如何檢索超過25使用Java的YouTube的意見

Question

 YouTubeService service = new YouTubeService("MyApp"); 
     String videoEntryUrl = "http://gdata.youtube.com/feeds/api/videos/nU9dinwMyHo"; 
     VideoEntry videoEntry = service.getEntry(new URL(videoEntryUrl), VideoEntry.class); 
     String commentUrl = videoEntry.getComments().getFeedLink().getHref(); 

     CommentFeed commentFeed = service.getFeed(new URL(commentUrl), CommentFeed.class); 
     for(CommentEntry comment : commentFeed.getEntries()) { 
     System.out.println(comment.getPlainTextContent()); 

這裏是我的代碼來檢索特定視頻的YouTube的評論。 我可以使用此代碼檢索約25條評論，但是如何檢索視頻中的所有評論？

Answer 1

我覺得是這樣的：

// Get a video entry 
String str = "http://gdata.youtube.com/feeds/api/videos/" + videoId; 
YouTubeQuery youtubeQuery = new YouTubeQuery(new URL(str)); 
String videoEntryUrl = youtubeQuery.getUrl().toString(); 
VideoEntry videoEntry = service.getEntry(new URL(videoEntryUrl),VideoEntry.class); 

// Get the comments url for this video 
if (videoEntry.getComments() != null) { 
    String commentUrl = videoEntry.getComments().getFeedLink().getHref(); 
    System.out.println(commentUrl); 

    // Get the comment feed; use a new query 
    YouTubeQuery youtubeQuery = new YouTubeQuery(new URL(commentUrl); 
    youtubeQuery.setMaxResults(50); 


    youtubeQuery.setStartIndex(1); 
    String commentUrlFeed = youtubeQuery.getUrl().toString(); 
    CommentFeed commentFeed = service.getFeed(new URL(commentUrlFeed),CommentFeed.class); 
    // The response should contain an url for the next feed, if any, already with an updated start-index. 

    for (int i = 0; i < commentFeed.getEntries().size() 
      && commentFeed.getEntries().get(i) != null; i++) { 

     String author=commentFeed.getEntries().get(i).getAuthors().get(0).getUri().substring(41) 
     String commentId=commentFeed.getEntries().get(i).getId().substring(47); 
     String comment=commentFeed.getEntries().get(i).getPlainTextContent(); 
     } 
    } 
    // Loop thru next comment feed call, if more can be expected. 
    // Use updated url from the response or set start-index = start-index + max-results. 
}