Search.execute（）YouTube的API不会工作

我正在写一个简单的Youtube应用程序的个人建议.. 代码将无法正常工作后，Search.execute（）应用程序“工作”，但没有做任何事。这是我的代码：Search.execute（）YouTube的API不会工作

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中，因为应用程序不会崩溃。

来源

2015-02-23 John Doe

所以当你调用search.execute（）时，你正在返回一个SearchListResponse。该应用程序之后什么都不做，因为你没有告诉应用程序对返回值做些什么。

您需要对该SearchListResponse做些事情。所以你应该取消注释search.execute（）下方的行。

来源

2015-02-23 22:31:27 Rohan

Search.execute（）YouTube的API不会工作

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