How to create a Restful web service with input parameters?

Question

I am creating restful web service and i wanted to know how do we create a service with input parameters and also how to invoke it from a web browser.

For example

Answer 1

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.

Answer 2

You can try this... put parameters as :
http://localhost:8080/WebApplication11/webresources/generic/getText?arg1=hello in your browser...

package newpackage;

import javax.ws.rs.core.Context;
import javax.ws.rs.core.UriInfo;
import javax.ws.rs.PathParam;
import javax.ws.rs.Produces;
import javax.ws.rs.Consumes;
import javax.ws.rs.DefaultValue;


import javax.ws.rs.GET;
import javax.ws.rs.Path;
import javax.ws.rs.PUT;
import javax.ws.rs.QueryParam;

@Path("generic")
public class GenericResource {

    @Context
    private UriInfo context;

    /**
     * Creates a new instance of GenericResource
     */
    public GenericResource() {
    }

    /**
     * Retrieves representation of an instance of newpackage.GenericResource

     * @return an instance of java.lang.String
     */
    @GET
    @Produces("text/plain")
    @Consumes("text/plain")
    @Path("getText/")
    public String getText(@QueryParam("arg1")
            @DefaultValue("") String arg1) {

       return  arg1 ;  }

    @PUT
    @Consumes("text/plain")
    public void putText(String content) {





    }
}

Answer 3

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

Answer 4

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

Answer 5

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 ;

}