EJB3中的嵌入式对象继承

用户拥有嵌入密码。我有两种密码：用户定义或不是。因此，我有一个超类密码，及其子类GeneratedPassword。建筑确实是有争议的。

这里有 “签名”：

@Entity 
@NamedQueries({ //... }) 
@Table(name="UserAccount") 
public class UserAccount implements Serializable { 

    @Id 
    @Email 
    private String email; 

    @Embedded 
    private Password password; 


    public UserAccount(String email) { 
     this.email = email; 
     this.password = new GeneratedPassword(); 
    } 

    // ... 
} 

@Embeddable 
public class Password implements Serializable { 

    private String encryptedPassword; 

    // ... 
} 

@Embeddable 
public class GeneratedPassword extends Password { 

    private String tmpPassword; 
    // ... 
}

问题是我有一个奇怪的例外（奇怪，因为我不明白......）：

Caused by: javax.persistence.EntityExistsException: 
Exception Description: No subclass matches this class [class entities.user.GeneratedPassword] for this Aggregate mapping with inheritance. 
Mapping: org.eclipse.persistence.mappings.AggregateObjectMapping[password] 
Descriptor: RelationalDescriptor(entities.user.UserAccount --> [DatabaseTable(UserAccount)])

第2部分：

Caused by: Exception [EclipseLink-126] (Eclipse Persistence Services - 2.3.0.v20110604-r9504): org.eclipse.persistence.exceptions.DescriptorException 
Exception Description: No subclass matches this class [class entities.user.GeneratedPassword] for this Aggregate mapping with inheritance. 
Mapping: org.eclipse.persistence.mappings.AggregateObjectMapping[password] 
Descriptor: RelationalDescriptor(entities.user.UserAccount --> [DatabaseTable(UserAccount)])

因此，我从这些例外中了解到，GeneratedPassword不被识别为实体。但是如果我使用密码类，evrything工作正常！所以我回到了不理解的状态......

任何人都知道如何在层次结构中使用可嵌入的实体？那甚至是问题？

来源

2011-12-02 Thomas