How to create a Restful web service with input parameters?

Be careful. For this you need @GET (not @PUT).

If you want query parameters, you use @QueryParam.

public Todo getXML(@QueryParam("summary") String x, 
                   @QueryParam("description") String y)

But you won't be able to send a PUT from a plain web browser (today). If you type in the URL directly, it will be a GET.

Philosophically, this looks like it should be a POST, though. In REST, you typically either POST to a common resource, /todo, where that resource creates and returns a new resource, or you PUT to a specifically-identified resource, like /todo/<id>, for creation and/or update.

You can. Try something like this:

@Path("/todo/{varX}/{varY}")
@Produces({"application/xml", "application/json"})
public Todo whatEverNameYouLike(@PathParam("varX") String varX,
    @PathParam("varY") String varY) {
        Todo todo = new Todo();
        todo.setSummary(varX);
        todo.setDescription(varY);
        return todo;
}

Then call your service with this URL;
http://localhost:8088/JerseyJAXB/rest/todo/summary/description

another way to do is get the UriInfo instead of all the QueryParam

Then you will be able to get the queryParam as per needed in your code

@GET
@Path("/query")
public Response getUsers(@Context UriInfo info) {

    String param_1 = info.getQueryParameters().getFirst("param_1");
    String param_2 = info.getQueryParameters().getFirst("param_2");


    return Response ;

}