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我正在使用 twitter4j ，我能够成功登录到我的 twitter 帐户，但是当我尝试发布一些内容时，会引发以下异常。

我的代码

try
        {

        Bitmap bmpFinal = getBitmapFromURL(Imagename.trim());
        Log.d("ImagePath",Imagename);

        Configuration conf = new ConfigurationBuilder()
        .setOAuthConsumerKey(TWITTER_CONSUMER_KEY)
        .setOAuthConsumerSecret(TWITTER_CONSUMER_SECRET)
        .setOAuthAccessToken(access_token)
        .setOAuthAccessTokenSecret(access_token_secret)
        .setMediaProviderAPIKey("320954b8bf13ffd7463a4dbe667b9dd2")
        .build();

        Uri tempUri = getImageUri(getApplicationContext(), bmpFinal);

        File finalFile = new File(getRealPathFromURI(tempUri));// THIS IS IMPORTANT. TWITPIC NEEDS THE ACTUAL / ABSOLUTE PATH OF THE IMAGE. JUST THE URI DOES NOT WORK!!!!

        ImageUpload upload = new ImageUploadFactory(conf).getInstance(MediaProvider.TWITPIC);

        String url = upload.upload(finalFile);
        Log.e("TWITTER URL RESPONSE", url);
        Log.d("TWITTER URL RESPONSE", url);
        // END OF FIRST STEP:

        // SECOND STEP IS TO UPLOAD TO TWITTER

        ConfigurationBuilder builder = new ConfigurationBuilder();

        builder.setOAuthConsumerKey(TWITTER_CONSUMER_KEY);
        builder.setOAuthConsumerSecret(TWITTER_CONSUMER_SECRET);

        AccessToken accessToken = new AccessToken(access_token, access_token_secret);
        Twitter twitter = new TwitterFactory(builder.build()).getInstance(accessToken);

        String finalStatusWithURL = null;

        if (status.trim().length() > 0) {

            finalStatusWithURL = status + ":\n " + url;

        } else {
            finalStatusWithURL = url;
        }
        twitter4j.Status response = twitter.updateStatus(finalStatusWithURL);
        Log.e("TWITTER RESPONSE", response.getText());
        }
        catch(Exception e)
        {
            Log.d("Exception ",""+e);
        }
        return null;
    }

Logcat 异常

06-14 12:51:31.117: D/Exception(3302): No authentication challenges found
06-14 12:51:31.117: D/Exception(3302): Relevant discussions can be found on the Internet at:
06-14 12:51:31.117: D/Exception(3302):  http://www.google.co.jp/search?q=ea09dc6d or
06-14 12:51:31.117: D/Exception(3302):  http://www.google.co.jp/search?q=051c1797
06-14 12:51:31.117: D/Exception(3302): TwitterException{exceptionCode=[ea09dc6d-051c1797 0237e8c8-9e66d70f], statusCode=-1, message=null, code=-1, retryAfter=-1, rateLimitStatus=null, version=3.0.2}
4

1 回答 1

1 
    twitter = new TwitterFactory().getInstance();
    //Retrieve access token                     
    authProvider.retrieveAccessToken(authConsumer,verifier);

    //Initialize access token of twitter4j
    AccessToken mAccessToken = new AccessToken( authConsumer.getToken(),authConsumer.getTokenSecret());

    //optional to save it for future use                    
    ProjectUtil.setAccessTokenTwitter((MainActivity)mScreen, mAccessToken.getToken());
    ProjectUtil.setAccessSecretTwitter((MainActivity)mScreen, mAccessToken.getTokenSecret());

    //Set twitter4j.Twitter Oauth                       
    twitter.setOAuthConsumer(AppConstants.CONSUMER_KEY, AppConstants.CONSUMER_SECRET);
    twitter.setOAuthAccessToken(mAccessToken);
    //message                       
    StatusUpdate ad=new StatusUpdate("my messgae");

    //Image from sdcard    
    File file = new File(Environment.getExternalStorageDirectory()+"/myimage.png");
    InputStream inputStream = new FileInputStream(file);
    ad.setMedia("myMedia",inputStream);

    twitter4j.Status response = twitter.updateStatus(ad);



    Log.d("Status", response.getText());
于 2013-06-14T08:11:08.523 回答