Question

我正在为个人提议写一个简单的Youtube应用程序.. 在Search.execute（）应用程序“正常工作”但没有做任何事情之后，代码将无法运行。这是我的代码：

import android.os.Bundle;
import android.app.Fragment;
import android.view.LayoutInflater;
import android.view.View;
import android.view.ViewGroup;
import android.widget.ImageView;
import android.widget.Toast;

import com.google.api.client.http.HttpRequest;
import com.google.api.client.http.HttpRequestInitializer;
import com.google.api.client.http.javanet.NetHttpTransport;
import com.google.api.client.json.jackson.*;
import com.google.api.services.youtube.YouTube;
import com.google.api.services.youtube.model.ResourceId;
import com.google.api.services.youtube.model.SearchListResponse;
import com.google.api.services.youtube.model.SearchResult;
import com.squareup.picasso.Picasso;

import java.io.IOException;
import java.io.InputStream;
import java.util.Iterator;
import java.util.List;

/**
 * A simple {@link Fragment} subclass.
 */
public class fYoutube extends Fragment{
    /*Global Variables*/
    static final long NUMBER_OF_VIDEOS = 10; // To show only 10 videos at a time.
//    static final String PROPERTIES_FILENAME = "youtube.properties";
    private YouTube youTube;
    ImageView imageView;
    String apiKey = "AIzaSyD47jgw5kPCDUOoAvWPs-4jLFGW6ZinYrI";
    public fYoutube() {
        // Required empty public constructor
    }

    @Override
    public View onCreateView(LayoutInflater inflater, ViewGroup container,
                             Bundle savedInstanceState) {
        // Inflate the layout for this fragment
        return inflater.inflate(R.layout.fragment_youtube, container, false);
    }

    @Override
    public void onActivityCreated(Bundle savedInstanceState) {
        super.onActivityCreated(savedInstanceState);
    }

    public void getQuery(String query){
        if (query.length()<1) query="Youtube"; // Default search word is 'Youtube'.
        try {
            youTube = new YouTube.Builder(new NetHttpTransport(), new JacksonFactory(), new HttpRequestInitializer() {
                @Override
                public void initialize(HttpRequest httpRequest) throws IOException {
                }
            }).setApplicationName(getString(R.string.app_name)).build();
            YouTube.Search.List search = youTube.search().list("id,snippet");


            search.setKey(apiKey);
            search.setQ(query); //What to search for.
            search.setType("video");
            search.setFields("items(id/kind,id/videoId,snippet/title,snippet/thumbnails/default/url)");
            search.setMaxResults(NUMBER_OF_VIDEOS);

            SearchListResponse searchListResponse = search.execute(); << After this line, the app won't do anything..
            List<SearchResult> searchResultList = searchListResponse.getItems();
            displayThumbnails(searchResultList.iterator(),query);

           // if(searchResultList!=null){

  //          }
        }catch(IOException e){}

    }

    public void displayThumbnails(Iterator<SearchResult> iteratorSearchResults, String query){
        if(!iteratorSearchResults.hasNext()){ Toast.makeText(getActivity(),"No results for: "+query,Toast.LENGTH_LONG).show();}
        while(iteratorSearchResults.hasNext()){
            SearchResult singleVideo = iteratorSearchResults.next();
            ResourceId resourceId = singleVideo.getId();
            if(resourceId.getKind().equals("youtube#video")){
                try {
                    String img_url = "http://img.youtube.com/vi/" + resourceId + "/default.jpg";
                    imageView = (ImageView) getActivity().findViewById(R.id.thumbnail1);
                    Picasso.with(getActivity()).load(img_url).placeholder(R.drawable.ic_launcher).into(imageView);
                }catch (Exception e){e.printStackTrace();}
            }
        }
    }
}

我不知道该怎么做.. 没有添加logcat，因为应用程序没有崩溃。

Answer 1

因此，当您调用search.execute（）时，您将返回SearchListResponse。之后该应用程序不执行任何操作，因为您没有告诉应用程序对返回的值执行某些操作。

您需要对该SearchListResponse执行某些操作。所以你应该取消注释search.execute（）下面的行。

Search.execute（）Youtube API无法正常工作

1 个答案: