Hibernate继承映射的“Could not format discriminator value to SQL string”错误解决方法

借助于Hibernate强大的O/R Mapping能力，我们能够通过discriminator轻易地将一颗继承树映射到一个表中，通过discriminator确定具体映射的子类。

在设置@hibernate.discriminator column="type" type="integer"后，启动Hibernate报错：
Could not format discriminator value to SQL string

搜索Hibernate官方文档后发现问题，原来Hibernate默认的discriminator的type是String，当设置discriminator的type为integer后，需要为父类也设置 class
table="TABLE_NAME" discriminator-value="not null"，否则，Hibernate默认的discriminator-value是完整的类名，在转换String到int时造成NumberFormatException。

即：

<class name="com.sjr.bean.PeopleT" table="J_PEOPLE2" schema="SXBBKF" discriminator-value="not null">  --这里加上
<id name="id" type="long">
<column name="ID" precision="22" scale="0" />
<generator class="native"></generator>
</id>

<discriminator column="type" type="int"/>   --必须直接跟在id后面

<property name="name" type="string">
<column name="NAME" length="20" />
</property>
<property name="age" type="int">
<column name="AGE" precision="22" scale="0" />
</property>

<subclass name="com.sjr.bean.Student" discriminator-value='1'>   --实际是从字符串转化为整型的
<property name="school"></property>
</subclass>

<subclass name="com.sjr.bean.Staff" discriminator-value='2'>
<property name="company"></property>
</subclass>

最后运行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>