No Persistence provider for EntityManager named

Question

I have my persistence.xml with the same name using TopLink under the META-INF directory. Then, I have my code calling it with:

Answer 1

After <persistence-unit name="agisdb">, define the persistence provider name:

<provider>org.hibernate.ejb.HibernatePersistence</provider>

Answer 2

In an OSGi-context, it's necessary to list your persistence units in the bundle's MANIFEST.MF, e.g.

JPA-PersistenceUnits: my-persistence-unit

Otherwise, the JPA-bundle won't know your bundle contains persistence units.

See http://wiki.eclipse.org/EclipseLink/Examples/OSGi/Developing_with_EclipseLink_OSGi_in_PDE .

Answer 3

You have to use the absolute path of the file otherwise this will not work. Then with that path we build the file and pass it to the configuration.

@Throws(HibernateException::class)
fun getSessionFactory() : SessionFactory {
    return Configuration()
                .configure(getFile())
                .buildSessionFactory()
    }


private fun getFile(canonicalName: String): File {
    val absolutePathCurrentModule = System.getProperty("user.dir")
    val pathFromProjectRoot = absolutePathCurrentModule.dropLastWhile { it != '/' }
    val absolutePathFromProjectRoot = "${pathFromProjectRoot}module-name/src/main/resources/$canonicalName"
    println("Absolute Path of secret-hibernate.cfg.xml: $absolutePathFromProjectRoot")
    return File(absolutePathFromProjectRoot)
}

GL

Source

Answer 4

Hibernate 5.2.5
Jar Files Required in the class path. This is within a required folder of Hibernate 5.2.5 Final release. It can be downloaded from http://hibernate.org/orm/downloads/

1. antlr-2.7.7
2. cdi-api-1.1
3. classmate-1.3.0
4. dom4j-1.6.1
5. el-api-2.2
6. geronimo-jta_1.1_spec-1.1.1
7. hibernate-commons-annotation-5.0.1.Final
8. hibernate-core-5.2.5.Final
9. hibernate-jpa-2.1-api-1.0.0.Final
10. jandex-2.0.3.Final
11. javassist-3.20.0-GA
12. javax.inject-1
13. jboss-interceptor-api_1.1_spec-1.0.0.Beta1
14. jboss-logging-3.3.0.Final
15. jsr250-api-1.0

Create an xml file "persistence.xml" in

YourProject/src/META-INF/persistence.xml

persistence.xml

<?xml version="1.0" encoding="UTF-8"?>

<persistence 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"
         version="2.1">

<persistence-unit name="sample">

    <class>org.pramod.data.object.UserDetail</class>
    <exclude-unlisted-classes>true</exclude-unlisted-classes>
    <properties>
        <property name="hibernate.connection.driver_class" value="com.mysql.jdbc.Driver"/>
        <property name="hibernate.dialect" value="org.hibernate.dialect.MySQLDialect"/>
        <property name="hibernate.connection.url" value="jdbc:mysql://localhost:3306/hibernate_project"/>
        <property name="hibernate.connection.username" value="root"/>
        <property name="hibernate.connection.password" value="root"/>
        <property name="hibernate.hbm2ddl.auto" value="create-drop"/>
        <property name="hibernate.show_sql" value="true"/>
        <property name="hibernate.format_sql" value="false"/>

        <property name="hibernate.cache.use_second_level_cache" value="false"/>

        <property name="hibernate.archive.autodetection" value="true"/>
    </properties>
</persistence-unit>

1. please note down the information mentioned in the < persistance > tag and version should be 2.1.
2. please note the name < persistance-unit > tag, name is mentioned as "sample". This name needs to be used exactly same while loading your

EntityManagerFactor = Persistance.createEntityManagerFactory("sample");. "sample" can be changed as per your naming convention.

Now create a Entity class. with name as per my example UserDetail, in the package org.pramod.data.object

UserDetail.java

package org.pramod.data.object;

import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.Id;
import javax.persistence.Table;

@Entity
@Table(name = "user_detail")
public class UserDetail {
    @Id
    @Column(name="user_id")
    private int id;
    @Column(name="user_name")
    private String userName;

    public int getId() {
        return id;
    }
    public void setId(int id) {
        this.id = id;
    }
    public String getUserName() {
        return userName;
    }
    public void setUserName(String userName) {
        this.userName = userName;
    }   
    @Override
    public String toString() {
       return "UserDetail [id=" + id + ", userName=" + userName + "]";
    }
}

Now create a class with main method.

HibernateTest.java

package org.pramod.hibernate;

import javax.persistence.EntityManager;
import javax.persistence.EntityManagerFactory;
import javax.persistence.Persistence;
import org.pramod.data.object.UserDetail;

public class HibernateTest {
    private static EntityManagerFactory entityManagerFactory;

    public static void main(String[] args)  {
        UserDetail user = new UserDetail();
        user.setId(1);
        user.setUserName("Pramod Sharma");

        try {
            entityManagerFactory = Persistence.createEntityManagerFactory("sample");
            EntityManager entityManager = entityManagerFactory.createEntityManager();
            entityManager.getTransaction().begin();
            entityManager.persist( user );
            entityManager.getTransaction().commit();
            System.out.println("successfull");
            entityManager.close();
        } catch (Exception e) {
            e.printStackTrace();
        }

    }

}

Output will be

UserDetail [id=1, userName=Pramod Sharma]
Hibernate: drop table if exists user_details
Hibernate: create table user_details (user_id integer not null, user_name varchar(255), primary key (user_id))
Hibernate: insert into user_details (user_name, user_id) values (?, ?)
successfull

Answer 5

In my case it was about mistake in two properties as below. When I changed them ‘No Persistence provider for EntityManager named’ disappered.

So you could try test connection with your properties to check if everything is correct.

<property name="javax.persistence.jdbc.url" value="...”/>
<property name="javax.persistence.jdbc.password" value="...”/>

Strange error, I was totally confused because of it.

Answer 6

I also had this error but the issue was the namespace uri in the persistence.xml.

I replaced http://xmlns.jcp.org/xml/ns/persistence to http://java.sun.com/xml/ns/persistence and the version 2.1 to 2.0.

It's now working.