Question

我目前正在学习JPA。在文档中，它指出只有当实体被其他JVM远程分离时，它才需要是Serializable。

但是，出于测试目的，我创建了我的实体作为我的持久化类（CDI）的内部私有类。当我尝试使用EntityManager持久化实体时。我得到一个例外如下：

Caused by: Exception [EclipseLink-7155] (Eclipse Persistence Services - 2.5.0.v20130507-3faac2b): org.eclipse.persistence.exceptions.ValidationException
Exception Description: The type [class minh.ea.common.Ultilities.DatabaseLogger] for the attribute [this$0] on the entity class [class minh.ea.common.Ultilities.DatabaseLogger$LogRecord] is not a valid type for a serialized mapping. The attribute type must implement the Serializable interface.

我理解这个例外，意味着我的实体属性需要是Serializable以及Class。那是什么原因，它被传递到哪里？

所有这些都在GlassFish 4.0 Container下运行。 JPA使用EclipseLink 2.1

我对持久性和内部实体的实现如下：

package minh.ea.common.Ultilities;

import java.util.Date;
import javax.enterprise.context.ApplicationScoped;
import javax.persistence.Entity;
import javax.persistence.EntityManager;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
import javax.persistence.PersistenceContext;
import javax.persistence.Temporal;
import javax.persistence.TemporalType;
import javax.transaction.Transactional;

/**
 *
 * @author Minh
 */

@ApplicationScoped
@minh.ea.common.Ultilities.qualifiers.Database
public class DatabaseLogger implements Logger {

    @PersistenceContext
    private EntityManager em;

    private final long MAX_SIZE=2097152;

    public DatabaseLogger(){

    }

    @Override
    @Transactional
    public void info(Object obj) {
        em.persist(new RecordEntry("INFO", new Date(), obj.toString()));
    }

    @Override
    @Transactional
    public void warn(Object obj) {
        em.persist(new RecordEntry("WARN", new Date(), obj.toString()));
    }

    @Override
    @Transactional
    public void error(Object obj) {
        em.persist(new RecordEntry("ERROR", new Date(), obj.toString()));
    }

    @Override
    @Transactional
    public void fatal(Object obj) {
        em.persist(new RecordEntry("FATAL", new Date(), obj.toString()));
    }

    @Entity
    public class LogRecord{

        @Id @GeneratedValue(strategy = GenerationType.IDENTITY)
        private int id;
        private String type;
        @Temporal(TemporalType.TIMESTAMP)
        private Date time;
        private String message;

        public String getType() {
            return type;
        }

        public void setType(String type) {
            this.type = type;
        }

        public Date getTime() {
            return time;
        }

        public void setTime(Date time) {
            this.time = time;
        }

        public String getMessage() {
            return message;
        }

        public void setMessage(String message) {
            this.message = message;
        }

        public LogRecord() {
        }

        public LogRecord(String type, Date time, String message) {
            this.type = type;
            this.time = time;
            this.message = message;
        }
    }

}

致以最诚挚的问候，

Answer 1

根据this reference，有效的Entity类必须是顶级类，这意味着您的内部类将无法工作。

尝试将其拉出到自己的Entity类中，就像它在适当的系统中一样，然后再次存在。

JPA，当必须实体实现Serializable时

1 个答案: