引起：java.lang.IllegalArgumentException：不是实体：

Question

I am working on an app with Hibernate 5.6, Spring 5.3 and we use xml-based configuration + Hibernate mapping files.hbm.xml for each entity. 我们没有注释 model 类。

从 Hibernate 3 迁移到 5.6 后，我们不得不使用 Springs 事务管理器，它目前运行良好。 但由于 Hibernate 标准 API 已被弃用，我尝试将一些标准 API 请求迁移到 CriteriaBuilder 并设置一些样板代码。

public List<ENTITY> findByCriteria(final Map<String, Object> criteriaMap, final List<String> fields, final Class<ENTITY> entityClass) {
    CriteriaBuilder cb = this.getSession().getCriteriaBuilder();
    CriteriaQuery<ENTITY> cq = cb.createQuery(entityClass);
    Root<ENTITY> root = cq.from(entityClass);
...

但是当我执行代码时，最后一行向我抛出了这个错误：

Caused by: java.lang.IllegalArgumentException: Not an entity: interface com.myapp.pm.common.business.PmPropertyVO
        at deployment.myapp-clienta.war//org.hibernate.metamodel.internal.MetamodelImpl.entity(MetamodelImpl.java:567)
        at deployment.myapp-clienta.war//org.hibernate.query.criteria.internal.QueryStructure.from(QueryStructure.java:128)
        at deployment.myapp-clienta.war//org.hibernate.query.criteria.internal.CriteriaQueryImpl.from(CriteriaQueryImpl.java:158)
        at deployment.myapp-clienta.war//com.myapp.integration.dao.impl.CommonHibernateDAOImpl.findByCriteria(CommonHibernateDAOImpl.java:554)
        ...

我的配置如下所示：

<bean id="sessionFactory" class="org.springframework.orm.hibernate5.LocalSessionFactoryBean">
        <property name="mappingLocations" value="classpath*:/com/*/*/**/integration/hbm/*.hbm.xml" />       
        <property name="hibernateProperties">
            <props>
                <prop key="hibernate.dialect">${hibernate.dialect}</prop>
                <prop key="hibernate.show_sql">${hibernate.show_sql}</prop>
                <prop key="hibernate.format_sql">${hibernate.format_sql}</prop>
                <prop key="hibernate.cglib.use_reflection_optimizer">${hibernate.cglib.use_reflection_optimizer}</prop>             
                <prop key="hibernate.jdbc.batch_size">${hibernate.jdbc.batch_size}</prop>
                <prop key="hibernate.connection.release_mode">${hibernate.connection.release_mode}</prop>                       
            </props>
        </property>
        <property name="dataSource" ref="dataSource"/>
    </bean>

我试图添加另一个属性，但这没有帮助：

<property name="packagesToScan" value="classpath*:/com/*/*/**/integration/hbm/*.hbm.xml" />

实体的映射文件：

<?xml version="1.0"?>
<!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.myapp.business.impl.PmPropertyVOImpl" table="pm_t_property" proxy="com.myapp.pm.common.business.PmPropertyVO">
        <id name="rowguid" column="rowguid" type="java.lang.Long">
            <generator class="sequence">
                <param name="sequence_name">pm_t_property_rowguid_seq</param>
            </generator>
        </id>               
        <property name="key" type="java.lang.String" column="py_key" not-null="true" /> 
        <property name="value" type="java.lang.String" column="py_value" not-null="true" /> 
        <property name="description" type="java.lang.String" column="py_description" not-null="false" /> 
        <property name="owner" type="java.lang.String" column="py_owner" not-null="true" />
        <property name="type" type="java.lang.String" column="py_type" not-null="false" />       
        
        <property name="creationDate" type="timestamp" column="py_creation_date" /> 
        <property name="creationUser" type="java.lang.String" column="py_creation_user" />  
        <property name="modificationDate" type="timestamp" column="py_modification_date" /> 
        <property name="modificationUser" type="java.lang.String" column="py_modification_user" />                                
    </class>    
</hibernate-mapping>

我的问题：使用 CriteriaBuilder 时是否必须使用 persistence.xml 文件？ 我们之前没有使用过这种配置，只是简单的 Hibernate 配置。

编辑：在 Alexander Kashpirovsky 发表评论后，我尝试更改此行以获取实现而不是接口 class。 但我得到的只是对 java.lang.class 的引用

CriteriaQuery cq = cb.createQuery(entityClass.getClass());

Answer 1

变量entityClass等于com.myapp.business.impl.PmPropertyVOImpl或com.myapp.pm.common.business.PmPropertyVO ？

必须是com.myapp.business.impl.PmPropertyVOImpl