First Hibernate Application

In this tutorial you will learn about how to create an application of Hibernate 4.

Here I am giving an example which will demonstrate you how can you make your application using Hibernate 4. To create an application I will use Eclipse IDE ( to see the basic requirement before creating an application with Hibernate 4 click here ).

Example :

Following steps that I have followed to create an application. These are as follows :

Step 1 : At first I have created a table named person in MySQL.

CREATE TABLE `person` ( 
`Id` int(10) NOT NULL, 
`Name` varchar(15) default NULL, 
PRIMARY KEY (`Id`) 
) ENGINE=InnoDB DEFAULT CHARSET=latin1 

Step 2 : Created a Java Project named coreHibernateExample (you can give the name as you wish).

File -> New -> project / Java Project -> giveProjectName -> Finish (in this example I have given coreHibernateExample).

Step 3 : Add the Hibernate jar files (to see how to add Hibernate's jar files click here )

Step 4 : Created a package named roseindia (you can give the name as you wish) under the src folder.

Select src -> Right Click ->New -> package -> givePackageName -> Finish (in this exmaple I have given roseindia).

Ste 5 : Created a file by following the name convention hibernate.cfg.xml (mandatory ) file under the src folder that the hibernate utilize to create a connection pool and the required environment setup.

Select src -> Right Click -> New -> File ->hibernate.cfg.xml -> Finish.

<?xml version='1.0' encoding='utf-8'?>
<!DOCTYPE hibernate-configuration PUBLIC
"-//Hibernate/Hibernate Configuration DTD//EN"
"http://www.hibernate.org/dtd/hibernate-configuration-3.0.dtd">

<hibernate-configuration>
<session-factory>
<property name="hibernate.connection.driver_class">com.mysql.jdbc.Driver</property>
<property name="hibernate.connection.url">jdbc:mysql://192.168.10.13:3306/data
</property>
<property name="hibernate.connection.username">root</property>
<property name="hibernate.connection.password">root</property>
<property name="hibernate.connection.pool_size">10</property>
<property name="show_sql">true</property>
<property name="dialect">org.hibernate.dialect.MySQLDialect</property>
<property name="hibernate.current_session_context_class">thread</property>

</session-factory>
</hibernate-configuration>

Step 6 : Created a POJO class (persistent class) named Person.java in the package roseindia.

package roseindia;

public class Person
{
int id;
String name;
public Person()
{ 
}
public Person(int id, String name) {
super();
this.id = id;
this.name = name;
}
public int getId() {
return id;
}
public void setId(int id) {
this.id = id;
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
}

Step 7 : Crated a file named ( you may follow the naming convention ClassName.hbm.xml )  person.hbm.xml under the src folder to map a Person Object to the database table named person

<?xml version='1.0'?>
<!DOCTYPE hibernate-mapping PUBLIC
"-//Hibernate/Hibernate Mapping DTD 3.0//EN"
"http://www.hibernate.org/dtd/hibernate-mapping-3.0.dtd">

<hibernate-mapping package="roseindia">
<class name="Person" table="person">
<id name="id" type="int" column="Id" >
<generator class="assigned"/>
</id>

<property name="name">
<column name="Name" />
</property>

</class>
</hibernate-mapping>

Step 8 : Developed a code by writing a java class named PersonDetail in the package roseindia which will persist the Person object in the person table.

package roseindia;

import org.hibernate.Session;
import org.hibernate.SessionFactory;
import org.hibernate.Transaction;
import org.hibernate.cfg.Configuration;
import org.hibernate.service.ServiceRegistry;
import org.hibernate.service.ServiceRegistryBuilder;

public class PersonDetail
{
private static SessionFactory sessionFactory;
private static ServiceRegistry serviceRegistry;

public static void main(String[] args)
{
Session session = null;
try 
{
try
{
Configuration cfg = new Configuration().addResource(
"person.hbm.xml").configure();
serviceRegistry = new ServiceRegistryBuilder().applySettings(
cfg.getProperties()).buildServiceRegistry();
sessionFactory = cfg.buildSessionFactory(serviceRegistry);
}
catch (Throwable ex)
{
System.err.println("Failed to create sessionFactory object."+ ex);
throw new ExceptionInInitializerError(ex);
}
session = sessionFactory.openSession();

Person person = new Person();
System.out.println("Inserting Record");
Transaction tx = session.beginTransaction();
person.setId(1);
person.setName("Roseindia");
session.save(person);
tx.commit();
System.out.println("Done");
} catch (Exception e) {
System.out.println(e.getMessage());
} finally {
session.close();
}
}
}

Execute the example : You can execute the application as follows

Go to your PersonDetail class Right Click -> Run As -> Java Application or,

Select from the menu bar tab Run -> Run or,

press the Ctrl button along with f11 i.e. CTRL + f11.

Output :

When you will run the application output will be as follows :

1. On console of Eclipse the output will be displayed as :

2. After Successfully execution of the code the record will be saved to the table that you had created in MySQL. as :

Download Source Code

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Comments

Pedro
July 13, 2012
Muchas gracias!!

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Very useful. Thanks to you!!!

Sanjay Sreedhar
July 18, 2012
SQL Exception

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Hi, After following the steps as in the tutorial, I am getting exceptions as shown in the below log Inserting Record Jul 18, 2012 11:31:52 PM org.hibernate.engine.jdbc.spi.SqlExceptionHelper logExceptions WARN: SQL Error: 0, SQLState: 08S01 Jul 18, 2012 11:31:52 PM org.hibernate.engine.jdbc.spi.SqlExceptionHelper logExceptions ERROR: Communications link failure Last packet sent to the server was 0 ms ago. Could not open connection Can someone please assist me with this?

Vicky
September 26, 2012
mapping resource

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if mapping file in other package than src, i am not able to load it in cfg.xml using <mapping resource="roseindia/person.hbm.xml"/>