Can anybody tell me how to make a restful web service with Jersey 2.0 by not using maven. I have searched everywhere and found tutorial for Jersey1.x versions but not for 2.0. Please help
We provide detail answere based on the user answer user2629427. we checked this on windows 7.
Requirement: (brackets indicate version which this example is tested)
tomcat (8 zip version)
jersey (2.x)
Unzip the tomcat & create a below folder structure in tomcat's 'webapps' folder (folder names are case sensitive).
abc
|___ WEB-INF
|____ classes
|____ lib
Put 'Hello.java' and 'MyApplication.java' into 'classes' folder and 'web.xml' into 'WEB-INF' folder.
web.xml
<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns="http://xmlns.jcp.org/xml/ns/javaee"
xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/javaee http://xmlns.jcp.org/xml/ns/javaee/web-app_3_1.xsd"
id="WebApp_ID"
version="3.1">
<servlet>
<servlet-name>rest</servlet-name>
<servlet-class>org.glassfish.jersey.servlet.ServletContainer</servlet-class>
<init-param>
<param-name>javax.ws.rs.Application</param-name>
<param-value>com.king.MyApplication</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>rest</servlet-name>
<url-pattern>/rest/*</url-pattern>
</servlet-mapping>
</web-app>
Myapplication.java
package com.king;
import org.glassfish.jersey.server.ResourceConfig;
public class MyApplication extends ResourceConfig {
public MyApplication() {
packages("com.king");
}
}
Hello.java
package com.king;
import javax.ws.rs.GET;
import javax.ws.rs.Path;
import javax.ws.rs.Produces;
import javax.ws.rs.core.MediaType;
#Path("/hello")
public class Hello {
#GET
#Produces(MediaType.TEXT_PLAIN)
public String sayPlainTextHello() {
return "Hello Jersey";
}
// This method is called if XML is request
#GET
#Produces(MediaType.TEXT_XML)
public String sayXMLHello() {
return "<?xml version=\"1.0\"?><hello>Hello Jersey</hello>";
}
// This method is called if HTML is request
#GET
#Produces(MediaType.TEXT_HTML)
public String sayHtmlHello() {
return "<html><title>Hi Jersey</title><body><h1>Hello Jersey this is laksys</body></h1></html>";
}
}
Unzip jersey and copy all jar files from api, ext, and lib (not folders) into your apps 'lib' folder.
Now compile the two java files using following command
D:\apache-tc-8\webapps\abc\WEB-INF\classes>javac -d . -cp ..\lib\javax.ws.rs-api-2.0.1.jar;..\lib\jersey-server.jar;..\l ib\jersey-common.jar *.java
Next run the tomcat server
D:\apache-tc-8\bin>startup
In browser address bar type this: http://localhost:8080/abc/rest/hello
I found the answer
package com.hellowebservice;
import javax.ws.rs.GET;
import javax.ws.rs.Path;
import javax.ws.rs.Produces;
import javax.ws.rs.core.MediaType;
#Path("/hello")
public class Hello {
#GET
#Produces(MediaType.TEXT_PLAIN)
public String sayPlainTextHello() {
return "Hello Jersey";
}
// This method is called if XML is request
#GET
#Produces(MediaType.TEXT_XML)
public String sayXMLHello() {
return "<?xml version=\"1.0\"?>" + "<hello> Hello Jersey" + "</hello>";
}
// This method is called if HTML is request
#GET
#Produces(MediaType.TEXT_HTML)
public String sayHtmlHello() {
return "<html> " + "<title>" + "Hello Jersey" + "</title>"
+ "<body><h1>" + "Hello Jersey" + "</body></h1>" + "</html> ";
}
}
Web.xml
<?xml version="1.0" encoding="UTF-8"?>
<web-app id="WebApp_ID" version="2.5"
xmlns="http://java.sun.com/xml/ns/javaee"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd">
<display-name>FirstRestWebService</display-name>
<servlet>
<display-name>Rest Servlet</display-name>
<servlet-name>RestServlet</servlet-name>
<servlet-class>org.glassfish.jersey.servlet.ServletContainer</servlet-class>
<init-param>
<param-name>javax.ws.rs.Application</param-name>
<param-value>com.hellowebservice.MyApplication</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>RestServlet</servlet-name>
<url-pattern>/rest/*</url-pattern>
</servlet-mapping>
</web-app>
MyApplication.java
package com.hellowebservice;
import org.glassfish.jersey.server.ResourceConfig;
public class MyApplication extends ResourceConfig {
public MyApplication() {
packages("com.hellowebservice");
}
}
run with localhost:8080/FirstRestWebService/rest/hello
Just to add to the previous answer. If you aren't using Maven and just building using Eclipse with a Dynamic Web Project and deploying to web app server like Tomcat.
Just download the Jersey JAX-RS 2.0 RI bundle Jersey Downloads, unzip and add all the jars in the lib, api and ext folders to your build path. (I tried without ext jars but got classnotfound when starting the server).
Also add all the jars to the Deployment Assembly of your Dynamic Web Project so they get automatically copied to the WEB-INF/lib directory when deployed to your web app server. Along with the code & web.xml in the above answer, you should have a RESTful api using Jersey 2 up and running.
Related
I'm trying implement a web service for two endpoints and getting this error
"404 Not Found: Invalid Request" when tried accessing the service after deploying onto the apache toncat 8.
Below are my web service implementation classes, sun-jaxws.xml and web.xml
WebImplementation1.java
package com.ws.soap.services;
import javax.jws.WebService;
#WebService(endpointInterface = "com.ws.soap.services.WebServiceImpl1")
public class WebServiceImpl1 {
public String printMessage() {
return "Hello from WebServiceImpl1 ";
}
}
WebServiceImplementation2.java
package com.ws.soap.services;
import javax.jws.WebService;
#WebService(endpointInterface = "com.ws.soap.services.WebServiceImpl2")
public class WebServiceImpl2 {
public String displayMessage() {
return "Hello from WebServiceImpl2 ";
}
}
sun-jaxws.xml
<?xml version="1.0" encoding="UTF-8"?>
<endpoints xmlns="http://java.sun.com/xml/ns/jax-ws/ri/runtime"
version="2.0">
<endpoint name="WebServiceImpl1" implementation="com.ws.soap.services.WebServiceImpl1"
url-pattern="/impl1" />
<endpoint name="WebServiceImpl2" implementation="com.ws.soap.services.WebServiceImpl2"
url-pattern="/impl2" />
</endpoints>
web.xml
<?xml version="1.0" encoding="UTF-8"?>
<web-app version="2.5" xmlns="http://java.sun.com/xml/ns/javaee"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/javaee
http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd">
<display-name>JAX-WS-Tomcat</display-name>
<listener>
<listener-class>
com.sun.xml.ws.transport.http.servlet.WSServletContextListener
</listener-class>
</listener>
<servlet>
<servlet-name>sayhello</servlet-name>
<servlet-class>com.sun.xml.ws.transport.http.servlet.WSServlet</servlet-class>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>sayhello</servlet-name>
<url-pattern>/impl1</url-pattern>
<url-pattern>/impl2</url-pattern>
</servlet-mapping>
<session-config>
<session-timeout>30</session-timeout>
</session-config>
</web-app>
Using the exact code supplied (plus the JAX-WS RI jars downloaded from https://jax-ws.java.net/), I was able to create a webapp and successfully access the service endpoints /impl1 and /impl2. Be advised the <url-pattern> and <endpoint ... url-pattern="/impl1"> directives state the resource path to the JAX-WS endpoints within the context path of the enclosing web application.
So, if the name of the webapp is MyWebServices (MyWebServices.war with no other files/code than described in the post, deployed to Tomcat 8) and you have <url-pattern>/impl1</url-pattern> in web.xml, and with a default Tomcat instance listening on port 8080, your web service endpoint would be http://localhost:800/MyWebServices/impl1 with the WSDL available via http://localhost:800/MyWebServices/impl1?wsdl
If you want to customize your context path of your webapp (e.g. you don't want /MyWebServices/... you can use the techniques described in this SO question.
For example, my local Tomcat 8 is running on port 8081:
I am getting FormDataContentDisposition object as NULL while implementing jersey file upload example as mentioned in below URL :
http://www.mkyong.com/webservices/jax-rs/file-upload-example-in-jersey/
Here is my sample code,
IMPORTS
import java.io.InputStream;
import javax.ws.rs.Consumes;
import javax.ws.rs.GET;
import javax.ws.rs.POST;
import javax.ws.rs.Path;
import javax.ws.rs.PathParam;
import javax.ws.rs.Produces;
import javax.ws.rs.core.Context;
import javax.ws.rs.core.MediaType;
import javax.ws.rs.core.Response;
import com.sun.jersey.core.header.FormDataContentDisposition;
import com.sun.jersey.multipart.FormDataParam;
Rest API code,
#POST
#Path("/uploadFiles")
#Produces({ MediaType.APPLICATION_JSON})
#Consumes({ MediaType.MULTIPART_FORM_DATA })
public Response uploadFiles(#FormDataParam("file") InputStream uploadedInputStream,
#FormDataParam("file") FormDataContentDisposition fileDetail)
throws IOException{
I am getting fileDetail object null in above code.
JAR versions used,
<dependency>
<groupId>com.sun.jersey</groupId>
<artifactId>jersey-core</artifactId>
<version>1.11</version>
</dependency>
<dependency>
<groupId>com.sun.jersey.contribs</groupId>
<artifactId>jersey-multipart</artifactId>
<version>1.19</version>
</dependency>
I am hitting my request from Chrome POSTMAN Client,
Here is the snapshot of my request from POSTMAN client...
I already gone through the link : File upload with Jersey : FormDataContentDisposition is null having same problem but the difference is that I am using REST Client instead of HTML/JSP to submit my request.
Can any one please help me to know why I am getting NULL object of FormDataContentDisposition,
Thanks in advance!
Edit:
I found the difference between example and my application implementation is that they have below configuration in web.xml
<servlet>
<servlet-name>jersey-serlvet</servlet-name>
<servlet-class>com.sun.jersey.spi.container.servlet.ServletContainer</servlet-class>
<init-param>
<param-name>com.sun.jersey.config.property.packages</param-name>
<param-value>com.mkyong.rest</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>jersey-serlvet</servlet-name>
<url-pattern>/rest/*</url-pattern>
</servlet-mapping>
and My Application configuration is,
#javax.ws.rs.ApplicationPath("/rest")
#javax.ws.rs.Path("application")
public class RestServices extends Application {
#Override
public Set<Class<?>> getClasses() {
HashSet<Class<?>> classes = new HashSet<Class<?>>();
classes.add(this.getClass());
classes.add(RestAPI.class);
return classes;
}
Can anybody tell me what is the exact difference between these two
configurations. Can Above both implementation will changes the way
things working?
Remove #javax.ws.rs.Path("application").
You don't need two kinds of Deployment Descriptor, use either Programmatic approach or web.xml;
My approach:
empty web.xml
<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns="http://java.sun.com/xml/ns/javaee"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/javaee
http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd"
version="3.0">
</web-app>
and config class:
#ApplicationPath("/rootpath")
public class YourApplication extends ResourceConfig {
// Constructor
public YourApplication() {
super (
//YOUR JAX-RS Service Classes
MyJaxRSService.class,
// This one corresponds to jersey-media-json-jackson in dependencies
JacksonFeature.class,
// This is one to : jersey-media-multipart
MultiPartFeature.class
);
}
}
update dependencies
<dependency>
<groupId>org.glassfish.jersey.containers</groupId>
<artifactId>jersey-container-servlet</artifactId>
<version>2.22.1</version>
</dependency>
<dependency>
<groupId>org.glassfish.jersey.media</groupId>
<artifactId>jersey-media-json-jackson</artifactId>
<version>2.22.1</version>
</dependency>
<dependency>
<groupId>org.glassfish.jersey.media</groupId>
<artifactId>jersey-media-multipart</artifactId>
<version>2.22.1</version>
</dependency>
Last and most important: forget such tutorials, it helps you start instantly but when you encounter problem you're stuck, and you'll just waste your time. Instead read official documentation and their real world examples, get some real insight into technology how and why it works.
Jersey Documentation: https://jersey.java.net/documentation/latest/index.html
In addition: I'd recommend this book by Bill Burke - RESTful Java with JAX-RS 2.0.
Good luck.
Though I am using Jersey 2.22.1, my web.xml has provided configuration for multipart data.
web.xml
<web-app version="2.5" xmlns="http://java.sun.com/xml/ns/javaee"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd">
<servlet>
<servlet-name>Jersey Web Application</servlet-name>
<servlet-class>org.glassfish.jersey.servlet.ServletContainer</servlet-class>
<init-param>
<param-name>jersey.config.server.provider.packages</param-name>
<param-value>com.xyz.rest.controller;com.xyz.rest.filters</param-value>
</init-param>
<init-param>
<param-name>com.sun.jersey.api.json.POJOMappingFeature</param-name>
<param-value>true</param-value>
</init-param>
<init-param>
<param-name>jersey.config.server.provider.classnames</param-name>
<param-value>org.glassfish.jersey.media.multipart.MultiPartFeature</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>Jersey Web Application</servlet-name>
<url-pattern>/api/*</url-pattern>
</servlet-mapping>
</web-app>
Reply back if still problem persist
I am novice to writing REST WebServices.
Currently I am trying to write a RESTful service using jersey-2.x and tomcat 8.0
However, when I try to deploy in eclipse, it gives me error as follows :
java.lang.NoSuchMethodError: javax.ws.rs.core.Application.getProperties()Ljava/util/Map;
What I did is :
wrote below classes :
#ApplicationPath("resources")
public class RestTestApplication extends Application
{
#Override
public Set<Class<?>> getClasses() {
final Set<Class<?>> classes = new HashSet<Class<?>>();
// register root resource
classes.add(HelloResource.class);
return classes;
}
}
#Path("sayhello")
public class HelloResource
{
#GET
#Produces("text/plain")
public String sayhello ()
{
return "Hi, How are you !!";
}
}
downloaded from http://repo1.maven.org/maven2/org/glassfish/jersey/bundles/jaxrs-ri/2.12/jaxrs-ri-2.12.zip
downloaded jsr311-api-1.1.2.r612.jar
copied all *.jar files from jaxrs-ri-2.12.zip and jsr311-api-1.1.2.r612.jar to WEB-INF/lib and also imported to build path.
Edited web.xml as below :
<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="http://xmlns.jcp.org/xml/ns/javaee" xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/javaee http://xmlns.jcp.org/xml/ns/javaee/web-app_3_1.xsd" id="WebApp_ID" version="3.1">
<display-name>RestWS</display-name>
<welcome-file-list>
<welcome-file>index.html</welcome-file>
<welcome-file>index.htm</welcome-file>
<welcome-file>index.jsp</welcome-file>
<welcome-file>default.html</welcome-file>
<welcome-file>default.htm</welcome-file>
<welcome-file>default.jsp</welcome-file>
</welcome-file-list>
<servlet>
<servlet-name>Rest Test</servlet-name>
<servlet-class>org.glassfish.jersey.servlet.ServletContainer</servlet-class>
<init-param>
<param-name>jersey.config.server.provider.packages</param-name>
<param-value>com.example</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>Rest Test</servlet-name>
<url-pattern>/resources/*</url-pattern>
</servlet-mapping>
</web-app>`
run as-->run on server
Getting the error mentioned above.
please let me know what I am doing wrong.
I think its because you have both jars from jaxrs-ri-2.12.zip and jsr311-api-1.1.2.r612.jar in your classpath. jsr311-api-1.1.2.r612.jar has the older implementation of JAX-RS API. Your Application class that your RestTestApplication extends from is from the jsr311-api-1.1.2.r612.jar; however at runtime the Application class from your jaxrs jar in jaxrs-ri-2.12.zip is being referred to. Removing the jsr311 jar from your WEBINF/lib should hopefully resolve the issue.
If you decompile the Application class from both the jars you will notice that the one in jsr311 jar doesn't have getProperties method and hence the java.lang.NoSuchMethodError error.
I have created a simple Spring MVC web application and trying to expose the services as SOAP based JAX-WS services using JAX-WS commons RI implementation.
After deploying my application on Tomcat 7, when I try accessing my web service, I get a message as 404 Not Found: Invalid Request. Below are my configurations, kindly help in resolving this.
web.xml
<web-app version="2.5" xmlns="http://java.sun.com/xml/ns/javaee"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd">
<!-- The definition of the Root Spring Container shared by all Servlets and Filters -->
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/spring/root-context.xml</param-value>
</context-param>
<!-- Creates the Spring Container shared by all Servlets and Filters -->
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
<!-- Processes SOAP Web Service requests -->
<servlet>
<servlet-name>jaxws-servlet</servlet-name>
<servlet-class>com.sun.xml.ws.transport.http.servlet.WSSpringServlet</servlet-class>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>jaxws-servlet</servlet-name>
<url-pattern>/ws/*</url-pattern>
</servlet-mapping>
<!-- Processes application requests -->
<servlet>
<servlet-name>appServlet</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/spring/appServlet/servlet-context.xml</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>appServlet</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>
</web-app>
WEB-INF/spring/appServlet/servlet-context.xml
<beans:bean id="customerService" class="com.home.service.CustomerService" />
<!-- Web Service definition -->
<beans:bean id="customerWS" class="com.home.ws.CustomerWS">
<beans:property name="customerService" ref="customerService" />
</beans:bean>
<wss:binding url="/ws/CustomerServ">
<wss:service>
<ws:service bean="#customerWS" />
</wss:service>
</wss:binding>
CustomerWS.java
#WebService
#SOAPBinding(style = Style.DOCUMENT, use = Use.LITERAL)
public class CustomerWS {
private CustomerService customerService;
#WebMethod
public Customer read(long id) {
return customerService.read(id);
}
public void setCustomerService(CustomerService customerService) {
this.customerService = customerService;
}
}
CustomerService.java
#Service
public class CustomerService {
public Customer read(long id) {
Customer cust = null;
System.out.println("CustomerService.read invoked");
return cust;
}
}
pom.xml - included the dependency of jaxws-spring
<dependency>
<groupId>org.jvnet.jax-ws-commons.spring</groupId>
<artifactId>jaxws-spring</artifactId>
<version>1.9</version>
</dependency>
There are no errors while building or deploying the application. When I access the URL, still I see no errors in the server log files. However, the browser displays the message - 404 Not Found: Invalid Request
URL I am trying is - http://localhost:8080/crrs/ws/CustomerServ?wsdl
If I access my HomeController, it works fine. Home page is loaded as expected.
Appreciate any help. Thanks in advance.
I'm trying to do the same thing. My code is almost the same, I'm just using #Name and #Inject instead of #Service.
Just added extends SpringBeanAutowiringSupport to the #WebService class and it's working
The servlet mappings in the web.xml seem to be the cause.
<servlet-mapping>
<servlet-name>jaxws-servlet</servlet-name>
<url-pattern>/ws/*</url-pattern>
</servlet-mapping>
[...]
<servlet-mapping>
<servlet-name>appServlet</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>
The spring DispatcherServlet named appServlet takes care of all urls after http://localhost:8080/crrs, even http://localhost:8080/crrs/ws/CustomerServ?wsdl.
The WSSpringServlet url-pattern cannot be reached.
So I started to read this tutorial about how to develop a Restful service with Jersey. I want to develop Rest service that sends the data from a MySQL database to an Android client. I read and followed the steps on the tutorial and made my own resource classes, but when I tried to run the service on Apache I got the following error here.
I'm just starting to experiment with web services and Rest, I have read the information related to the subject from that IBM site and I thought I got the hang of it, but I'm really lost as to why is not working.
My web.xml is as follows
<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="http://java.sun.com/xml/ns/javaee" xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" id="WebApp_ID" version="2.5">
<display-name>TesterRest</display-name>
<welcome-file-list>
<welcome-file>index.html</welcome-file>
</welcome-file-list>
<servlet>
<servlet-name>Jersey REST Service</servlet-name>
<servlet-class>com.sun.jersey.spi.container.servlet.ServletContainer</servlet-class>
<init-param>
<param-name>com.sun.jersey.config.property.packages</param-name>
<param-value>com.mx.ipn.escom.testerRest.resources</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>Jersey REST Service</servlet-name>
<url-pattern>/rest/*</url-pattern>
</servlet-mapping>
</web-app>
This is my resource class:
package com.mx.ipn.escom.testerRest.resources;
import java.sql.ResultSet;
import java.sql.SQLException;
import java.util.ArrayList;
import java.util.List;
import javax.ws.rs.GET;
import javax.ws.rs.Path;
import javax.ws.rs.Produces;
import javax.ws.rs.core.Context;
import javax.ws.rs.core.MediaType;
import javax.ws.rs.core.Request;
import javax.ws.rs.core.UriInfo;
import javax.xml.bind.annotation.XmlRootElement;
import com.mx.ipn.escom.testerRest.dao.TemaDao;
import com.mx.ipn.escom.testerRest.db.Connector;
import com.mx.ipn.escom.testerRest.modelo.Tema;
#XmlRootElement
#Path("/temas")
public class TemaResource {
#Context
UriInfo uriInfo;
#Context
Request request;
#GET
#Produces({MediaType.APPLICATION_JSON, MediaType.APPLICATION_XML})
public List<Tema> getTemas() throws SQLException{
TemaDao temaDao = new TemaDao();
List<Tema> temas=temaDao.getTemas();
temaDao.terminarSesion();
return temas;
}
}
My class for database connection works fine, so that one isn't the problem.
I'm using Eclipse 3.6 to develop and Apache Tomcat 6.
I'm completely new to JAXB, so if anyone can give me guidelines to what kind of annotations am I missing I would appreciate it.
Based on your screenshot I think you should update your web.xml to have the correct package name:
<servlet>
<servlet-name>Jersey REST Service</servlet-name>
<servlet-class>com.sun.jersey.spi.container.servlet.ServletContainer</servlet-class>
<init-param>
<param-name>com.sun.jersey.config.property.packages</param-name>
<param-value>com.mx.ipn.escom.testerRest.resources</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
The mx is missing from your configuration.