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Hibernate条件查询JPA条件查询

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我正在迁移到Hibernate 5.2，因为不推荐使用createCriteria，我们计划将查询重写为JPA样式 . 我正面临一个问题（java.lang.IllegalArgumentException：不是实体），其中相同的代码使用hibernate criteria方法 .

使用hibernate标准

Criteria criteria = getSession().createCriteria(className);

for (Map.Entry<String, Object> entry : matchingCriteria.entrySet()) {
    criteria.add(Expression.eq(entry.getKey(), entry.getValue()));
}
List objsMatchingCriteria = criteria.list();

获取JPA标准的“java.lang.IllegalArgumentException：非实体”

CriteriaBuilder cb = getSession().getCriteriaBuilder();
CriteriaQuery  cq = cbuilder.createQuery(refClassName);
Root root = cquery.from(refClassName);
cq.select(root);
List<Predicate> predicates = new ArrayList<>();
for (Map.Entry<String, Object> entry : matchingCriteria.entrySet()) {
    predicates.add(cb.equal(root.get(entry.getKey()), entry.getValue()));
}
cq.where(predicates.toArray(new Predicate[predicates.size()]));

List objsMatchingCriteria = 
    getSession().createQuery(cq).getResultList();

你能告诉我JPA代码有什么问题吗？

Update 当我们有基类（抽象）并且实际的实体类是具体的子类时，就会出现这个问题 . 早期的hibernate标准，即使我传递基类，检索结果也没有问题 .

但在迁移到JPA标准后，我无法传递基类 . 这可以在JPA中管理 .

提前致谢 .

java hibernate jpa

1 回答

  • 0

    如果你在hibernate中使用注释，那么你将在refClass上定义一个 @Entity . 就像你必须像这样映射实体：

    @Entity
@Table(name = "tableName")
public class ClassOfTable implements Serializable,Cloneable {

  public ClassOfTable () {
  }

  @Column(name = "column1")
  private Boolean member1;

  @Column(name = "column2")
  private String member2;

}

    并且您将在创建 sessionFactory 时传递此类 . 如果你在创建sessionFactory时没有传递这个类，那么它将抛出一个异常，该类不是一个实体 .

    StandardServiceRegistry standardRegistry =
        new StandardServiceRegistryBuilder().applySettings(createProperties(
            host, port, dbName, protocol, userName, password, minPoolSize, maxPoolSize)
            .getProperties()).build();

    MetadataSources sources = new MetadataSources(standardRegistry);
    annotatedClassNames.add(refClass.class);
    Metadata metaData = sources.getMetadataBuilder().build();
    return metaData.getSessionFactoryBuilder().build();

    或者，如果使用Hibernate配置文件，则必须在hibernate.cfg.xml中指定hbm.xml的映射文件名，而someName.hbm.xml包含refClass的映射 .

    然后你创建会话工厂 .

    Configuration configuration = new Configuration();
            configuration.configure("hibernate.cfg.xml");
            StandardServiceRegistryBuilder ssrb = new StandardServiceRegistryBuilder().applySettings(configuration.getProperties());
            SessionFactory sessionFactory = configuration.buildSessionFactory(ssrb.build());
            Session session = sessionFactory.openSession();

    你的hibernate.cfg.xml必须包含（映射resourece）条目，其中定义了refClass mappiing .

    <hibernate-configuration>
    <session-factory>
        <!-- JDBC connection settings -->
        <property name="connection.driver_class">com.mysql.jdbc.Driver</property>
        <property name="connection.url">jdbc:mysql://localhost/mediadb</property>
        <property name="connection.username">kodejava</property>
        <property name="connection.password">kodejava123</property>



        <!-- Mapping to hibernate mapping files -->
        <mapping resource="org/kodejava/example/hibernate/model/someName.hbm.xml"/>
    </session-factory>
</hibernate-configuration>

    并且您的hbm.xml必须具有refClass映射

    <?xml version = "1.0" encoding = "utf-8"?>
<!DOCTYPE hibernate-mapping PUBLIC 
"-//Hibernate/Hibernate Mapping DTD//EN"
"http://www.hibernate.org/dtd/hibernate-mapping-3.0.dtd"> 

<hibernate-mapping>
   <class name = "refClass" table = "some Class">

      <meta attribute = "class-description">
         This class contains the some detail. 
      </meta>

      <id name = "id" type = "int" column = "id">
         <generator class="native"/>
      </id>

      <property name = "firstName" column = "first_name" type = "string"/>
      <property name = "lastName" column = "last_name" type = "string"/>
      <property name = "salary" column = "salary" type = "int"/>

   </class>
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