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Creating Rest Services With Rest Easy In Web Application

Posted on the 01 May 2013 by Abhishek Somani @somaniabhi
RestEasy is implementation of JAX-RS API created by JBoss. This post is about how to create and expose rest services in your existing web application. First of all , add the dependency in your pom.xml for RestEasy.If you are not using jboss[jboss AS 7.1 is shipped with rest easy] as your application server , remove Scope tag from this.
                 <dependency>
			<groupId>org.jboss.resteasy</groupId>
			<artifactId>resteasy-jaxrs</artifactId>
			<version>2.3.2.Final</version>
			<scope>provided</scope> 
		</dependency>
Now create a root rest service. By default it returns Empty set of classes in getClasses method which means whole web Application will be scanned to find jax rs annotations.
package rest;

import java.util.Set;

import javax.ws.rs.core.Application;

public class RootRestService extends Application{
	
	public Set> getClasses()
	{
		return super.getClasses();
	}

}
This is our sample Rest Class .
package rest;
import java.io.IOException;
import java.util.HashMap;
import java.util.Map;

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 javax.ws.rs.core.Response.Status;



@Path("/application")
public class RestService {

	HttpClient cl = new HttpClient();
	@Path("/test")
	@GET
	@Produces(MediaType.TEXT_PLAIN)
	public Response test() {
		return Response.status(Status.OK).entity("working").build();
	}
}



Now in your web.xml , you have to define a servlet to dispatch your request to rest service. Provide url pattern for the dispatcher. in our case it is /rest/* , accordingly you have to define the prefix in context parameter . So if the url pattern is /rest/* then prefix will also be /rest.If you map the dispatcher to /* pattern , then you will not be able to access other resources in your web app like html ,css,jsp ,servlets etc . because every request will go through the rest easy dispatcher and you will get 404 error .So ,it's better to have a separate prefix for rest services in web app.
<context-param>
		<param-name>resteasy.servlet.mapping.prefix</param-name>
		<param-value>/rest</param-value>
	</context-param>
	
	
	
	<context-param>
		<param-name>resteasy.resources</param-name>
		<param-value>rest.RestService</param-value>
	</context-param> 
 
	<servlet>
		<servlet-name>resteasy-servlet</servlet-name>
		<servlet-class>
			org.jboss.resteasy.plugins.server.servlet.HttpServletDispatcher
         </servlet-class>
		<init-param>
			<param-name>javax.ws.rs.Application</param-name>
			<param-value>rest.RootRestService</param-value>
		</init-param>
	</servlet>
You may get error like this : Could not find resource for relative : /application/test of full path: http://localhost:8080/onlinehttpclient/rest/application/test To resolve this error , you have to define resteasy.resource context param with the full path of Rest class.
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