JAX-RS Jersey Rest service with JSON + Maven + Tomcat

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JAX-RS jerseyJAX-RS Jersey Rest API with JSON is one of the way of exposing web services API. I demonstrated with Maven as a build tool and Tomcat as a web server for this JAX-RS Jersey example.

1. Create Maven project for JAX-RS Jersey Rest

I used Maven as the build tool in this web project deployed in the Tomcat web server for this JAX-RS tutorial series of Jersey Rest API.

1. File -> New -> Others -> Maven ->Maven Project -> click Next -> choose maven-archetype-webapp
2. In Eclipse IDE create “Dynamic Web Project” and select Project -> write click on project -> select Configure -> Convert Maven Project.

2. Environment used as:

  • JDK 7
  • Eclipse Luna 4.4.1
  • Tomcat 7
  • Maven 3.2.3
  • Jersey 1.9

3. Maven pom.xml

Add dependency of jersey server specific jar of your suitable version. Maven pull required dependency from specified repository.

  Rest Jax-rs jersey Maven Webapp



4. web.xml , Deployment descriptor entry

We need to add ServletContainer entry in our Deployment structure i.e. web.xml here. And package we need to specify in init-param , where our services class resides. When web server tomcat start, the instance of ServletContainer loads into servlet container.



5. PersonJsonService.java for JAX-RS Jersey

package com.mysoftkey.jaxrs;

import javax.ws.rs.Consumes;
import javax.ws.rs.GET;
import javax.ws.rs.POST;
import javax.ws.rs.Path;
import javax.ws.rs.Produces;
import javax.ws.rs.core.MediaType;
import javax.ws.rs.core.Response;

import com.mysoftkey.jaxrs.model.Person;

 * This Service class demonstrate input as JSON payload to submit the request
 * and return a json output.
 * @author Ranjeet Jha
public class PersonJsonService {

  * This method is used to get Person object in JSON as it Produces application/json media type.
  * URL: http://localhost:8080/jersey/person/get
  * @return
 public Response getPersonJSONHandler() {
  	Person person = new Person();
	person.setName("ranjeet Jha");
	//return person;
	return Response.status(200).entity(person).build();

  * This method is used to add the person as payload JSON with the endpoint. 
  * URL: http://localhost:8080/jersey/person/add
  * jsonPayload: {"id":1, "name":"Ranjeet Jha"}
  * @param response
  * @return
 public Response addPersonHandler(Person person) {
	String result = "Person added successfully : " + person.getName();
	// write code to add person into db or in-memory.
	return Response.status(201).entity(result).build();

6. Model class Person.java

package com.mysoftkey.jaxrs.model;

import java.io.Serializable;

 * this is model class to encapsulate property of person.
 * @author Ranjeet Jha
public class Person implements Serializable {
	private int id;
	private String 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;

7. Run application

Add project into Tomcat configured within Eclipse , run Tomcat and open browser and type url as : http://localhost:8080/jersey/person/get
output on browser as :

name: "ranjeet Jha",
id: 30

For post JSON payload, open Advance Rest client or any other client and enter the following:

Service Enpoint: http://localhost:8080/jersey/person/add
Http method: POST
Content-Type: application/json

 {"id":1,="" "name":"ranjeet="" jha"}

Click submit and see output as follows with http status ok:

person added successfully : Ranjeet Jha

8. Download Source code

Download project by click here

9. References for JAX-RS Jersey tutorials

Jersey Official site
JAX-RS java
Oracle Java JEE

Happy learning for JAX-RS Jersey for XML example.

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