How to implement jaxrs Application without web.xml [duplicate]

Question

I am trying to deploy a really simple jaxrs application with without a web.xml config and cannot get it working. My URL I'd expect to access is serverandport/{appname}/rest/welcomes/hello and I think I must be missing something dead obvious.

Application

@ApplicationPath("/rest")
public class EngineApp extends Application {

    @Override
    public Set<Class<?>> getClasses() {
        Set<Class<?>> s = new HashSet<Class<?>>();
        s.add(RestTestImpl.class);
        return s;
    }

}

Resource

@Path("/welcomes")
public class RestTestImpl {
    @GET
    @Path("hello")
    public String sayPlainHello() {
        return "Hi from Rest";
    }
}

POM snippet

<groupId>org.apache.maven.plugins</groupId>
    <artifactId>maven-war-plugin</artifactId>
    <version>2.1.1</version>
    <configuration>
        <failOnMissingWebXml>false</failOnMissingWebXml>
    </configuration>

Edit: Further to the response below, I tried with an empty web.xml and also with the following web.xml. Bother also return 404, however the xml below states "Servlet javax.ws.rs.core.Application is not available":

web.xml

<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    version="3.0"
    xmlns="http://java.sun.com/xml/ns/javaee"
    xsi:schemaLocation="http://java.sun.com/xml/ns/javaee
      http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" >
    <servlet> 
        <servlet-name>javax.ws.rs.core.Application</servlet-name> 
        <load-on-startup>1</load-on-startup> 
    </servlet> 
    <servlet-mapping>
        <servlet-name>javax.ws.rs.core.Application</servlet-name>
        <url-pattern>/*</url-pattern>
    </servlet-mapping>

</web-app>

Answer 1

You need to deploy your application into a Servlet 3.0 compliant container to take advantage of this functionality. Try GlassFish 3.x or Tomcat 7.