【问题标题】:Twitpic Error - Invalid Username and PasswordTwitpic 错误 - 用户名和密码无效
【发布时间】:2012-06-08 19:06:31
【问题描述】:

我正在使用 twitpic 库在 twitter 上发布图像,但我得到 - 无效的 twitpic 用户名和密码错误。 这个有什么解决办法吗?? witpic的用户名和密码是什么?我只是用我的 Twitter ID 和密码来代替它。

这是我的代码

// Create file
    File picture = new File(Environment.getExternalStorageDirectory()+"/image.jpg");
    if(picture.exists())
    {
        System.out.println("Picture accesseds");
    }

    // Create TwitPic object and allocate TwitPicResponse object
    TwitPic tpRequest = new TwitPic("gauravarora90", "gaurav");
    TwitPicResponse tpResponse = null;

    // Make request and handle exceptions                           
    try {
            tpResponse = tpRequest.uploadAndPost(picture, "Hello World!!!");
    } catch (IOException e) {
            e.printStackTrace();
    } catch (TwitPicException e) {
            e.printStackTrace();
    }


    // If we got a response back, print out response variables
    if(tpResponse != null) {         
           tpResponse.dumpVars();
           System.out.println(tpResponse.getStatus());
           if(tpResponse.getStatus().equals("ok")){
                Toast.makeText(getApplicationContext(), "Photo posted on Twitter.",Toast.LENGTH_SHORT).show();
                //picture.delete();
           }
     }

请帮助我。提前致谢。

【问题讨论】:

    标签: android twitter


    【解决方案1】:

    试试这个代码:

    ConfigurationBuilder cb = new ConfigurationBuilder();
    
                 cb.setOAuthConsumerKey(TwitterClass.CONSUMER_KEY);
                 cb.setOAuthConsumerSecret(TwitterClass.CONSUMER_SECRET);
                 cb.setOAuthAccessToken(tw.getAccessToken());
                AccessToken a = new AccessToken(tw.getAccessToken(),tw.getSecretToken());
    
    
                 TwitterFactory tf = new TwitterFactory(cb.build());
                    t4jTwitter = tf.getInstance();
    
                    t4jTwitter.setOAuthAccessToken(a);
    
                  Configuration conf = new ConfigurationBuilder()                 
                        .setOAuthConsumerKey(TwitterClass.CONSUMER_KEY) 
                        .setOAuthConsumerSecret(TwitterClass.CONSUMER_SECRET) 
                        .setOAuthAccessToken(tw.getAccessToken()) 
                   .setOAuthAccessTokenSecret(tw.getSecretToken()) 
                        .build(); 
    
                    OAuthAuthorization auth = new OAuthAuthorization (conf, conf.getOAuthConsumerKey (), conf.getOAuthConsumerSecret (),
                            new AccessToken (conf.getOAuthAccessToken (), conf.getOAuthAccessTokenSecret ()));
    
                   ImageUpload upload = ImageUpload.getTwitpicUploader (TWITPIC_KEY, auth);
    
                    try {
                          //  String    url = upload.upload(new File("/mnt/sdcard/download/3029302084159350163.jpg"));
    
    
                                                    //result = 1;
    
    
                            URL myfileUrl=new URL(strFilePath);
    
                            HttpURLConnection conn = (HttpURLConnection) myfileUrl
                            .openConnection();
                    conn.setDoInput(true);
                    conn.connect();
                    InputStream is = conn.getInputStream();
    
    
                    //   String url = upload.upload(new File("/mnt/sdcard/download/3029302084159350163.jpg"), StaticData.strTwitte);
    
    
                       String   url = upload.upload(StaticData.strTwitte, is);
    
                       System.out.println("Image url"+url);
                            System.out.println("Image uploaded");
                            result_img=1;
                            twitter4j.Status status = t4jTwitter.updateStatus(StaticData.strTwitte+"\n"+url);
                            int statusId = (int)status.getId();
    
                                //Log.d(TAG, "Image uploaded, Twitpic url is " + url);          
                            } catch (Exception e) {        
                                //Log.e(TAG, "Failed to send image");
                                  result_img=0;
                                  System.out.println("Failed to send image");
                                e.printStackTrace();
                            }
    

    【讨论】:

    • 好。你回复得非常非常快..太好了,谢谢
    • 其实明天有人推荐给我这篇文章,因为我使用这段代码完成了这项任务,我之前没有看到这篇文章。
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