spring jpa jsf integration

nisu pathak

Greenhorn

Posts: 29

posted 3 years ago

hello friends hope all you are fine.

i am confuse that for web application is it necessary to use maven ? if i am adding all jars that required to develop web application .that scenerio is not good? i didnt get why to use Maven? please help me and also help me for spring jsf jpa integration

i want to integrate spring jsf and jpa.

index.xhtml

<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN"
"http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml"
xmlns:f="http://java.sun.com/jsf/core"
xmlns:h="http://java.sun.com/jsf/html">

<h:head>
<title> Spring jsf jpa integration </title>
</h:head>

<h:body>

<h

utputText id="id" value="ID : "/>
<h:message for="txtid"/>
<h:inputText id="txtID" value="#{UserBeanManager.id}"/>

<h

utputText id="txtNAME" value="Name:"/>
<h:inputText id="txtName" value="#{UserBeanManager.name}"/>
<h

utputText id="txtPASSWORD" value="password:"/>
<h:inputText id="txtName" value="#{UserBeanManage.password}"/>

<h:commandButton value="#{UserBeanManager.adduser}"></h:commandButton>
</h:body>

</html>

Model class :

Dao Layer

Dao implementation

package org.integration.dao;

import java.util.List;

import javax.persistence.EntityManager;
import javax.persistence.Persistence;
import javax.persistence.PersistenceContext;
import javax.persistence.Query;

import org.integration.model.User;
import org.springframework.stereotype.Repository;
import org.springframework.transaction.annotation.Transactional;

@Repository
@Transactional
public class UserDaoImpl implements UserDao
{

@PersistenceContext(name="SpringJpajsf")
	EntityManager entitymanager=Persistence.createEntityManagerFactory("SpringJpajsf").createEntityManager();
	
	@Override  @Transactional(readOnly=true)
	public void createUser(User user)
	{

entitymanager.persist(user);
		
		
	}

@Override @Transactional(readOnly=true)
	public void deleteUser(int id) 
	{
	
		User user=entitymanager.find(User.class,1);
		
		entitymanager.getTransaction().begin();
		
	    entitymanager.remove(user);
		
	    entitymanager.getTransaction().commit();
		
	}

@Override
	public void updateUser(int id) 
	{
		User u =new User();
		entitymanager.getTransaction().begin();
		u.setId(id);
		entitymanager.getTransaction().commit();
		
	}

@Override  @Transactional(readOnly=true)
	public List<User> getUsers() 
	{
		Query q=entitymanager.createQuery("select a from User a"); 
		List<User> data=q.getResultList();
		return data;
	}

}

//Service Layer

//Service Layer implementation
package org.integration.service;

Bean manager:

package org.integration.beanmanager;

import javax.faces.bean.ManagedProperty;

import org.integration.model.User;
import org.integration.service.UserService;

public class UserBeanManager 
{
	
	private int id; 
	private String name;
	private String password;
	
	public UserBeanManager()
	{
		
	}
	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;
	}

public String getPassword() {
		return password;
	}

public void setPassword(String password) {
		this.password = password;
	}

public UserService getUserservice() {
		return userservice;
	}

public void setUserservice(UserService userservice) {
		this.userservice = userservice;
	}

@ManagedProperty(value="#{UserService}")
	private UserService userservice;
	
	//all service method have been defined 
	public String  addUser()
	{
		User u= new User();
		u.setId(1);
		u.setUsername("nisarg");
		u.setPassword("pathak");
		
		userservice.createUser(u);
		
		return "Success";
	}

}

Please help me

thanks

Consider Paul's rocket mass heater.

spring jpa jsf integration

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