java – org. hibernate. Persistent objectexception: passing detached entities to persistence – using JPA

I'm creating a simple application that just inserts a row into the table (if the table doesn't exist, create it) using java JPA

I've seen it on Google. Here, I still don't understand what the anomaly is Even if there were the same problem as me, I didn't get a solution I'm still new to Java JPA, so please keep me naked

I attached some code to run it

Back to this question, this is the exception and stacktrace I got:

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

Here is my code:

Main shift:

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();
            }
        }
    }
}

And humans:

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() {
    }
}

This is my persistence XML file

<?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 –––––––

I just changed the provider to eclipse link without further changes I'm confused now Why does it work with eclipse link, but using hibernate will produce exceptions?

Solution

Try the following code, which will then allow you to set the ID manually

Simply use the @ ID annotation to define which attribute is the identifier of your entity If you do not want hibernate to generate this property for you, you do not need to use the @ generatedvalue annotation

Assign – allows the application to assign an identifier to an object before calling save() This is the default policy if no < generator > element is specified

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 lastName) {
        this.id = id;
        this.name = name;
        this.lastName = lastName;
    }

    public Person() {
    }
}