Question

我正在创建一个简单的应用程序，只需使用Java JPA向表中插入行（如果表不存在，创建它）。

我为它的可运行示例附加了一些代码。

这是我得到的异常和stacktrace：

EXCEPTION -- > org.hibernate.PersistentObjectException: detached entity passed to persist: view.Person
javax.persistence.PersistenceException: org.hibernate.PersistentObjectException: detached entity passed to persist: view.Person
    at org.hibernate.jpa.spi.AbstractEntityManagerImpl.convert(AbstractEntityManagerImpl.java:1763)
    at org.hibernate.jpa.spi.AbstractEntityManagerImpl.convert(AbstractEntityManagerImpl.java:1677)
    at org.hibernate.jpa.spi.AbstractEntityManagerImpl.convert(AbstractEntityManagerImpl.java:1683)
    at org.hibernate.jpa.spi.AbstractEntityManagerImpl.persist(AbstractEntityManagerImpl.java:1187)
    at view.TestJPA.main(TestJPA.java:34)
Caused by: org.hibernate.PersistentObjectException: detached entity passed to persist: view.Person
    at org.hibernate.event.internal.DefaultPersistEventListener.onPersist(DefaultPersistEventListener.java:139)
    at org.hibernate.event.internal.DefaultPersistEventListener.onPersist(DefaultPersistEventListener.java:75)
    at org.hibernate.internal.SessionImpl.firePersist(SessionImpl.java:811)
    at org.hibernate.internal.SessionImpl.persist(SessionImpl.java:784)
    at org.hibernate.internal.SessionImpl.persist(SessionImpl.java:789)
    at org.hibernate.jpa.spi.AbstractEntityManagerImpl.persist(AbstractEntityManagerImpl.java:1181)
    ... 1 more

这是我的代码：

主要课程：

package view;

import javax.persistence.EntityManager;
import javax.persistence.EntityManagerFactory;
import javax.persistence.EntityTransaction;
import javax.persistence.Persistence;

public class TestJPA {

    public static void main(String[] args) {

        Person p = new Person(1, "Peter", "Parker");

        EntityManagerFactory entityManagerFactory = Persistence.createEntityManagerFactory("TesePersistentUnit");
        EntityManager entityManager = entityManagerFactory.createEntityManager();

        EntityTransaction transaction = entityManager.getTransaction();
        try {
            transaction.begin();

            entityManager.persist(p);
            entityManager.getTransaction().commit();
        } 
        catch (Exception e) {
            if (transaction != null) {
                transaction.rollback();
            }
            System.out.println("EXCEPTION -- > " + e.getMessage());
            e.printStackTrace();
        } 
        finally {
            if (entityManager != null) {
                entityManager.close();
            }
        }
    }
}

Person类：

package view;

import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
import javax.persistence.Table;

@Entity
@Table(name = "People")
public class Person {

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    private int id;

    private String name;
    private String lastName;

    public Person(int id, String name, String lastName) {
        this.id = id;
        this.name = name;
        this.lastName = lastName;
    }

    public Person() {
    }
}

这是我的persistence.xml文件

<?xml version="1.0" encoding="UTF-8"?>
<persistence version="2.1" xmlns="http://xmlns.jcp.org/xml/ns/persistence" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/persistence http://xmlns.jcp.org/xml/ns/persistence/persistence_2_1.xsd">
    <persistence-unit name="TesePersistentUnit" transaction-type="RESOURCE_LOCAL">
        <provider>org.hibernate.ejb.HibernatePersistence</provider>
        <class>view.Person</class>
        <properties>
            <!-- SQL dialect -->
            <property name="hibernate.dialect" value="org.hibernate.dialect.MySQLDialect"/>

            <property name="javax.persistence.jdbc.url" value="jdbc:mysql://localhost:3306/tese_tabelas?zeroDateTimeBehavior=convertToNull"/>
            <property name="javax.persistence.jdbc.user" value="root"/>
            <property name="javax.persistence.jdbc.driver" value="com.mysql.jdbc.Driver"/>
            <property name="javax.persistence.jdbc.password" value=""/>

            <!-- Create/update tables automatically using mapping metadata -->
            <property name="hibernate.hbm2ddl.auto" value="update"/>
        </properties>
    </persistence-unit>
</persistence>

----------------------- EDIT -------------------- -------

我刚刚将提供程序更改为EclipseLink，并且没有进一步更改，它正在运行。我现在很困惑。为什么它适用于EclipseLink，但是使用Hibernate会产生异常？

Answer 1

这样做的原因是你已经在Person类中声明了使用auto策略生成的id，JPA尝试在持久化实体时插入id本身。但是，在constructor中，您手动设置了id变量。由于ID是手动分配的，并且persistence context中的实体不存在，这会导致JPA认为您正在尝试保留与持久性上下文分离的实体，从而导致异常。< / p>

要修复它，请不要在构造函数中设置id。

@Id
@GeneratedValue(strategy = GenerationType.AUTO)
private int id;



public Person(int id, String name, String lastName) {
       // this.id = id;
        this.name = name;
        this.lastName = lastName;
  }

Answer 2

尝试使用以下代码，然后您可以手动设置ID。

只需使用@Id注释，即可以定义哪个属性是实体的标识符。如果您不希望hibernate为您生成此属性，则不需要使用@GeneratedValue注释。

已分配 - 允许应用程序在调用save()之前为对象分配标识符。如果未指定<generator>元素，则这是默认策略。

package view;

import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
import javax.persistence.Table;

@Entity
@Table(name = "People")
public class Person {
    @Id
    //@GeneratedValue(strategy = GenerationType.AUTO) // commented for manually set the id
    private int id;

    private String name;
    private String lastName;

    public Person(int id, String name, String lastName) {
        this.id = id;
        this.name = name;
        this.lastName = lastName;
    }

    public Person() {
    }
}

Answer 3

在类定义中，您具有由持久性提供程序选择的策略（表，序列等）生成的带注释的id，但您正在通过构造函数初始化Person对象的id。我认为留下id null可以解决你的问题。

Answer 4

你可以像这样使用这个控制器：

Person p = new Person(0, "Peter", "Parker");

然后当JPA持续到数据库时，它会自动插入AUTO_INCREMENT策略。

org.hibernate.PersistentObjectException：传递给持久异常的分离实体

4 个答案: