如何通过Hibernate

我目前想我的Java类层次结构映射到使用Hibernate我的数据库，但我一直用下面的（子）类没有保存正确的鉴别值：如何通过Hibernate

我有一个叫MailAccount类其中有3个属性（见下面的代码）：

public class MailAccount{ 
    long id; 
    IncomingMailServer incomingServer; 
    OutgoingMailServer outgoingServer; 

    public MailAccount(){ 
     super(); 
    } 

    // Getter and setter omitted 
}

服务器类层次结构看起来像这样：

MailServer.java

public abstract class MailServer { 
    String password; 
    String host; 
    String username; 
    String port; 

    // Getter and setter omitted 
}

IncomingMailServer.java

public abstract class IncomingMailServer extends MailServer { 
}

OutgoingMailServer.java

public abstract class OutgoingMailServer extends MailServer { 
}

Pop3Server.java

public class Pop3Server extends IncomingMailServer{ 
    public Pop3Server(){ 
     super(); 
    } 
}

ImapServer.java

public class ImapServer extends IncomingMailServer{ 
    public ImapServer(){ 
     super(); 
    } 
}

SmtpServer.java

public class SmtpServer extends OutgoingMailServer{ 
    public SmtpServer(){ 
     super(); 
    } 
}

性质incomingServer和outgoingServer中当然MailAccount.java仅持有的实例任Pop3Server，ImapServer（对于incomingServer）或SMTPSERVER（用于outgoingServer）。

我使用下面的映射配置：

MailAccount.hbm.xml

<hibernate-mapping package="test.account"> 
    <class name="MailAccount" table="MAILACCOUNTS" dynamic-update="true"> 

     <id name="id" column="MAIL_ACCOUNT_ID"> 
      <generator class="native" /> 
     </id> 

     <one-to-one name="incomingServer" cascade="all"> 
     </one-to-one> 
     <one-to-one name="outgoingServer" cascade="all"> 
     </one-to-one> 
    </class> 
</hibernate-mapping>

MailServer.hbm.xml

<hibernate-mapping> 
    <class name="test.server.MailServer" table="MAILSERVER" abstract="true"> 

     <id name="id" type="long" access="field"> 
      <column name="MAIL_SERVER_ID" /> 
      <generator class="native" /> 
     </id> 

     <discriminator column="SERVER_TYPE" type="string"/> 

     <property name="password" column="PASSWORD" /> 
     <property name="host" column="HOST" /> 
     <property name="username" column="USERNAME" /> 
     <property name="port" column="PORT" /> 

     <one-to-one name="mailAccount" class="test.account.MailAccount" foreign-key="MAIL_SERVER_ID"></one-to-one> 

     <subclass name="test.server.incoming.ImapServer" extends="test.server.incoming.IncomingMailServer" discriminator-value="IMAP_SERVER"> 
     </subclass> 

     <subclass name="test.server.incoming.Pop3Server" extends="test.server.incoming.IncomingMailServer" discriminator-value="POP3_SERVER"> 
     </subclass> 

     <subclass name="test.server.outgoing.SmtpServer" extends="test.server.outgoing.OutgoingMailServer" discriminator-value="SMTP_SERVER"> 
     </subclass> 

     <subclass name="test.server.incoming.IncomingMailServer" extends="test.server.MailServer" abstract="true" discriminator-value="INCOMING_SERVER"> 
     </subclass> 

     <subclass name="test.server.outgoing.OutgoingMailServer" extends="test.server.MailServer" abstract="true" discriminator-value="OUTGOING_SERVER"> 
     </subclass> 

    </class> 
</hibernate-mapping>

问题：每当我告诉Hibernate来保存麦的实例lAccount，像这样：

session = getSession(); 
transaction = session.beginTransaction(); 
session.save(mailAccount); 
transaction.commit();

.. Hibernate正确存储所有内容，除了表MailServer中的鉴别器列SERVER_TYPE以外。在此列，休眠应该将任“IMAP_SERVER”，“POP3_SERVER”或“SMTP_SERVER”，而是将其保存或者“INCOMING_SERVER”或“OUTGOING_SERVER”。

当我尝试从数据库中加载这个实体时，发生了一个异常（当然），因为Hibernate试图实例化两个抽象类型的“IncomingMailServer”或“OutgoingMailServer”类型的对象。所以，我如何才能休眠保存正确的类型？

例子：如果incomingServer持有Pop3Server实例的属性，则应该的Hiberante存储到我的数据库，当我加载根据MailAccount回来了，我想休眠重新Pop3Server的一个实例。

注：我是Hibernate的新手，这些是我的第一步，所以请温柔:-)。我很清楚，我.hmb.xml文件可能会显得凌乱，所以如果你有改进建议，继续:-)

来源

2011-05-03 Timo

我不没有使用.hbm文件的经验，但指定抽象类的鉴别器值有什么意义？你不应该只为叶子指定这些吗？ – 2011-05-03 12:36:57

嗨，谢谢你的回复！当我删除行'和''时，我得到以下异常：'org.hibernate.MappingException：持久化类不知道：test.server.incoming.IncomingMailServer'。如果我只删除“discriminator-value =”INCOMING_SERVER“'和discriminator-value =”OUTGOING_SERVER“属性，那么Hibernate将保存完整的类名'test.server.incoming.IncomingMailServer'和'test.server.outgoing。 OutgoingMailServer'到鉴别器列SERVER_TYPE – Timo 2011-05-03 12:57:28

<subclass>元素嵌套应类似于继承层次：

<subclass name="test.server.incoming.IncomingMailServer" extends="test.server.MailServer" abstract="true" discriminator-value="INCOMING_SERVER"> 
    <subclass name="test.server.incoming.ImapServer" extends="test.server.incoming.IncomingMailServer" discriminator-value="IMAP_SERVER"> 
    </subclass> 
    <subclass name="test.server.incoming.Pop3Server" extends="test.server.incoming.IncomingMailServer" discriminator-value="POP3_SERVER"> 
    </subclass> 
</subclass> 

<subclass name="test.server.outgoing.OutgoingMailServer" extends="test.server.MailServer" abstract="true" discriminator-value="OUTGOING_SERVER"> 
    <subclass name="test.server.outgoing.SmtpServer" extends="test.server.outgoing.OutgoingMailServer" discriminator-value="SMTP_SERVER"> 
    </subclass> 
</subclass>

来源

2011-05-03 13:46:57 axtavt

太棒了！有效。非常感谢！ – Timo 2011-05-03 13:56:18

如何通过Hibernate

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