Question

使用以下代码和twitter4j库管理最终将我自己的推文发送到处理。我现在一直在努力调整代码来吸引特定用户的推文，没有任何运气，无论网上有多少人发布了“工作代码”。有人可以指导我，并告诉我究竟需要改变什么？谢谢！

import twitter4j.util.*;
import twitter4j.*;
import twitter4j.management.*;
import twitter4j.api.*;
import twitter4j.conf.*;
import twitter4j.json.*;
import twitter4j.auth.*;
import java.util.*;

List<Status>statuses = null;

TwitterFactory twitterFactory;
Twitter twitter;

void setup() {     
  size(100, 100);    
  background(0); 

  connectTwitter();    
  getTimeline();  
}  

void draw() {     
  background(0);
}  

// Initial connection
void connectTwitter() {  
  ConfigurationBuilder cb = new ConfigurationBuilder();  
  cb.setOAuthConsumerKey("XXXX");
  cb.setOAuthConsumerSecret("XXXX");
  cb.setOAuthAccessToken("XXXX");
  cb.setOAuthAccessTokenSecret("XXXX"); 

  twitterFactory = new TwitterFactory(cb.build());    
  twitter = twitterFactory.getInstance();  

  println("connected");
} 

// Get your tweets
void getTimeline() {     
  try {        
    statuses = twitter.getUserTimeline();
  }   
  catch(TwitterException e) {         
    println("Get timeline: " + e + " Status code: " + e.getStatusCode());
  }     
  for (Status status:statuses) {               
    println(status.getUser().getName() + ": " + status.getText());
  }
}

编辑 - 用于获取用户推文的修订代码。不产生任何错误或结果......

void getUserTimeLine(long stephenfry) {
  try {
    ResponseList<Status> statuses = twitter.getUserTimeline(stephenfry);
  }
  catch(TwitterException e) {
    println("Get timeline: " + e + " Status code: " + e.getStatusCode());
  }
    for (Status status : statuses) {
        System.out.println(status.getText());
    }
}

Answer 1

您只需在代码中添加以下内容即可检索任何用户时间表 -

void getUserTimeLine(long userID/*You can also use screenName*/) {
    ResponseList<Status> statuses = twitter.getUserTimeline(userID/*You can also use screenName*/);
    for (Status status : statuses) {
        System.out.println(status.getText());
    }
}

让用户推文成功处理 - Twitter4j

1 个答案: