借助于Hibernate強大的O/R Mapping能力，我們能夠通過discriminator輕易地將一顆繼承樹映射到一個表中，通過discriminator確定具體映射的子類。
在設置@hibernate.discriminator column="type" type="integer"后，啟動Hibernate報錯：
Could not format discriminator value to SQL string
http://dev.csdn.net/article/82/82734.shtm
http://forum.hibernate.org/viewtopic.php?t=974225
搜索Hibernate官方文檔后發現問題，原來Hibernate默認的discriminator的type是String，當設置discriminator的type為integer后，需要為父類也設置@hibernate.class table="TABLE_NAME" discriminator-value="not null"，否則，Hibernate默認的discriminator-value是完整的類名，在轉換String到int時造成NumberFormatException。
最后運行XDoclet，生成hbm文件：
<?xml version="1.0" encoding="iso-8859-1"?>
<!DOCTYPE hibernate-mapping PUBLIC
"-//Hibernate/Hibernate Mapping DTD 3.0//EN"
"http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd">
<hibernate-mapping>
<class name="com.crackj2ee.example.AbstractClass" table="TABLE_NAME" discriminator-value="not null">
<id name="id" column="id" type="java.lang.Long" unsaved-value="null">
<generator class="increment"/>
</id>
<discriminator column="type" not-null="true" type="integer"/>
<subclass name="com.crackj2ee.example.SubClass1" discriminator-value="1">
<subclass name="com.crackj2ee.example.SubClass2" discriminator-value="2">
</class>
</hibernate-mapping>