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Home/ Questions/Q 4118886
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Editorial Team
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Editorial Team
Asked: May 20, 20262026-05-20T23:01:10+00:00 2026-05-20T23:01:10+00:00

I’m working on a little app and using GWT to build it. I just

  • 0

I’m working on a little app and using GWT to build it.
I just tried making a request to a remote server which will return a response as JSON.
I’ve tried using the overlay types concept but I couldn’t get it working. I’ve been changing the code around so its a bit off from where the Google GWT tutorials left.

JavaScriptObject json;
    public JavaScriptObject executeQuery(String query) {
        String url = "http://api.domain.com?client_id=xxxx&query=";
        RequestBuilder builder = new RequestBuilder(RequestBuilder.GET,
                URL.encode(url + query));
        try {
            @SuppressWarnings("unused")
            Request request = builder.sendRequest(null, new RequestCallback() {
                public void onError(Request request, Throwable exception) {
                    // violation, etc.)
                }

                public void onResponseReceived(Request request,
                        Response response) {
                    if (200 == response.getStatusCode()) {
                        // Process the response in response.getText()
                        json =parseJson(response.getText());
                    } else {

                    }
                }
            });
        } catch (RequestException e) {
            // Couldn't connect to server
        }
        return json;
    }

    public static native JavaScriptObject parseJson(String jsonStr) /*-{
        return eval(jsonStr );
        ;
    }-*/;

In the chrome’s debugger I get umbrellaexception, unable to see the stack trace and GWT debugger dies with NoSuchMethodError… Any ideas, pointers?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-20T23:01:10+00:00Added an answer on May 20, 2026 at 11:01 pm
    1. Use JsonUtils#safeEval() to evaluate the JSON string instead of calling eval() directly.
    2. More importantly, don’t try to pass the result of an asynchronous call (like RequestBuilder#sendRequest() back to a caller using return – use a callback:

      public void executeQuery(String query,
                               final AsyncCallback<JavaScriptObject> callback)
      {
        ...
        try {
          builder.sendRequest(null, new RequestCallback() {
            public void onError(Request request, Throwable caught) {
              callback.onFailure(caught);
            }
      
            public void onResponseReceived(Request request, Response response) {
              if (Response.SC_OK == response.getStatusCode()) {
                try {
                  callback.onSuccess(JsonUtils.safeEval(response.getText()));
                } catch (IllegalArgumentException iax) {
                  callback.onFailure(iax);
                }
              } else {
                // Better to use a typed exception here to indicate the specific
                // cause of the failure.
                callback.onFailure(new Exception("Bad return code."));
              }
            }
          });
        } catch (RequestException e) {
          callback.onFailure(e);
        }
      }
      
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