Question

所以我试图将鉴别器值赋予PesquisaFiltroGeral类中的字段属性：

 <class name="PesquisaFiltroGeral" table="PESQUISA_FILTRO_GERAL" abstract="true">
    <id name="id" column="ID">
        <generator class="native"/>
    </id>

    <discriminator column="tipo" type="string"/>
    <property name="fieldType" column="tipo" insert="false" update="false"/>

    <subclass name="com.domain.entity.PesquisaFiltroProcess"  discriminator-value="PROCESS_TYPE">
        <property name="field" column="PROCFIELD">
            <type name="org.hibernate.type.EnumType">
                <param name="useNamed">true</param>
                <param name="identifierMethod">getValue</param>
                <param name="valueOfMethod">getFromValue</param>
                <param name="enumClass">com.domain.enums.PesquisaFieldsProcesso</param>
            </type>
        </property>
     </subclass>
    <!--DOCUMENT FIELD - SUBCLASS 2-->
    <subclass name="com.domain.entity.PesquisaFiltroDocument" discriminator-value="DOCUMENT_TYPE">
        <property name="field" column="DOCFIELD">
            <type name="org.hibernate.type.EnumType">
                <param name="identifierMethod">getValue</param>
                <param name="valueOfMethod">getFromValue</param>
                <param name="enumClass">com.domain.enums.PesquisaFieldsDocumento</param>
            </type>
        </property>
    </subclass>
 </class>

但是当我尝试从db中获取此类型的对象时，fieldType为null。有什么理由吗？

Answer 1

出于某种原因，将属性名称更改为与该列相同的属性：

<discriminator column="tipo" type="string"/>
<property name="tipo" column="tipo" insert="false" update="false"/>

Hibernate-使用鉴别器作为类属性

1 个答案: