Twitter4j v2.2.6流式API與地理定位

Question

我使用twitter4j v2.2.6的流API，並有一些過濾問題。我可以獲得樣本流，但是當我嘗試添加地理位置過濾器時，流輸出不會改變（我仍然收到沒有地理位置和geoloc但位於我的邊框外的推文）。這裏是我的一些代碼：

private static FilterQuery getBoundingBoxFilter() { 
    // New Delhi India 
    double lat = 28.6; 
    double lon = 77.2; 
    double lon1 = lon - .5; 
    double lon2 = lon + .5; 
    double lat1 = lat - .5; 
    double lat2 = lat + .5; 

    double bbox[][] = {{lon1, lat1}, {lon2, lat2}};   
    FilterQuery filtro = new FilterQuery(); 
    return filtro.locations(bbox);   
} 

public static void streamIt() { 
    TwitterStream twitterStream = new TwitterStreamFactory().getInstance(); 
    FilterQuery fq = getBoundingBoxFilter(); 
    StatusListener listener = new StatusListener() { 
     @Override 
     public void onStatus(Status status) { 
      GeoLocation gl = status.getGeoLocation(); 
      StringBuilder msg = new StringBuilder(); 
      if (gl != null) { 
       msg.append("Lat/Lon: ").append(gl.getLatitude()).append(",").append(gl.getLongitude()).append(" - "); 
       msg.append(status.getText()); 
       LOG.info(msg.toString()); 
      } else { 
       msg.append(status.getText()); 
       LOG.info(msg.toString()); 
      } 
     } 

     @Override 
     public void onDeletionNotice(StatusDeletionNotice statusDeletionNotice) { 
      System.out.println("Got a status deletion notice id:" + statusDeletionNotice.getStatusId()); 
     } 

     @Override 
     public void onTrackLimitationNotice(int numberOfLimitedStatuses) { 
      System.out.println("Got track limitation notice:" + numberOfLimitedStatuses); 
     } 

     @Override 
     public void onScrubGeo(long userId, long upToStatusId) { 
      System.out.println("Got scrub_geo event userId:" + userId + " upToStatusId:" + upToStatusId); 
     } 

     @Override 
     public void onException(Exception ex) { 
      if (!(ex instanceof IllegalStateException)) { 
       ex.printStackTrace(); 
      } 
     } 

     @Override 
     public void onStallWarning(StallWarning sw) { 
      throw new UnsupportedOperationException("Not supported yet."); 
     } 
    }; 
    twitterStream.addListener(listener); 
    if (fq != null) { 
     twitterStream.filter(fq); 
    }   
    twitterStream.sample();   
} 

那麼，我在做什麼錯了？有任何想法嗎？

問候，

添

Answer 1

刪除以下行，然後再試一次。

twitterStream.sample();