TABLE `player` (
  `id` varchar(50) NOT NULL,
  `name` varchar(50) default NULL,
  `style` varchar(50) default NULL,
  PRIMARY KEY  (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=gb2312;

实体结构：

package uniontest;

public abstract class Person ...{
  private String id;
  private String name;
public String getId() ...{
    return id;
}
public void setId(String id) ...{
    this.id = id;
}
public String getName() ...{
    return name;
}
public void setName(String name) ...{
    this.name = name;
}
}


package uniontest;

public class Programmer extends Person ...{
  private String language;

public String getLanguage() ...{
    return language;
}

public void setLanguage(String language) ...{
    this.language = language;
}
}


package uniontest;

public class Player extends Person ...{
   private String style;

public String getStyle() ...{
    return style;
}

public void setStyle(String style) ...{
    this.style = style;
}
}


package uniontest;

public class Player1 extends Player ...{
   private String platform;

public String getPlatform() ...{
    return platform;
}

public void setPlatform(String platform) ...{
    this.platform = platform;
}
}

映射文件：

<?xml version="1.0" encoding="utf-8"?>
<!DOCTYPE hibernate-mapping PUBLIC "-//Hibernate/Hibernate Mapping DTD 3.0//EN"
"http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd">
<!-- 
    Mapping file autogenerated by MyEclipse - Hibernate Tools
-->
<hibernate-mapping package="uniontest">
    <class name="Person" abstract="true" >
       <id name="id">
         <generator class="uuid.hex"></generator>
       </id>
      
      <property name="name"></property>
      <union-subclass name="Programmer">
         <property name="language" column="lang"></property>
      </union-subclass>
      <union-subclass name="Player">
         <property name="style" column="style"></property>
         <union-subclass name="Player1">
           <property name="platform" column="platform"></property>
         </union-subclass>
      </union-subclass>
    </class>
</hibernate-mapping>

测试代码：

public static void main(String[] args) ...{
        Configuration cfg=new Configuration();
        cfg.configure();
        SessionFactory sf=cfg.buildSessionFactory();
        Session session=sf.openSession();
        Transaction t=session.beginTransaction();
        Programmer p=new Programmer();
        p.setName("gaoxiang");
        p.setLanguage("java");
        
        Player p1=new Player();
        p1.setName("xiaoyu");
        p1.setStyle("piano");
        
        Player1 p2=new Player1();
        p2.setName("xiaomao");
        p2.setStyle("qin");
        p2.setPlatform("eclipse");
        
        session.save(p);
        session.save(p1);
        session.save(p2);
        t.commit();
        Query query=session.createQuery("from Person");
        List a=query.list();
        for(int i=0;i<a.size();i++)...{
            Person c=(Person)a.get(i);
            System.out.println(c.getName());
        }
        
        Query query2=session.createQuery("from Player");
        List a1=query2.list();
        for(int i=0;i<a1.size();i++)...{
            Player c=(Player)a1.get(i);
            System.out.println(c.getName());
        }
        System.out.println("success");

    }

运行结果：
Hibernate: insert into Programmer (name, lang, id) values (?, ?, ?)
Hibernate: insert into Player (name, style, id) values (?, ?, ?)
Hibernate: insert into Player1 (name, style, platform, id) values (?, ?, ?, ?)
Hibernate: select person0_.id as id0_, person0_.name as name0_, person0_.lang as lang1_, person0_.style as style2_, person0_.platform as platform3_, person0_.clazz_ as clazz_ from ( select platform, null as lang, style, name, id, 3 as clazz_ from Player1 union select null as platform, lang, null as style, name, id, 1 as clazz_ from Programmer union select null as platform, null as lang, style, name, id, 2 as clazz_ from Player ) person0_
xiaomao
gaoxiang
xiaoyu
Hibernate: select player0_.id as id0_, player0_.name as name0_, player0_.style as style2_, player0_.platform as platform3_, player0_.clazz_ as clazz_ from ( select null as platform, style, name, id, 2 as clazz_ from Player union select platform, style, name, id, 3 as clazz_ from Player1 ) player0_
xiaoyu
xiaomao
success