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我正在尝试使用 Virgo 和 EclipseLink 并实现基于 Greenpage 项目的应用程序。

我已经实现了一个层次结构，但我遇到了一个奇怪的错误（使用 EclipseLink）：

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.

等级制度：

Entity <- NamedEntity <- Person <- ... - 扩展 Person 的其他类

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：Kevin Bowersox 发布的答案（非常感谢！）帮助......部分;）。问题变得更深了。我们现在是：Entity <- NamedEntity <- Attribute <- 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;

}

还有一个例外：

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.
4

2 回答 2

3

将注释从访问器移至字段声明，并将字段可访问性更改为受保护。

@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;
    }
}
于 2013-03-10T18:38:06.870 回答
-1

好的，Kevin Bowersox 的回答让我找到了正确的答案：我更改了所有受保护的字段并将注释更改为字段。这导致良好的工作;)。

于 2013-03-10T20:11:34.187 回答