EclipseLink层次结构错误

Question

我正在尝试与处女座和EclipseLink合作并实施基于Greenpage项目的应用程序。

我实现了一个层次，但我得到（使用的EclipseLink）一个奇怪的错误：

Internal Exception: Exception [EclipseLink-7161] (Eclipse Persistence Services - 2.0.0.v20091127-r5931): org.eclipse.persistence.exceptions.ValidationException 
Exception Description: Entity class [class model.Person] has no primary key specified. It should define either an @Id, @EmbeddedId or an @IdClass. If you have defined PK using any of these annotations then make sure that you do not have mixed access-type (both fields and properties annotated) in your entity class hierarchy. 

层次：

实体< - NamedEntity < - 人< - ...... - 其他类，扩展人

import java.io.Serializable; 

import javax.persistence.GeneratedValue; 
import javax.persistence.GenerationType; 
import javax.persistence.Id; 
import javax.persistence.MappedSuperclass; 

@MappedSuperclass 
public abstract class Entity implements Serializable { 

    private static final long serialVersionUID = 1L; 

    private Long id; 

    @Id 
    @GeneratedValue(strategy=GenerationType.AUTO) 
    public Long getId() { 
     return id; 
    } 

    public void setId(Long id) { 
     this.id = id; 
    } 
} 


import javax.persistence.MappedSuperclass; 

@MappedSuperclass 
public abstract class NamedEntity extends Entity { 
    private static final long serialVersionUID = 1L; 

    private String name; 

    public String getName() { 
     return name; 
    } 

    public void setName(String name) { 
     this.name = name; 
    } 

    @Override 
    public String toString() { 
     return name; 
    } 
} 



import javax.persistence.DiscriminatorColumn; 
import javax.persistence.Entity; 
import javax.persistence.Inheritance; 
import javax.persistence.InheritanceType; 

import pl.com.mgr.model.NamedEntity; 

@Entity 
@Inheritance(strategy = InheritanceType.SINGLE_TABLE) 
@DiscriminatorColumn(name = "discriminator") 
public abstract class Person extends NamedEntity { 
    private static final long serialVersionUID = 1L; 

} 

编辑：由凯文Bowersox（非常感谢！）发布的答复...部分;）。问题进一步深入。我们现在是：实体< - NamedEntity < - 属性< - LecturerAttribute

import javax.persistence.Column; 
import javax.persistence.JoinColumn; 
import javax.persistence.ManyToOne; 
import javax.persistence.MappedSuperclass; 

import model.NamedEntity; 

@MappedSuperclass 
public abstract class Attribute<T extends NamedEntity> extends NamedEntity { 
    private static final long serialVersionUID = 1L; 

    private T entity; 
    private String value; 

    @ManyToOne 
    @JoinColumn(name = "entity", nullable = false) 
    public T getEntity() { 
     return entity; 
    } 

    public void setEntity(T entity) { 
     this.entity = entity; 
    } 

    @Column(nullable = false) 
    public String getValue() { 
     return value; 
    } 

    public void setValue(String value) { 
     this.value = value; 
    } 


} 




import javax.persistence.Entity; 

@Entity 
public class LecturerAttribute extends Attribute<Lecturer> { 
    private static final long serialVersionUID = 1L; 

} 

和异常：

Internal Exception: Exception [EclipseLink-7161] (Eclipse Persistence Services - 2.0.0.v20091127-r5931): org.eclipse.persistence.exceptions.ValidationException 
Exception Description: Entity class [class model.LecturerAttribute] has no primary key specified. It should define either an @Id, @EmbeddedId or an @IdClass. If you have defined PK using any of these annotations then make sure that you do not have mixed access-type (both fields and properties annotated) in your entity class hierarchy. 

Answer 1

移动从访问到字段声明注释改变域无障碍保护。

@MappedSuperclass 
public abstract class Entity implements Serializable { 

    private static final long serialVersionUID = 1L; 

    @Id 
    @GeneratedValue(strategy=GenerationType.AUTO) 
    protected Long id; 

    public Long getId() { 
     return id; 
    } 

    public void setId(Long id) { 
     this.id = id; 
    } 
} 

Answer 2

好的，凯文鲍尔斯科克回答指示我正确的答案：我改变所有字段保护和更改注释字段。这导致良好的工作;）。