本文介绍了我怎样才能得到一个特定URL的鸣叫?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我要一些资源,可以使我捕捉鸣叫只能在特定的链接。
假设我会捕捉其中包含一个链接 http://aaa.com
我该怎么办呢?
谢谢


解决方案

阅读有关Twitter的API ,但更好地库是。

在这里你有code样品:


  • 添加新的鸣叫:

      Twitter的微博= TwitterFactory.getSingleton();
    状态状态= twitter.updateStatus(latestStatus);
    的System.out.println(已成功更新状态[+ status.getText()+。);


  • 歌厅时间表:

      Twitter的微博= TwitterFactory.getSingleton();
       清单<状态>状态= twitter.getHomeTimeline();
      的System.out.println(显示时间轴回家。);
       对于(状态状态:状态){
          。的System.out.println(status.getUser()的getName()+:+
                           status.getText());
    }


  • 登录微博使用OAuth(code for Java中,编辑此):

      Twitter的微博= TwitterFactory.getSingleton();
    twitter.setOAuthConsumer([消费键],[消费秘密]);
    RequestToken requestToken = twitter.getOAuthRequestToken();
    的accessToken的accessToken = NULL;
    BR的BufferedReader =新的BufferedReader(新的InputStreamReader(System.in));
    而(空==的accessToken){
      的System.out.println(打开以下网址和访问权限授予您的帐户:);
      的System.out.println(requestToken.getAuthorizationURL());
      System.out.print(输入PIN码(如果aviailable)或只需点击进入[密码]:);
      串销= br.readLine();
      尝试{
         如果(pin.length()大于0){
           的accessToken = twitter.getOAuthAccessToken(requestToken,脚);
         }其他{
           的accessToken = twitter.getOAuthAccessToken();
         }
      }赶上(TwitterException TE){
        如果(401 == te.getStatus code()){
          的System.out.println(无法获得访问令牌。);
        }其他{
          te.printStackTrace();
        }
      }
    }
    //坚持到的accessToken以供将来参考。
    storeAccessToken(twitter.verifyCredentials()的getId(),的accessToken);
    状态状态= twitter.updateStatus(参数[0]);
    的System.out.println(已成功更新状态[+ status.getText()+。);
    System.exit(0);

    }

    私有静态无效storeAccessToken(INT USEID,的accessToken的accessToken){
           //商店accessToken.getToken()
           //商店accessToken.getTokenSecret()


    • 越来越鸣叫:

        Twitter的微博= TwitterFactory.getSingleton();
      查询查询=新的查询(来源:twitter4j yusukey);
      QueryResult中结果= twitter.search(查询);
      对于(状态状态:result.getStatuses()){
          的System.out.println(@+ status.getUser()getScreenName()+:+ status.getText());
      }



我希望我帮助。

来源:

I want some resource which can lead me to capture tweets for only a specific links.Suppose i will capture all tweets which contain a link http://aaa.comhow can i do it ?thanks

解决方案

Read about Twitter API here, but better library is Twitter4J.

Here you have code samples to :

  • add new tweet :

    Twitter twitter = TwitterFactory.getSingleton();
    Status status = twitter.updateStatus(latestStatus);
    System.out.println("Successfully updated the status to [" + status.getText() + "].");
    

  • geting timeline :

       Twitter twitter = TwitterFactory.getSingleton();
       List<Status> statuses = twitter.getHomeTimeline();
      System.out.println("Showing home timeline.");
       for (Status status : statuses) {
          System.out.println(status.getUser().getName() + ":" +
                           status.getText());
    }
    

  • Log in Twitter using OAuth (code for Java, edit this):

    Twitter twitter = TwitterFactory.getSingleton();
    twitter.setOAuthConsumer("[consumer key]", "[consumer secret]");
    RequestToken requestToken = twitter.getOAuthRequestToken();
    AccessToken accessToken = null;
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    while (null == accessToken) {
      System.out.println("Open the following URL and grant access to your account:");
      System.out.println(requestToken.getAuthorizationURL());
      System.out.print("Enter the PIN(if aviailable) or just hit enter.[PIN]:");
      String pin = br.readLine();
      try{
         if(pin.length() > 0){
           accessToken = twitter.getOAuthAccessToken(requestToken, pin);
         }else{
           accessToken = twitter.getOAuthAccessToken();
         }
      } catch (TwitterException te) {
        if(401 == te.getStatusCode()){
          System.out.println("Unable to get the access token.");
        }else{
          te.printStackTrace();
        }
      }
    }
    //persist to the accessToken for future reference.
    storeAccessToken(twitter.verifyCredentials().getId() , accessToken);
    Status status = twitter.updateStatus(args[0]);
    System.out.println("Successfully updated the status to [" + status.getText() + "].");
    System.exit(0);
    

    }

    private static void storeAccessToken(int useId, AccessToken accessToken){ //store accessToken.getToken() //store accessToken.getTokenSecret()

    • getting tweets :

      Twitter twitter = TwitterFactory.getSingleton();
      Query query = new Query("source:twitter4j yusukey");
      QueryResult result = twitter.search(query);
      for (Status status : result.getStatuses()) {
          System.out.println("@" + status.getUser().getScreenName() + ":" + status.getText());
      }
      

I hope I helped.

Source : http://twitter4j.org/en/code-examples.html

这篇关于我怎样才能得到一个特定URL的鸣叫?的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!

07-11 20:49