11

当一个实体映射到另一个在其子类上有直接实现的实体时,我遇到了问题。请参阅下面的示例映射:

@Entity
class Location {
      @OneToOne
      @JoinColumn(...)
      private Person person;
}

@Entity
@Inheritance(strategy=InheritanceType.SINGLE_TABLE)
@DiscriminatorColumn(name="person_type",discriminatorType=DiscriminatorType.STRING)
abstract class Person {
}

@Entity
@DiscriminatorValue("M")
class Man extends Person {
    ...
}

@Entity
@DiscriminatorValue("W")
class Woman extends Person {
    ...
}

现在,这就是我的数据库表中的内容:

位置表:id=1,person_id=1 人表:id=1,person_type="M"

当我使用实体管理器检索位置时,hibernate 会抛出一个异常,指出我无法实例化抽象类或接口。

Location location = entityManager.find(Location.class, 1L);

Hibernate 抛出这个错误:

javax.persistence.PersistenceException: org.hibernate.InstantiationException: Cannot instantiate abstract class or interface: Person
at org.hibernate.ejb.AbstractEntityManagerImpl.throwPersistenceException(AbstractEntityManagerImpl.java:630)
at org.hibernate.ejb.AbstractEntityManagerImpl.find(AbstractEntityManagerImpl.java:195)
at ......
at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:39)
at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:25)
at java.lang.reflect.Method.invoke(Method.java:585)
at org.junit.internal.runners.TestMethod.invoke(TestMethod.java:59)
at org.junit.internal.runners.MethodRoadie.runTestMethod(MethodRoadie.java:98)
at org.unitils.UnitilsJUnit4TestClassRunner$TestListenerInvokingMethodRoadie.runTestMethod(UnitilsJUnit4TestClassRunner.java:174)
at org.junit.internal.runners.MethodRoadie$2.run(MethodRoadie.java:79)
at org.junit.internal.runners.MethodRoadie.runBeforesThenTestThenAfters(MethodRoadie.java:87)
at org.unitils.UnitilsJUnit4TestClassRunner$TestListenerInvokingMethodRoadie.runBeforesThenTestThenAfters(UnitilsJUnit4TestClassRunner.java:156)
at org.junit.internal.runners.MethodRoadie.runTest(MethodRoadie.java:77)
at org.junit.internal.runners.MethodRoadie.run(MethodRoadie.java:42)
at org.unitils.UnitilsJUnit4TestClassRunner.invokeTestMethod(UnitilsJUnit4TestClassRunner.java:95)
at org.junit.internal.runners.JUnit4ClassRunner.runMethods(JUnit4ClassRunner.java:51)
at org.unitils.UnitilsJUnit4TestClassRunner.access$000(UnitilsJUnit4TestClassRunner.java:44)
at org.unitils.UnitilsJUnit4TestClassRunner$1.run(UnitilsJUnit4TestClassRunner.java:62)
at org.junit.internal.runners.ClassRoadie.runUnprotected(ClassRoadie.java:27)
at org.junit.internal.runners.ClassRoadie.runProtected(ClassRoadie.java:37)
at org.unitils.UnitilsJUnit4TestClassRunner.run(UnitilsJUnit4TestClassRunner.java:68)
at org.eclipse.jdt.internal.junit4.runner.JUnit4TestReference.run(JUnit4TestReference.java:49)
at org.eclipse.jdt.internal.junit.runner.TestExecution.run(TestExecution.java:38)
at org.eclipse.jdt.internal.junit.runner.RemoteTestRunner.runTests(RemoteTestRunner.java:467)
at org.eclipse.jdt.internal.junit.runner.RemoteTestRunner.runTests(RemoteTestRunner.java:683)
at org.eclipse.jdt.internal.junit.runner.RemoteTestRunner.run(RemoteTestRunner.java:390)
at org.eclipse.jdt.internal.junit.runner.RemoteTestRunner.main(RemoteTestRunner.java:197)
4

7 回答 7

4

这对我有用,使用休眠作为持久性提供程序。

@OneToOne(cascade = {CascadeType.PERSIST,CascadeType.MERGE}) 
于 2012-09-28T09:12:58.660 回答
2

来自Java EE 6 教程 - 实体继承

非实体超类中的任何映射或关系注释都将被忽略。

因此,您必须注释Person类以@Entity将其与Locationvia关联,这似乎是正确的@OneToOne

来自@MappedSuperclass javadoc

使用 MappedSuperclass 注释指定的类可以以与实体相同的方式映射,除了映射将仅应用于其子类,因为映射的超类本身不存在表。

所以你不能@MappedSuperclass在 Person 上使用,然后用 映射它@OneToOne,因为不会有 Person 表。

似乎您使用的 JPA 注释是正确的。您是否尝试过@Martin Klinke 对Person班级虚拟鉴别器值的建议?

于 2011-03-09T13:57:03.500 回答
1

我有一个类似的错误消息,结构如下:

@Entity
@Table(name = "PERSON")
@Inheritance(strategy = InheritanceType.SINGLE_TABLE)
@DiscriminatorColumn(name = "TYPE", discriminatorType = DiscriminatorType.STRING)
public abstract class Person {

  @Id
  @GeneratedValue
  private Long     id;

}

有一个具体的子实例

@Entity
@DiscriminatorValue("REALPERSON")
public class RealPerson extends Person{ etc... }

和一个具有引用抽象类的字段的类:

public class SomeClass() {
  @OneToOne
  private Person person;
}

以下更改为我解决了这个问题:

  1. 更改@OneToOne@OneToOne(cascade = CascadeType.ALL)
  2. 更改@GeneratedValue@GeneratedValue(strategy = GenerationType.TABLE)

更新:我还发现在将 RealPerson 对象链接到 SomeClass 之前我没有保存它。所以现在我先保存实例,cascade不再需要该属性

于 2012-03-01T15:13:59.983 回答
1

我发现如果 Entity 类实现,这种问题会自行解决Serializable

于 2011-03-09T23:39:07.053 回答
0

正如我在上面的评论中所指出的,我已经对 EclipseLink 进行了同样的尝试,并且它可以工作。

创建一些测试数据后,我已经清除了数据库中人员条目的鉴别器值,现在在尝试加载关联位置时遇到了类似的异常。EclipseLink 的消息更具描述性:

Exception Description: Missing class for indicator field value [] of type [class java.lang.String].
Descriptor: RelationalDescriptor(com.mklinke.webtest.domain.Person --> [DatabaseTable(PERSON)])
at org.eclipse.persistence.exceptions.DescriptorException.missingClassForIndicatorFieldValue(DescriptorException.java:937)
at org.eclipse.persistence.descriptors.InheritancePolicy.classFromValue(InheritancePolicy.java:355)
at org.eclipse.persistence.descriptors.InheritancePolicy.classFromRow(InheritancePolicy.java:342)
at org.eclipse.persistence.internal.descriptors.ObjectBuilder.buildObject(ObjectBuilder.java:485)
at org.eclipse.persistence.internal.descriptors.ObjectBuilder.buildObject(ObjectBuilder.java:456)
at org.eclipse.persistence.queries.ObjectLevelReadQuery.buildObject(ObjectLevelReadQuery.java:723)
at org.eclipse.persistence.queries.ReadObjectQuery.registerResultInUnitOfWork(ReadObjectQuery.java:766)
at org.eclipse.persistence.queries.ReadObjectQuery.executeObjectLevelReadQuery(ReadObjectQuery.java:451)
at org.eclipse.persistence.queries.ObjectLevelReadQuery.executeDatabaseQuery(ObjectLevelReadQuery.java:1080)
at org.eclipse.persistence.queries.DatabaseQuery.execute(DatabaseQuery.java:808)
...

映射似乎有效,因此除非数据“损坏”(如您所说,在您的情况下并非如此),否则它可能是 Hibernate 中的错误或至少不同的行为。

于 2011-03-08T22:21:52.580 回答
0

尝试将@ForceDiscriminator 添加到 Person。如果没有这个注解,Hibernate 经常尝试实例化父类,忽略应该告诉它实例化哪个子类的鉴别器。

于 2011-06-13T15:22:32.460 回答
0

我运行与此类似的代码,但有一些不同之处。首先,我将 Abstract 类放在接口后面。其次,我为@OnetoOne 映射明确定义了targetEntity。一个例子是:

@Entity
class Location {
    @OneToOne(targetEntity = AbstractPerson.class)
    @JoinColumn(...)
    private Person person;
}

public interface Person {
}

@Entity
@Inheritance(strategy=InheritanceType.SINGLE_TABLE)
@DiscriminatorColumn(name="person_type",discriminatorType=DiscriminatorType.STRING)
abstract class AbstractPerson implements Person {
}

在这里,为了清楚起见,我将您的 Person 抽象类重命名为“AbstractPerson”。花了很多时间来让它工作,我希望它能解决你的问题。

于 2011-03-29T11:59:57.650 回答