Lightweight Class In Hibernate3

Lightweight Class In Hibernate3


Suppose I have the following persistent class:
假如有这样一个持久化类：

public class Document implements Node {
   private Long _key;
   private String _name;
   private Calendar _created;
   private Calendar _updated;
   private Folder _folder;
   private Clob _text;
   public String getKey() { return _key; }
   public void setKey(Long key) { _key = key; }
   public String getName() { return _name; }
   public void setName(String name) { _name = name; }
   public Calendar getCreated() { return _created; }
   public void setCreated(Calendar created) { _created = created; }
   public Calendar getUpdated() { return _updated; }
   public void setUpdated(Calendar updated) { _updated = updated; }
   public Folder getFolder() { return _folder; }
   public void setFolder(Folder folder) { _folder = folder; }
   public Clob getText() { return _text; }
   public void setText(Clob text) { _text = text; }
}

这个类中的所有属性都与数据库中的DOCUMENTS表中的某一列对应。
但是实例化这个类中的_text字段要消耗很大的内存。所以如果我要对这个表进行操作比如列出所有document的名字，或者给某个docuemnt改名。那么我又不想load出Clob类型的_text这个属性。那么怎么做呢？当然，有很多方法，而下面的方法是Hibernate官方网站
推荐的方法：
我们可以把这个持久化类分为"lightweight" superclass和 "heavyweight" subclass
如下：

public class DocumentInfo implements Node {
   private Long _key;
   private String _name;
   private Calendar _created;
   private Calendar _updated;
   private Folder _folder;
   private Clob _text;
   public String getKey() { return _key; }
   public void setKey(Long key) { _key = key; }
   public String getName() { return _name; }
   public void setName(String name) { _name = name; }
   public Calendar getCreated() { return _created; }
   public void setCreated(Calendar created) { _created = created; }
   public Calendar getUpdated() { return _updated; }
   public void setUpdated(Calendar updated) { _updated = updated; }
   public Folder getFolder() { return _folder; }
   public void setFolder(Folder folder) { _folder = folder; }
}

public class Document extends DocumentInfo {
   private Clob _text;
   public Clob getText() { return _text; }
   public void setText(Clob text) { _text = text; }
}

We use the following mapping:

<class name="DocumentInfo" table="DOCUMENTS">
   <id name="key" type="long" column="ID">
       <generator class="native"/>
   </id>
   <property name="name"/>
   <property name="created"/>
   <property name="updated"/>
   <many-to-one name="folder"/>
</class>

<class name="Document" table="DOCUMENTS" polymorphism="explicit">
   <id name="key" type="long" column="ID">
       <generator class="native"/>
   </id>
   <property name="name"/>
   <property name="created"/>
   <property name="updated"/>
   <many-to-one name="folder"/>
   <property name="text"/>
</class>

ok，如果我们要得到一个不包含_text属性的持久化对象，可以这样：

from DocumentInfo
from Node
from java.lang.Object

同样由于我们在mapping 文件中设置了polymorphism="explicit"，所以如果我们希望得到包含_text属性的持久化对象
只要这样（注意hibernate2不支持）

from d in class Document
DocumentInfo info = (DocumentInfo) session.load(DocumentInfo.class, new Long(1));
Document doc = (Document) session.load(Document.class, new Long(1));

如果你希望同时查询出同一个id的DocumentInfo 和 Document两个对象，你需要在load DocumentInfo 之后
使用session.evict（）或者在mapping文件中设置polymorphism="explicit"