In the previous article, we have discussed Hibernate 5 - Save an Entity Example. In this article, we will create a simple Hibernate application to demonstrate how to persist an entity into a database.
persist(Object object) Method
Make a transient instance persistent. This operation cascades to associated instances if the association is mapped with cascade="persist"
Below diagram shows the snippet of persisting an entity in a database:
Let's start developing step by step Hibernate application using Maven as project management and build tool.
Technologies and tools used
- Hibernate 5.3.7.Final
- IDE - Eclipse Noen
- Maven 3.5.3
- JavaSE 1.8
- MySQL - 8.0.13
Development Steps
- Create a Simple Maven Project
- Project Directory Structure
- Add jar Dependencies to pom.xml
- Creating the JPA Entity Class(Persistent class)
- Create a Hibernate configuration file - hibernate.cfg.xml
- Create a Hibernate utility class
- Create the Main class and Run an Application
1. Create a Simple Maven Project
Use How to Create a Simple Maven Project in Eclipse article to create 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-xml-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)
Let's create a Student persistent class that is mapped to a database table.
A simple Persistent class should follow some rules:
- A no-arg constructor: It is recommended that you have a default constructor at least package visibility so that hibernate can create the instance of the Persistent class by newInstance() method.
- Provide an identifier property: It is better to assign an attribute as id. This attribute behaves as a primary key in a database.
- Declare getter and setter methods: The Hibernate recognizes the method by getter and setter method names by default.
- Prefer non-final class: Hibernate uses the concept of proxies, that depends on the persistent class. The application programmer will not be able to use proxies for lazy association fetching.
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 Hibernate configuration file - hibernate.cfg.xml
The configuration file contains information about the database and mapping file. Conventionally, its name should be hibernate.cfg.xml.
You can use Hibernate 5 Java Configuration Example instead of a hibernate.cfg.xml file.
Let's create an XML file named as hibernate.cfg.xml under resources folder and write the following code in it.
<!DOCTYPE hibernate-configuration PUBLIC
"-//Hibernate/Hibernate Configuration DTD 3.0//EN"
"http://www.hibernate.org/dtd/hibernate-configuration-3.0.dtd">
<hibernate-configuration>
<session-factory>
<!-- JDBC Database connection settings -->
<property name="connection.driver_class">com.mysql.cj.jdbc.Driver</property>
<property name="connection.url">jdbc:mysql://localhost:3306/hibernate_db?useSSL=false</property>
<property name="connection.username">root</property>
<property name="connection.password">root</property>
<!-- JDBC connection pool settings ... using built-in test pool -->
<property name="connection.pool_size">1</property>
<!-- Select our SQL dialect -->
<property name="dialect">org.hibernate.dialect.MySQL5Dialect</property>
<!-- Echo the SQL to stdout -->
<property name="show_sql">true</property>
<!-- Set the current session context -->
<property name="current_session_context_class">thread</property>
<!-- Drop and re-create the database schema on startup -->
<property name="hbm2ddl.auto">create-drop</property>
<!-- dbcp connection pool configuration -->
<property name="hibernate.dbcp.initialSize">5</property>
<property name="hibernate.dbcp.maxTotal">20</property>
<property name="hibernate.dbcp.maxIdle">10</property>
<property name="hibernate.dbcp.minIdle">5</property>
<property name="hibernate.dbcp.maxWaitMillis">-1</property>
<mapping class="net.javaguides.hibernate.entity.Student" />
</session-factory>
</hibernate-configuration>
6. Create a Hibernate utility Class
Create a helper class to bootstrap hibernate SessionFactory. In most Hibernate applications, the SessionFactory should be instantiated once during application initialization. The single instance should then be used by all code in a particular process, and any Session should be created using this single SessionFactory. The SessionFactory is thread-safe and can be shared; a Session is a single-threaded object. Let's create HibernateUtil.java class to configure SessionFactory as a singleton and use throughout the application.
The bootstrapping API is quite flexible, but in most cases, it makes the most sense to think of it as a 3 step process:
- Build the StandardServiceRegistry
- Build the Metadata
- Use those 2 to build the SessionFactory
package net.javaguides.hibernate.util;
import org.hibernate.SessionFactory;
import org.hibernate.boot.Metadata;
import org.hibernate.boot.MetadataSources;
import org.hibernate.boot.registry.StandardServiceRegistry;
import org.hibernate.boot.registry.StandardServiceRegistryBuilder;
public class HibernateUtil {
private static StandardServiceRegistry registry;
private static SessionFactory sessionFactory;
public static SessionFactory getSessionFactory() {
if (sessionFactory == null) {
try {
// Create registry
registry = new StandardServiceRegistryBuilder().configure().build();
// Create MetadataSources
MetadataSources sources = new MetadataSources(registry);
// Create Metadata
Metadata metadata = sources.getMetadataBuilder().build();
// Create SessionFactory
sessionFactory = metadata.getSessionFactoryBuilder().build();
} catch (Exception e) {
e.printStackTrace();
if (registry != null) {
StandardServiceRegistryBuilder.destroy(registry);
}
}
}
return sessionFactory;
}
public static void shutdown() {
if (registry != null) {
StandardServiceRegistryBuilder.destroy(registry);
}
}
}
7. Create the main App class and Run an Application
Here is main App class which is used to connect MySQL database and persist Student object in a database table. Let's test Hibernate application to connect MySQL database.
package net.javaguides.hibernate; import org.hibernate.Session; import org.hibernate.Transaction; import net.javaguides.hibernate.entity.Student; import net.javaguides.hibernate.util.HibernateUtil; public class App { public static void main(String[] args) { Student student = new Student("Ramesh", "Fadatare", "rameshfadatare@javaguides.com"); persistStudent(student); } public static void persistStudent(Student student) { Transaction transaction = null; try (Session session = HibernateUtil.getSessionFactory().openSession()) { // start a transaction transaction = session.beginTransaction(); // save the student object session.persist(student); // commit transaction transaction.commit(); } catch (Exception e) { if (transaction != null) { transaction.rollback(); } e.printStackTrace(); } } }
Output
GitHub Repository
The complete source code of this article available on my GitHub Repository - https://github.com/RameshMF/Hibernate-ORM-Tutorials
Conclusion
In this article, we have created a step by step hibernate application to demonstrate how to persist an entity into a database using Session.persist() method.
You can learn more about Hibernate ORM Framework at Hibernate Tutorial
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