Twitter REST API 1.1 Java

Question

您好我正在尝试使用Twitter的REST API而不使用解析Sdk。 我从解析twitter类获得了令牌和访问令牌密码，任何人都可以告诉我我做错了什么。

我正在使用twitter的REST API 1.1。

private void usingNetwork() { 
    // TODO Auto-generated method stub 
    new AsyncTask<Void, Void, String>() { 

     @Override 
     protected String doInBackground(Void... params) { 
      // TODO Auto-generated method stub 

      try { 
       URL ur = new URL(
             "https://api.twitter.com/1.1/statuses/user_timeline.json? screen_name=suresh_bora&include_entities=true"); 
       HttpURLConnection conn = (HttpURLConnection) ur 
         .openConnection(); 
       conn.addRequestProperty("Content-Type", 
         "application/x-www-form-urlencoded"); 
       conn.addRequestProperty(
         "Authorization", 
         "OAuth oauth_consumer_key=" 
           + ParseTwitterUtils.getTwitter() 
             .getConsumerKey() 
           + ",oauth_token=" 
           + ParseTwitterUtils.getTwitter() 
             .getAuthToken() 
           + ",oauth_nonce=kYjzVBB8Y0ZFdfdfabxSWbWovY3uYSQ2pTgmZeNu2VS4cg," + 
           "oauth_signature_method=HMAC-SHA1," + 
           "oauth_timestamp="+ new Timestamp(date.getSeconds()) + 
           ",oauth_version=1.0,"+ 
           "oauth_signature="+ParseTwitterUtils.getTwitter().getAuthTokenSecret()+""); 
       conn.setDoInput(true); 
       conn.setDoOutput(true); 
       conn.setRequestMethod("GET"); 

       readStream(conn.getInputStream()); 
      } catch (MalformedURLException e) { 
       // TODO Auto-generated catch block 
       e.printStackTrace(); 
      } catch (IOException e) { 
       e.printStackTrace(); 
      } 
      return null; 

     } 

     @Override 
     protected void onPostExecute(String result) { 
      // TODO Auto-generated method stub 
      super.onPostExecute(result); 
     } 
    }.execute(); 
} 

}

Answer 1

，帮助我：

“SSL绝对需要 这种认证方式安全的前提是使用SSL因此，所有请求（包括获取和使用。 （（c）https://dev.twitter.com/docs/auth/application-only-auth）

所以你必须使用HttpsURLConnection而不是Http。 之后，检查是否得到令牌，它看起来像“AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA％2FAAAAAAAAAAAAAAAAAAAA％3DAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA”。

下面是一个例子：

private final static String getTokenURL = "https://api.twitter.com/oauth2/token"; 
private static String bearerToken; 

/** 
* @param args 
*/ 
public static void main(String[] args) { 

    // encodeKeys(APIKEY, APISECRET); 

    new Thread(new Runnable() { 

     @Override 
     public void run() { 
      try { 

       bearerToken = requestBearerToken(getTokenURL); 
       fetchTimelineTweet(twitURL); 

      } catch (IOException e) { 
       System.out.println("IOException e"); 
       e.printStackTrace(); 
      } 
     } 
    }).start(); 

} 

// Encodes the consumer key and secret to create the basic authorization key 
private static String encodeKeys(String consumerKey, String consumerSecret) { 
    try { 
     String encodedConsumerKey = URLEncoder.encode(consumerKey, "UTF-8"); 
     String encodedConsumerSecret = URLEncoder.encode(consumerSecret, 
       "UTF-8"); 

     String fullKey = encodedConsumerKey + ":" + encodedConsumerSecret; 
     byte[] encodedBytes = Base64.encodeBase64(fullKey.getBytes()); 

     return new String(encodedBytes); 
    } catch (UnsupportedEncodingException e) { 
     return new String(); 
    } 
} 

// Constructs the request for requesting a bearer token and returns that 
// token as a string 
private static String requestBearerToken(String endPointUrl) 
     throws IOException { 
    HttpsURLConnection connection = null; 
    String encodedCredentials = encodeKeys(APIKEY, APISECRET); 

    System.out.println("encodedCredentials "+encodedCredentials); 
    try { 
     URL url = new URL(endPointUrl); 
     connection = (HttpsURLConnection) url.openConnection(); 
     System.out.println(connection); 
     connection.setDoOutput(true); 
     connection.setDoInput(true); 
     connection.setRequestMethod("POST"); 
     connection.setRequestProperty("Host", "api.twitter.com"); 
     connection.setRequestProperty("User-Agent", "anyApplication"); 
     connection.setRequestProperty("Authorization", "Basic " 
       + encodedCredentials); 
     connection.setRequestProperty("Content-Type", "application/x-www-form-urlencoded;charset=UTF-8"); 
     connection.setRequestProperty("Content-Length", "29"); 
     connection.setUseCaches(false); 

     writeRequest(connection, "grant_type=client_credentials"); 

     // Parse the JSON response into a JSON mapped object to fetch fields 
     // from. 
     JSONObject obj = (JSONObject) JSONValue.parse(readResponse(connection)); 

     if (obj != null) { 
      String tokenType = (String) obj.get("token_type"); 
      String token = (String) obj.get("access_token"); 

      return ((tokenType.equals("bearer")) && (token != null)) ? token 
        : ""; 
     } 
     return new String(); 
    } catch (MalformedURLException e) { 
     throw new IOException("Invalid endpoint URL specified.", e); 
    } finally { 
     if (connection != null) { 
      connection.disconnect(); 
     } 
    } 
} 

// Fetches the first tweet from a given user's timeline 
private static String fetchTimelineTweet(String endPointUrl) 
     throws IOException { 
    HttpsURLConnection connection = null; 

    try { 
     URL url = new URL(endPointUrl); 
     connection = (HttpsURLConnection) url.openConnection(); 
     connection.setDoOutput(true); 
     connection.setDoInput(true); 
     connection.setRequestMethod("GET"); 
     connection.setRequestProperty("Host", "api.twitter.com"); 
     connection.setRequestProperty("User-Agent", "anyApplication"); 
     connection.setRequestProperty("Authorization", "Bearer " + bearerToken); 
     connection.setUseCaches(false); 

     // Parse the JSON response into a JSON mapped object to fetch fields 
     // from. 
     JSONArray obj = (JSONArray) JSONValue.parse(readResponse(connection)); 
     System.out.println("JSON obj = "+obj); 

     if (obj != null) { 
      String tweet = ((JSONObject) obj.get(0)).get("text").toString(); 
      System.out.println(tweet); 
      return (tweet != null) ? tweet : ""; 
     } 
     return new String(); 
    } catch (MalformedURLException e) { 
     throw new IOException("Invalid endpoint URL specified.", e); 
    } finally { 
     if (connection != null) { 
      connection.disconnect(); 
     } 
    } 
} 

// Writes a request to a connection 
private static boolean writeRequest(HttpURLConnection connection, 
     String textBody) { 
    try { 
     BufferedWriter wr = new BufferedWriter(new OutputStreamWriter(
       connection.getOutputStream())); 
     wr.write(textBody); 
     wr.flush(); 
     wr.close(); 

     return true; 
    } catch (IOException e) { 
     return false; 
    } 
} 

// Reads a response for a given connection and returns it as a string. 
private static String readResponse(HttpURLConnection connection) { 
    try { 
     StringBuilder str = new StringBuilder(); 

     BufferedReader br = new BufferedReader(new InputStreamReader(
       connection.getInputStream())); 
     String line = ""; 
     while ((line = br.readLine()) != null) { 
      str.append(line + System.getProperty("line.separator")); 
     } 
     return str.toString(); 
    } catch (IOException e) { 
     return new String(); 
    } 
} 

}

Answer 2

要添加到这一点，你可以找到Defuera上面贴的代码示例和一些更多的信息在：http://www.coderslexicon.com/demo-of-twitter-application-only-oauth-authentication-using-java/

我是什么做错的事情是把持票人标记编码为base 64，你不需要这样做。

一旦拥有不记名令牌，您可以打开另一个HttpsUrlConnection到Twitter并处理响应。

我个人所做的只是获取一个对象并将其作为json返回给前端。

如果试图如果要使用org.simple.json投射到特定的JSON类型，例如，微博返回任一JSON数组或JSON对象，这取决于哪个REST URL调用。所以，如果你试图施放此来的JSONObject或JSONArray，它会抛出一个异常，如果你选择了错误类型：

private Object requestTwitterApi(String endPointUrl, Map<String, String[]> params) 
     throws IOException { 

    HttpsURLConnection connection = null; 

    // Generate query 
    String query = generateQueryUrl(params); 
    // Append query to URL 
    endPointUrl += query; 
    // endPointUrl is now: https://api.twitter.com/1.1/search/tweets.json?q=%40twitterapi for example 

    try { 
     URL url = new URL(endPointUrl); 
     connection = (HttpsURLConnection) url.openConnection(); 
     LOGGER.debug("TwitterApiUrl: " + connection); 
     connection.setDoOutput(true); 
     connection.setRequestMethod("GET"); 
     connection.setRequestProperty("Host", "api.twitter.com"); 
     connection.setRequestProperty("User-Agent", "anyApplication"); 
     connection.setRequestProperty("Authorization", "Bearer " + bearerToken); // bearer token is exactly what was returned from Twitter 
     connection.setUseCaches(false); 

     // Parse the JSON response into a JSON mapped object 
     JSONParser parser = new JSONParser(); 
     Object obj = null; 
     try { 
      obj = parser.parse(readResponse(connection)); 
     } catch (ParseException e) { 
      LOGGER.debug("Exception parsing JSON from Twitter"); 
      e.printStackTrace(); 
     } 
     // then just return the object and I use struts-json-plugin to convert to json and return to frontend. This will allow you to return JSON objects or arrays without issue 
     return obj != null ? obj : ""; 

    } catch (MalformedURLException e) { 
     LOGGER.error("Invalid endpoint URL specified : " + e); 
     throw new IOException("Invalid endpoint URL specified.", e); 
    } finally { 
     if (connection != null) { 
      connection.disconnect(); 
     } 
    } 
} 

我希望这可以帮助别人，这结束了一整天带我刚刚使用Twitter API ...（我们正在做定制分析）Facebook比较了30分钟。