用于JPA回调的Hibernate事件侦听器

Question

如何启用处理JPA回调的Hibernate事件侦听器？

目前我正在使用Hibernate 4和SessionFactory配置,但是当我持久化对象时,JPA回调没有正常运行.

任何建议都是最受欢迎的.

源代码

临时实体类:

package com.esp.entity;

import javax.persistence.Entity;
import javax.persistence.EntityListeners;
import javax.persistence.GeneratedValue;
import javax.persistence.Id;
import javax.persistence.PostLoad;
import javax.persistence.Table;

import com.esp.aaa.TempVal;

@Entity
@EntityListeners(value=TempVal.class)
@Table(name="TEMP")
public class Temp {
    private int id;
    private String name;
    private String email;
    private int roll;

    @Id @GeneratedValue
    public int getId() {
        return id;
    }
    public void setId(int id) {
        this.id = id;
    }
    public String getName() {
        return name;
    }
    public void setName(String name) {
        this.name = name;
    }
    public String getEmail() {
        return email;
    }
    public void setEmail(String email) {
        this.email = email;
    }
    public int getRoll() {
        return roll;
    }
    public void setRoll(int roll) {
        this.roll = roll;
    }
    @PostLoad
    public void load(){
        System.out.println("post load called here");
    }
}

TempVal类:

package com.esp.aaa;

import javax.persistence.PrePersist;

public class TempVal {
    @PrePersist
    public void validate(Object temp){
        System.out.println("Object will persist now");
    }
}

MainClass类:

package com.esp.aaa;

import org.hibernate.Session;
import com.esp.entity.Temp;
import com.esp.utility.HibernateUtils;

public class MainClass {
    public static void main(String args[]) {
        HibernateUtils.createSessionFactory();
        Session session=HibernateUtils.getSessionFactory().getCurrentSession();
        session.beginTransaction();

        Temp temp=new Temp();

        temp.setEmail("abc@gmail.com");
        temp.setName("Lucky");
        temp.setRoll(1112);

        session.save(temp);
        System.out.println("Object persist successfully");

        session.getTransaction().commit();
        HibernateUtils.shutdown();
    }
}

Hibernate配置:

<hibernate-configuration>
    <session-factory>
        <property name="connection.driver_class">com.mysql.jdbc.Driver</property>
        <property name="connection.url">jdbc:mysql://localhost:3306/demo</property>
        <property name="connection.username">root</property>
        <property name="connection.password">root</property>

        <property name="dialect">org.hibernate.dialect.MySQLDialect</property>
        <property name="show_sql">true</property>
        <property name="hbm2ddl.auto">update</property>
        <property name="hibernate.current_session_context_class">thread</property>

        <mapping class="com.esp.entity.Temp"/>
    </session-factory>
</hibernate-configuration>

程序输出

程序输出如下:

Hibernate: insert into TEMP (email, name, roll) values (?, ?, ?)
Object persist successfully

预期的产出是:

Object will persist now
Hibernate: insert into TEMP (email, name, roll) values (?, ?, ?)
Object persist successfully

Answer 1

这个问题基本上都是一样的.

事实证明,这些JPA实体监听器注释仅在您EntityManager在Hibernate 中使用时才起作用(这是可以理解的).因此,如果您想使用这些注释,您应该放弃SessionFactory,并使用JPA投诉EntityManager或EntityManagerFactory.

我也推荐这种方法,因为如果你正在尝试使用JPA注释,那么选择纯粹的JPA解决方案是合乎逻辑的,而不必将自己与特定于Hibernate的解决方案联系起来 - 即SessionFactory.

Answer 2

添加一个名为org.hibernate.integrator.spi.Integrator的文件,其中包含一行

org.hibernate.jpa.event.spi.JpaIntegrator

在您的META-INF/services文件夹中.这将为基于sessionFactory的配置启用JPA生命周期注释,包括@EntityListeners.