Jersey 2.x does not contain WebResource and resource class. What can I use instead?

Question

I am trying to create a web API using Jersey. I am trying to run a method similar to this:

WebResource r = c.resource("http://localhost:8080/Jersey/rest/contacts");

However Jersey 2.x does not have a WebResource or Resource class. So what class can I use instead in order to have the uri http://localhost:8080/Jersey/rest/contacts as a parameter? This will be ran in a ContactClient class

Answer 1

Have a look at the Client API from the Jersey documentation. With Jersey 2.x you instead want to use WebTarget . For example

Client client = ClientBuilder.newClient();
WebTarget target = client.target(url);
Response response = target.request().get();

See the documentation I linked to for much more information and examples.

Answer 2

: JAX-RS 2.0 introduces a new client API so that you can make http requests to your remote RESTful web services. ：JAX-RS 2.0 引入了一个新的客户端 API，以便您可以向远程RESTful Web 服务发出http 请求。

It is a 'fluent' request building API with really 3 main classes:

1. Client,
2. WebTarget, and
3. Response.

Client client = Client.create();
  WebResource webResource = client.resource(restURL).path("myresource/{param}");
  String result = webResource.pathParam("param", "value").get(String.class);

Client client = ClientFactory.newClient();
 WebTarget target = client.target(restURL).path("myresource/{param}");
 String result = target.pathParam("param", "value").get(String.class);

Client client = Client.create();
 WebResource webResource = client.resource(restURL);
 ClientResponse response = webResource.post(ClientResponse.class, "payload");

Client client = ClientFactory.newClient();
 WebTarget target = client.target(restURL);
 Response response = target.request().post(Entity.text("payload"), Response.class);