Question

我正在尝试创建“收藏夹”课程。我搜索我从用户那里获得的标题，并将视频添加到收藏夹类。但是，我总是得到不止一个结果。（我得到的最大结果是10）。我怎样才能为每个标题获得1个结果？

这是on create：

    @Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.favorites_layout);
    getSupportActionBar().setTitle("Favorites");
    initializeViews();

    extras = getIntent().getExtras();
    this.vidTitle = extras.getString("title");
    this.vidID = extras.getString("id");
    //Checking where to add the new Video
    for(int i=0;i<favorites.length;i++){
        if(favorites[i]==null){
            favorites[i] = vidTitle;
        }
        break; // Break so it won't add same video to all array
    }
    AppUtils.showToast("Loading Favorites");
    getVideo2();
}

这是得到10个结果的代码：

    public void getVideo(){
    AppUtils.showToast("Loading Favorites");
    mServiceTask = new ServiceTask(SEARCH_VIDEO);
    mServiceTask.setmServerResponseListener(this);
    for (int i=0; i<favorites.length;i++) {
        if(favorites[i]!=null){
            mServiceTask.execute(new Object[]{favorites[i]}); <--- Problem here
            break;
        }
        else{
            break;
        }
    }
}

ServiceTask：

public class ServiceTask extends AsyncTask<Object, Void, Object[]> implements ServiceTaskInterface{
private static final String TAG = ServiceTask.class.getSimpleName();
private ServerResponseListener mServerResponseListener = null;
private int mRequestCode = 0;

public void setmServerResponseListener(ServerResponseListener mServerResponseListener){
    this.mServerResponseListener = mServerResponseListener;
}

public ServiceTask(int iReqCode){
    mRequestCode = iReqCode;
}

@Override
protected void onPreExecute() {
    super.onPreExecute();
    mServerResponseListener.prepareRequest(mRequestCode);
}

@Override
protected Object[] doInBackground(Object... params) {
    if(params == null)
        throw  new NullPointerException("Parameters to the async task can never be null");
    mServerResponseListener.goBackground();
    Object[] resultDetails = new Object[2];
    resultDetails[0] = mRequestCode;

    switch (mRequestCode){
        case AppConstants.SEARCH_VIDEO:
            try {
                resultDetails[1] = loadVideos((String) params[0]);
                break;
            }catch (Exception e){
                AppUtils.showToast("BLABLABLA");}

    }
    return  resultDetails;
}

@Override
protected void onPostExecute(Object[] result) {
    super.onPostExecute(result);
    mServerResponseListener.completedRequest(result);
}
//Loading the videos, with help of Google's code.
private List<SearchResult> loadVideos(String queryTerm){
    try{
        YouTube youTube = new YouTube.Builder(transport,jsonFactory,new HttpRequestInitializer() {
            @Override
            public void initialize(HttpRequest httpRequest) throws IOException {}
        }).setApplicationName(YoutubeApplication.appName()).build();
        YouTube.Search.List search = youTube.search().list("id,snippet");
        search.setKey(AppConstants.KEY);
        search.setQ(queryTerm);

        //Only including videos
        search.setType("video");

        search.setFields("items(id/kind,id/videoId,snippet/title,snippet/description,snippet/thumbnails/default/url,snippet/thumbnails/medium/url)");
        search.setMaxResults(AppConstants.NUMBER_OF_VIDEOS_RETURNED);

        //Call the API to print results
        SearchListResponse searchListResponse = search.execute();
        List<SearchResult> searchResultList = searchListResponse.getItems();
        if(searchResultList != null){
            return searchResultList;
        }
    }catch (GoogleJsonResponseException e){
        System.err.println("There was a service error: " + e.getDetails().getCode() + " : "
                + e.getDetails().getMessage());
    }catch (IOException e){
        System.err.println("There was an IO error: " + e.getCause() + " : " + e.getMessage());
    }catch (Throwable t){
        t.printStackTrace();
    }
    return null;
}

}

我如何才能获得1个结果，但仍然可以获得10个结果（如10个不同的视频）？

Answer 1

首先，评论您的def logTry[T](what: Try[T], block: T => String): Unit = { what match { case Success(res) => println(block(res)) case Failure(t) => println(???) // how to do this } }行或您的循环无用。然后在列表中收集非空对象，并将其放在break;方法中execute。

toArray

<强>更新

关于LinkedList<Object> list = new LinkedList<Object>(); for (int i=0; i<favorites.length; i++) { if(favorites[i]!=null){ //mServiceTask.execute(new Object[]{favorites[i]}); //<--- Problem here list.add(favorites[i]); //break; } else{ //break; } } mServiceTask.execute(list.toArray());课程。在ServiceTask中，您只使用doInBackground - 第一个数组元素。所以应该有：

params[0]

获得太多结果

1 个答案: