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

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">