Uri not Absolute exception getting while calling Restful Webservice

Suresh Atta picture Suresh Atta · Feb 13, 2013 · Viewed 138.5k times · Source

The below code snippet is using to call my web service using restful API.

ClientConfig config = new DefaultClientConfig();
    Client client = Client.create(config);
    String uri= "https://127.0.0.1:8443/cas-server-webapp-3.5.0/login";
    WebResource resource = client.resource(URLEncoder.encode(uri));
      MultivaluedMap<String, String> queryParams = new MultivaluedMapImpl();
       queryParams.add("username", "suresh");
       queryParams.add("password", "suresh");
       resource.queryParams(queryParams); 
       ClientResponse response = resource.type(
            "application/x-www-form-urlencoded").get(ClientResponse.class);
    String en = response.getEntity(String.class);
    System.out.println(en); 

And getting this exception while running the above code

com.sun.jersey.api.client.ClientHandlerException: java.lang.IllegalArgumentException: URI is not absolute

    at com.sun.jersey.client.urlconnection.URLConnectionClientHandler.handle(URLConnectionClientHandler.java:151)
    at com.sun.jersey.api.client.Client.handle(Client.java:648)
    at com.sun.jersey.api.client.WebResource.handle(WebResource.java:680)

I googled many articles and did'nt get where i am doing wrong .

Side note :cas-server-webapp-3.5.0 war deployed on my machine in Apache tomacat7

Answer

Bob Flannigon picture Bob Flannigon · Feb 13, 2013

An absolute URI specifies a scheme; a URI that is not absolute is said to be relative.

http://docs.oracle.com/javase/8/docs/api/java/net/URI.html

So, perhaps your URLEncoder isn't working as you're expecting (the https bit)?

    URLEncoder.encode(uri)