JPA OneToMany列表找不到应该被继承的mappedBy属性。

Question

JPA OneToMany列表找不到应该被继承的mappedBy属性。

5

我们目前正在处理一些需求，需要将一些类似的实体（汽车图片、宠物图片、假日照片等）添加到数据库中，这些实体都属于同一个所有者（人）。我们不直接链接字节数组，而是使用引用。以后可能会添加更多类型的图片，为了保持复杂度低，我们想直接将其链接到Java类，这样我们就可以使用instance of和类似的东西。我们想使用@Inheritance来创建一个超类PictureRef，其中包含公共属性并链接到人。然后有另一个实体Person，它将具有这些子类的列表。这是一个OneToMany关系，具有一个mappedBy属性。这个mappedBy属性是未知的，所以JPA返回给我们这个错误：

Caused by: org.hibernate.AnnotationException: mappedBy reference an unknown
target entity property:  de.company.project.somepackages.PictureRef.person 
in de.company.project.somepackages.Person.picturesOfCars

我认为下面的代码可以最好地说明。为了提高可读性，我删除了其他属性、getter/setter和id序列。

1）所有的JPA实体都是从一个抽象实体派生出来的，该实体包含id和审计值。这个类与其他子类一起正常工作，所以我认为这个类不会引起问题。

AbstractEntity类

@MappedSuperclass
public abstract class AbstractEntity implements Serializable {
@Id
// Sequence definition removed
private Long id;
// other values are following //
}

2) 然后我们有一个超类，必须包含共同的属性。所有派生类都必须写在一个表中（因为它们看起来非常相似）。此外，我们还将有直接与此实体一起工作的业务逻辑，例如按ID加载或删除。

类PictureRef

@Entity
@Table(name = "t_picture_ref")
@Inheritance(strategy=InheritanceType.SINGLE_TABLE)
@DiscriminatorColumn(name = "picture_type")
// Sequence Definition removed
public class PictureRef extends AbstractEntity {
// common attributes, e.g. name or link to file //

@ManyToOne
@JoinColumn(name = "person_id")
private Person person;
}

3) 然后至少有两个子类。即将有更多内容。它们将包含仅适用于此类型图片的属性。

类CarPictureRef

@Entity
@DiscriminatorValue("car")
public class CarPictureRef extends PictureRef {

@Column(name="licence_plate_visible")
private boolean licensePlateVisible;
}

类 HolidayPictureRef

@Entity
@DiscriminatorValue("holiday")
public class HolidayPictureRef extends PictureRef {

@Column(name="weather_condition")
private String weatherCondition;
}

4) 然后有一个拥有/上传所有这些图片的人。这个人为每种类型的图片都有一个列表，因为应用程序会根据它们进行不同的处理。每个列表包含具体的子类，但是对于mappedBy属性，我们使用超类PictureRef中的person。也许这种继承不可能？

类 Person

@Entity
@Table(name = "t_person")
// Sequence Definition removed
public class Person extends AbstractEntity {

@OneToMany(mappedBy = "person", targetEntity = CarPictureRef.class)
private List<CarPictureRef> picturesOfCars;

@OneToMany(mappedBy = "person", targetEntity = HolidayPictureRef.class)
private List< HolidayPictureRef> picturesOfHolidays;

// a lot of other fields following //
}

一种解决方法可能是将所有带属性的内容都存储在一个表格中(我们本来就想这样做)，然后也只在一个实体PictureRef中存储。然后我们将在后端编写应用程序逻辑来评估pictureType并为相应的业务情况创建新类。但这似乎有些丑陋 - 我期望有一个 JPA 解决方案来处理这种常见用例？也许我们只是漏掉了一个或多个注释？

为了完整起见，我添加了完整的堆栈跟踪。我们使用的是 Hibernate 4.3.8.Final，错误发生在部署到 WildFly 8.2.0.Final 时。

21:19:24,283 ERROR [org.jboss.msc.service.fail] (ServerService Thread Pool -- 88) MSC000001: Failed to start service jboss.persistenceunit."Example-1.0-SNAPSHOT.war#ExamplePU": org.jboss.msc.service.StartException in service jboss.persistenceunit."Example-1.0-SNAPSHOT.war#ExamplePU": javax.persistence.PersistenceException: [PersistenceUnit: ExamplePU] Unable to build Hibernate SessionFactory
at org.jboss.as.jpa.service.PersistenceUnitServiceImpl$1$1.run(PersistenceUnitServiceImpl.java:172) [wildfly-jpa-8.2.0.Final.jar:8.2.0.Final]
at org.jboss.as.jpa.service.PersistenceUnitServiceImpl$1$1.run(PersistenceUnitServiceImpl.java:117) [wildfly-jpa-8.2.0.Final.jar:8.2.0.Final]
at java.security.AccessController.doPrivileged(Native Method) [rt.jar:1.8.0_25]
at org.wildfly.security.manager.WildFlySecurityManager.doChecked(WildFlySecurityManager.java:474) [wildfly-security-manager-1.0.0.Final.jar:1.0.0.Final]
at org.jboss.as.jpa.service.PersistenceUnitServiceImpl$1.run(PersistenceUnitServiceImpl.java:182) [wildfly-jpa-8.2.0.Final.jar:8.2.0.Final]
at java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1142) [rt.jar:1.8.0_25]
at java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:617) [rt.jar:1.8.0_25]
at java.lang.Thread.run(Thread.java:745) [rt.jar:1.8.0_25]
at org.jboss.threads.JBossThread.run(JBossThread.java:122) [jboss-threads-2.1.1.Final.jar:2.1.1.Final]
Caused by: javax.persistence.PersistenceException: [PersistenceUnit: ExamplePU] Unable to build Hibernate SessionFactory
at org.hibernate.jpa.boot.internal.EntityManagerFactoryBuilderImpl.persistenceException(EntityManagerFactoryBuilderImpl.java:1239) [hibernate-entitymanager-4.3.7.Final.jar:4.3.7.Final]
at org.hibernate.jpa.boot.internal.EntityManagerFactoryBuilderImpl.access$600(EntityManagerFactoryBuilderImpl.java:120) [hibernate-entitymanager-4.3.7.Final.jar:4.3.7.Final]
at org.hibernate.jpa.boot.internal.EntityManagerFactoryBuilderImpl$4.perform(EntityManagerFactoryBuilderImpl.java:855) [hibernate-entitymanager-4.3.7.Final.jar:4.3.7.Final]
at org.hibernate.jpa.boot.internal.EntityManagerFactoryBuilderImpl$4.perform(EntityManagerFactoryBuilderImpl.java:845) [hibernate-entitymanager-4.3.7.Final.jar:4.3.7.Final]
at org.hibernate.boot.registry.classloading.internal.ClassLoaderServiceImpl.withTccl(ClassLoaderServiceImpl.java:398) [hibernate-core-4.3.7.Final.jar:4.3.7.Final]
at org.hibernate.jpa.boot.internal.EntityManagerFactoryBuilderImpl.build(EntityManagerFactoryBuilderImpl.java:844) [hibernate-entitymanager-4.3.7.Final.jar:4.3.7.Final]
at org.jboss.as.jpa.hibernate4.TwoPhaseBootstrapImpl.build(TwoPhaseBootstrapImpl.java:44) [jipijapa-hibernate4-3-1.0.1.Final.jar:]
at org.jboss.as.jpa.service.PersistenceUnitServiceImpl$1$1.run(PersistenceUnitServiceImpl.java:154) [wildfly-jpa-8.2.0.Final.jar:8.2.0.Final]
... 8 more
Caused by: org.hibernate.AnnotationException: mappedBy reference an unknown target entity property:  
de.company.project.somepackages.PictureRef.person in de.company.project.somepackages.Person.picturesOfCars
at org.hibernate.cfg.annotations.CollectionBinder.bindStarToManySecondPass(CollectionBinder.java:768) [hibernate-core-4.3.7.Final.jar:4.3.7.Final]
at org.hibernate.cfg.annotations.CollectionBinder$1.secondPass(CollectionBinder.java:728) [hibernate-core-4.3.7.Final.jar:4.3.7.Final]
at org.hibernate.cfg.CollectionSecondPass.doSecondPass(CollectionSecondPass.java:70) [hibernate-core-4.3.7.Final.jar:4.3.7.Final]
at org.hibernate.cfg.Configuration.originalSecondPassCompile(Configuration.java:1697) [hibernate-core-4.3.7.Final.jar:4.3.7.Final]
at org.hibernate.cfg.Configuration.secondPassCompile(Configuration.java:1426) [hibernate-core-4.3.7.Final.jar:4.3.7.Final]
at org.hibernate.cfg.Configuration.buildSessionFactory(Configuration.java:1846) [hibernate-core-4.3.7.Final.jar:4.3.7.Final]
at org.hibernate.jpa.boot.internal.EntityManagerFactoryBuilderImpl$4.perform(EntityManagerFactoryBuilderImpl.java:852) [hibernate-entitymanager-4.3.7.Final.jar:4.3.7.Final]
... 13 more

21:19:24,291 ERROR [org.jboss.as.controller.management-operation] (management-handler-thread - 2) JBAS014613: Operation ("deploy") failed - address: ([("deployment" => "Example-1.0-SNAPSHOT.war")]) - failure description: {"JBAS014671: Failed services" => {"jboss.persistenceunit.\"Example-1.0-SNAPSHOT.war#ExamplePU\"" => "org.jboss.msc.service.StartException in service jboss.persistenceunit.\"Example-1.0-SNAPSHOT.war#ExamplePU\": javax.persistence.PersistenceException: [PersistenceUnit: ExamplePU] Unable to build Hibernate SessionFactory
Caused by: javax.persistence.PersistenceException: [PersistenceUnit: ExamplePU] Unable to build Hibernate SessionFactory
Caused by: org.hibernate.AnnotationException: mappedBy reference an unknown target entity property: de.company.project.somepackages.PictureRef.person in de.company.project.somepackages.Person.picturesOfCars"}}
21:19:24,292 ERROR [org.jboss.as.server] (management-handler-thread - 2) JBAS015870: Deploy of deployment "Example-1.0-SNAPSHOT.war" was rolled back with the following failure message: 
{"JBAS014671: Failed services" => {"jboss.persistenceunit.\"Example-1.0-SNAPSHOT.war#ExamplePU\"" => "org.jboss.msc.service.StartException in service jboss.persistenceunit.\"Example-1.0-SNAPSHOT.war#ExamplePU\": javax.persistence.PersistenceException: [PersistenceUnit: ExamplePU] Unable to build Hibernate SessionFactory
Caused by: javax.persistence.PersistenceException: [PersistenceUnit: ExamplePU] Unable to build Hibernate SessionFactory
Caused by: org.hibernate.AnnotationException: mappedBy reference an unknown target entity property: Caused by: org.hibernate.AnnotationException: mappedBy reference an unknown target entity property: de.company.project.somepackages.PictureRef.person in de.company.project.somepackages.Person.picturesOfCars"}}

- Christian

2个回答

-1

对于类似的情况，可以通过使用@JoinColumn而不是mappedBy以及@where（clause="discriminator value"）来解决它。然而，奇怪的是，尽管这个问题自2008年以来就存在，但为什么它仍然没有得到解决。查看此链接

- AlHassan

网页内容由stack overflow 提供, 点击上面的

可以查看英文原文，
原文链接

- NickJI · Accepted Answer

您碰到了一个在JPA规范中没有详细介绍的领域。JPA规范2.11节声明：

实体可以从另一个实体类继承。实体支持继承、多态关联和多态查询。

具体的支持方式并未定义。

当您试图将@OneToMany映射到继承@ManyToOne关系的子类时......

public class Person extends AbstractEntity {

    @OneToMany(mappedBy = "person", targetEntity = CarPictureRef.class)
    private List<CarPictureRef> picturesOfCars;

    @OneToMany(mappedBy = "person", targetEntity = HolidayPictureRef.class)
    private List< HolidayPictureRef> picturesOfHolidays;

    ...

}

Hibernate 抱怨

mappedBy reference an unknown target entity property

所以，当使用mappedBy解析到子类时，Hibernate不允许继承字段(person在此情况下)。然而，在关系定义中有足够的信息来解决这些关系问题，EclipseLink做得非常好。

可以将@ManyToOne从超类(PictureRef)转移到具体子类(CarPictureRef等)。

@Entity
public class CarPictureRef extends PictureRef {

    @ManyToOne
    @JoinColumn(name = "person_id")
    private Person person;

    ...

}

使用这种解决方案，您可以让Person中的@OneToMany映射保持不变。然而，在Hibernate中，这并不起作用，因为Hibernate会查找任何具有正确person_ID的子实体（在此情况下是pictureRef子类），然后抱怨找到的类型是错误的。

org.hibernate.WrongClassException: Object [id=55] was not of the specified subclass.

再次提醒，Hibernate虽然可以通过使用鉴别器列来解决这个问题，但它并没有这样做，而EclipseLink则可以。

解决这个问题的方法是为每个子类指定一个不同的连接列，这意味着每当出现新的子类时，您都需要修改您的表格，我想这不是很理想。

使用Hibernate，您可以针对@OneToMany关系中的PictureRef超类进行定位，从而使Person实体变成：

@Entity
@Table(name = "t_person")
// Sequence Definition removed
public class Person extends AbstractEntity {

    @OneToMany(mappedBy = "person", targetEntity = PictureRef.class)
    private List<PictureRef> pictures;

    ...

}

然后您将拥有一个PictureRefs列表，其中包含任何子类。您可以测试列表中的项并设置临时字段来持有所需的每个子类型。因此，在这种程度上，Hibernate支持多态关联，但是eclipseLink做得更多，因此请注意可移植性问题。