JPA Configuration with Hibernate Example

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In this post, we will show you how to create or configure a simple JPA application with Hibernate.
The Java Persistence API (a.k.a. JPA) is a Java specification for managing, persisting, and accessing data between objects and relational databases. Hibernate is an ORM (Object Relational Mapping) tool that implements JPA specifications.
Below article describes Hibernate native Bootstrapping - 
>> Hibernate 5 XML Configuration Example 
>> Hibernate 5 Java Configuration Example

Technologies and tools used

  • Hibernate 5.3.7.Final
  • JPA 2.1
  • IDE - Eclipse Noen
  • Maven 3.5.3
  • JavaSE 1.8
  • MySQL - 8.0.13
Let's start developing step by step Hibernate application using Maven as project management and build tool.

Development Steps

  1. Create a Simple Maven Project
  2. Project Directory Structure
  3. Add jar Dependencies to pom.xml
  4. Creating the JPA Entity Class(Persistent class)
  5. Create a JPA configuration file
  6. Create a JPA helper class
  7. Create the Main class and Run an Application

1. Create a Simple Maven Project

Use the How to Create a Simple Maven Project in Eclipse article to create a simple Maven project in Eclipse IDE.

2. Project Directory Structure

The project directory structure for your reference - 

3. Add jar Dependencies to pom.xml

<project
    xmlns="http://maven.apache.org/POM/4.0.0"
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
 xsi:schemaLocation="http://maven.apache.org/POM/4.0.0 http://maven.apache.org/xsd/maven-4.0.0.xsd">
    <modelVersion>4.0.0</modelVersion>
    <parent>
        <groupId>net.javaguides.hibernate</groupId>
        <artifactId>hibernate-tutorial</artifactId>
        <version>0.0.1-SNAPSHOT</version>
    </parent>
    <artifactId>hibernate-jpa-config-example</artifactId>
    <properties>
        <project.build.sourceEncoding>UTF-8</project.build.sourceEncoding>
    </properties>
    <dependencies>
        <!-- https://mvnrepository.com/artifact/mysql/mysql-connector-java -->
        <dependency>
            <groupId>mysql</groupId>
            <artifactId>mysql-connector-java</artifactId>
            <version>8.0.13</version>
        </dependency>
        <!-- https://mvnrepository.com/artifact/org.hibernate/hibernate-core -->
        <dependency>
            <groupId>org.hibernate</groupId>
            <artifactId>hibernate-core</artifactId>
            <version>5.3.7.Final</version>
        </dependency>
    </dependencies>
    <build>
        <sourceDirectory>src/main/java</sourceDirectory>
        <plugins>
            <plugin>
                <artifactId>maven-compiler-plugin</artifactId>
                <version>3.5.1</version>
                <configuration>
                    <source>1.8</source>
                    <target>1.8</target>
                </configuration>
            </plugin>
        </plugins>
    </build>
</project>

4. Creating the JPA Entity Class(Persistent class)

Create a Student entity class under net.javaguides.hibernate.entity package as follows.
package net.javaguides.hibernate.entity;

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

@Entity
@Table(name = "student")
public class Student {

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Column(name = "id")
    private int id;

    @Column(name = "first_name")
    private String firstName;

    @Column(name = "last_name")
    private String lastName;

    @Column(name = "email")
    private String email;

    public Student() {

    }

    public Student(String firstName, String lastName, String email) {
        this.firstName = firstName;
        this.lastName = lastName;
        this.email = email;
    }

    public int getId() {
        return id;
    }

    public void setId(int id) {
        this.id = id;
    }

    public String getFirstName() {
        return firstName;
    }

    public void setFirstName(String firstName) {
        this.firstName = firstName;
    }

    public String getLastName() {
        return lastName;
    }

    public void setLastName(String lastName) {
        this.lastName = lastName;
    }

    public String getEmail() {
        return email;
    }

    public void setEmail(String email) {
        this.email = email;
    }

    @Override
    public String toString() {
        return "Student [id=" + id + ", firstName=" + firstName + ", lastName=" + lastName + ", email=" + email + "]";
    }
}

5. Create a JPA configuration file

Let's create an XML file named persistence.xml under the src/main/java/META-INF folder and write the following code in it.
<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="PERSISTENCE">
        <description> Hibernate JPA Configuration Example</description>
        <provider>org.hibernate.jpa.HibernatePersistenceProvider</provider>
        <class>net.javaguides.hibernate.entity.Student</class>
        <properties>
            <property name="javax.persistence.jdbc.driver"
    value="com.mysql.cj.jdbc.Driver" />
            <property name="javax.persistence.jdbc.url"
    value="jdbc:mysql://localhost:3306/hibernate_db" />
            <property name="javax.persistence.jdbc.user" value="root" />
            <property name="javax.persistence.jdbc.password"
    value="root" />
            <property name="hibernate.show_sql" value="true" />
            <property name="hibernate.hbm2ddl.auto" value="create-drop" />
        </properties>
    </persistence-unit>
</persistence>

6. Create a JPA helper class

Let's create a helper class to bootstrap a JPA EntityManagerFactory.
package net.javaguides.hibernate.util;

import javax.persistence.EntityManagerFactory;
import javax.persistence.Persistence;

public class JPAUtil {
    private static final String PERSISTENCE_UNIT_NAME = "PERSISTENCE";
    private static EntityManagerFactory factory;

    public static EntityManagerFactory getEntityManagerFactory() {
        if (factory == null) {
            factory = Persistence.createEntityManagerFactory(PERSISTENCE_UNIT_NAME);
        }
        return factory;
    }

    public static void shutdown() {
        if (factory != null) {
            factory.close();
        }
    }
}

7. Create a main class and run an application

Here is the main class to persist Student object using the EntityManager.persist() method.

MainApp.java

package net.javaguides.hibernate;

import javax.persistence.EntityManager;

import net.javaguides.hibernate.entity.Student;
import net.javaguides.hibernate.util.JPAUtil;

public class App {
    public static void main(String[] args) {

        EntityManager entityManager = JPAUtil.getEntityManagerFactory().createEntityManager();
        entityManager.getTransaction().begin();

        Student student = new Student("Ramesh", "Fadatare", "rameshfadatare@javaguides.com");
        entityManager.persist(student);
        entityManager.getTransaction().commit();
        entityManager.close();

        JPAUtil.shutdown();
    }

}

Output



GitHub Repository

The complete source code of this article is available on my GitHub Repository - https://github.com/RameshMF/Hibernate-ORM-Tutorials

Conclusion

In this post, we have shown you how to create or configure a simple JPA application with Hibernate.

Learn more about JPA at JPA Tutorial

References


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Comments

  1. Swathi Muppalla1 August 2020 at 22:34

    How to Create a Simple Maven Project in Eclipse....this link giving 404 error. Please check sir!

    ReplyDelete

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