twitter4j的问题及处理

Question

我试图让处理显示包含我的关键字的最新推文。我卡住了。我不确定如何检查它是否在更新。我知道它可以拉出最近的推特，但到目前为止，我唯一可以查看最新推特的方法是重新启动我的程序。抱歉太乱了。

import twitter4j.conf.*;
import twitter4j.*;
import twitter4j.auth.*;
import twitter4j.api.*;

import java.util.*;

Twitter twitter; // creating the twitter object
String searchString = "newborn"; // twitter search query
List<Status> tweets;// List of Status object to hold tweets
List<Status> tweets2;
int currentTweet;// 
int nextTweet;

void setup ()
{
  size(800,600);

  ConfigurationBuilder cb = new ConfigurationBuilder(); // New config obj, auth.

  cb.setOAuthConsumerKey("");
  cb.setOAuthConsumerSecret("");
  cb.setOAuthAccessToken("");
  cb.setOAuthAccessTokenSecret("");

  TwitterFactory tf = new TwitterFactory(cb.build()); // init. twit obj  by retr inst from twit fact

  twitter = tf.getInstance(); //here, twit gets isnt is the inst
  getTweets2();
  getNewTweets(); //  func to get the tweets

  currentTweet = 0;
  nextTweet = 1;
  //Status status = tweets.get(currentTweet);

  thread("refreshTweets");





}

void draw()
{
  background(0);

  //currentTweet = currentTweet + 1;
  //nextTweet = nextTweet + 1;

  if (currentTweet >= tweets.size())
  {
    currentTweet = 0;
  }
  if (nextTweet >= tweets.size())
  {
    nextTweet = 1;
  }

  Status status = tweets.get(currentTweet); // retrieve current tweet from list of tweets
  Status status2 = tweets.get(nextTweet);
  Status status3 = tweets2.get(0);

  Status status4 = status;

  fill(200);

  if(tweets.get(0).getCreatedAt() != tweets2.get(0).getCreatedAt()) {
  //text(status.getText(),random(width),random(height), 300, 200);
  text("current tweet:" + status.getCreatedAt() + ".....if" ,250,250, 300, 200);
  //text("next tweet:" + status2.getCreatedAt(),250,350, 300, 200);
  //text("10th tweet:" + status3.getCreatedAt(),250,450, 300, 200);
  }else {
    text("current tweet:" + status2.getCreatedAt() + ".....else",250,250, 300, 200);
  }
  delay(2500);

  getNewTweets();

  //System.out.println(status2.getText());
}

void getNewTweets() 
{
  try
  {
    //we try to get the tweetuses!
    Query query = new Query(searchString);

    QueryResult result = twitter.search(query);

    tweets = result.getTweets();

  }
  catch (TwitterException te)
  {
    //tweetus aborted? 
    System.out.println("failed to search tweets: " + te.getMessage());
    System.exit(-1);

  }
}

  void getTweets2() //this neat thing gets those tweets 
{
  try
  {
    //we try to get the tweetuses!
    Query query = new Query(searchString);

    QueryResult result = twitter.search(query);

    tweets2 = result.getTweets();

  }
  catch (TwitterException te)
  {
    //tweetus aborted? 
    System.out.println("failed to search tweets: " + te.getMessage());
    System.exit(-1);

  }




}

void refreshTweets()
{
  while(true)
  {
    getNewTweets();
    println("Updated Tweets");
    delay(5000);
  }
}

Answer 1

注意：您已经将您的秘密令牌公开发布到internet上，所以您应该尽快更改它们！仅仅编辑你的问题而不包括它们是不够的，因为它们仍然会在修订历史中。

你在这个程序中有很多额外的东西。你可以简化很多。你所需要做的就是：

第1步：编写获取tweet（或tweet的ArrayList）的函数。在草图级变量中跟踪该tweet。您已经使用getnewtweets()函数实现了这一点，但是可以去掉其他函数。

import twitter4j.conf.*;
import twitter4j.*;
import twitter4j.auth.*;
import twitter4j.api.*;

Twitter twitter;
String searchString = "cat";
Status currentTweet;

void setup ()
{
  size(800, 600);

  ConfigurationBuilder cb = new ConfigurationBuilder(); // New config obj, auth.

  cb.setOAuthConsumerKey("");
  cb.setOAuthConsumerSecret("");
  cb.setOAuthAccessToken("");
  cb.setOAuthAccessTokenSecret("");

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

  getNewTweets();
}

void draw()
{

  //get new tweet once every 10 seconds
  if (frameCount % (60*10) == 0) {
    getNewTweets();
  }

  background(0);
  fill(200);

  text("current tweet:" + currentTweet.getText(), 250, 250, 300, 200);
}

void getNewTweets() 
{

  println("Getting new tweet...");
  try
  {
    Query query = new Query(searchString);

    QueryResult result = twitter.search(query);
    currentTweet = result.getTweets().get(0);
  }
  catch (TwitterException te)
  {
    te.printStackTrace();
  }
}