CODEWITHSUNDEEP
javaurljersey
Denys Romaniuk
Denys Romaniuk

Reputation: 139

java jersey get full URL

I need to do a proxy API service with Jersey. I need to have full request URL in jersey method. I don't want to specify all possible parameters.

For example:

@GET
@Produces({MediaType.APPLICATION_XML, MediaType.APPLICATION_JSON})
@Path("/media.json")
public  String getMedia( ){
    // here I want to get the full request URL like /media.json?param1=value1&param2=value2
}

How can I do it?

Upvotes: 11

Views: 21124

Answers (4)

Alexmelyon
Alexmelyon

Reputation: 1239

In Jersey 2.x (note that it uses HttpServletRequest object):

@GET
@Path("/test")
public Response test(@Context HttpServletRequest request) {
    String url = request.getRequestURL().toString();
    String query = request.getQueryString();
    String reqString = url + "?" + query;
    return Response.status(Status.OK).entity(reqString).build();
}

Upvotes: 19

Buddhika Alwis
Buddhika Alwis

Reputation: 371

try UriInfo as follow ,

    @POST
    @Consumes({ MediaType.APPLICATION_JSON})
    @Produces({ MediaType.APPLICATION_JSON})
    @Path("add")
    public Response addSuggestionAndFeedback(@Context UriInfo uriInfo, Student student) {

            System.out.println(uriInfo.getAbsolutePath());

      .........
    }

OUT PUT:- https://localhost:9091/api/suggestionandfeedback/add

you can try following options also

enter image description here

Upvotes: 7

Jack Daniel
Jack Daniel

Reputation: 2467

you can use Jersey filters.

public class HTTPFilter implements ContainerRequestFilter {

private static final Logger logger = LoggerFactory.getLogger(HTTPFilter.class);

    @Override
    public void filter(ContainerRequestContext containerRequestContext) throws IOException {

        logger.info(containerRequestContext.getUriInfo().getPath() + " endpoint called...");
        //logger.info(containerRequestContext.getUriInfo().getAbsolutePath() + " endpoint called...");

    }
}

After that u must register it in http configuration file or just extend ResourceConfig class. This is how u can register it in http config class

public class HTTPServer {

    public static final Logger logger = LoggerFactory.getLogger(HTTPServer.class);

    public static void init() {

        URI baseUri = UriBuilder.fromUri("http://localhost/").port(9191).build();
        ResourceConfig config = new ResourceConfig(Endpoints.class, HTTPFilter.class);
        HttpServer server = JdkHttpServerFactory.createHttpServer(baseUri, config);

        logger.info("HTTP Server started");

    }

}

Upvotes: 0

Azee
Azee

Reputation: 1829

If you need a smart proxy, you can get parameters, filter them and create a new url.

@GET
@Produces({MediaType.APPLICATION_XML, MediaType.APPLICATION_JSON})
@Path("/media.json")
public  String getMedia(@Context HttpServletRequest hsr){
    Enumeration parameters = hsr.getParameterNames();
    while (parameters.hasMoreElements()) {
        String key = (String) parameters.nextElement();
        String value = hsr.getParameter(key);
    //Here you can add values to a new string: key + "=" + value + "&"; 
    }

}

Upvotes: 4

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