android - 无法在 AsyncTask 中访问 “findViewById”

Question

https://stackoverflow.com/a/14164318/3575963

在这里，他用了

View myFragmentView = inflater.inflate(R.layout.fragment_a, container, false);

来自父类。我的 parent 也没有 inflater，它是一张 map 。

并且从异步任务 doitbackground 中，我返回了 5 个数组列表以在 postexecute 中创建 textview。

但我做不到，因为我无法使用 findviewbyid，因为我无法获得 Activity 。我得到了上下文，但它没有做任何事情。

这是我的执行后

  protected void onPostExecute(Wrapper wrap){

     TextView name = new TextView (mContext);
     TextView type = new TextView (mContext);
     TextView location = new TextView (mContext);
     TextView distance = new TextView (mContext);

     List<Double> dist = new ArrayList();
     List<String> loc = new ArrayList();
     List<String> nme = new ArrayList();
     List<String> typ = new ArrayList();
     List<Calendar> start = new ArrayList();
     List<Calendar> endd = new ArrayList();

     dist = wrap.getDist();
     loc = wrap.getLocation();
     nme = wrap.getName();
     typ = wrap.getType();
     start = wrap.getsDate();
     endd = wrap.geteDate();
     int idx = -1;
     LinearLayout shw_evnt = (LinearLayout) shw_evnt.findViewById(R.id.);

      for(Double dis:dist){
         idx++;
         name.setText(nme.get(idx));
         type.setText(typ.get(idx));
         location.setText(loc.get(idx));
         distance.setText(dist.get(idx).toString()+" meters");

这部分有问题

 LinearLayout shw_evnt = (LinearLayout) shw_evnt.findViewById(R.id.);

我尝试了其他方法，但没有用。我将使用以前未使用过的另一种布局。

show_events.xml:

  <?xml version="1.0" encoding="utf-8"?>
<LinearLayout xmlns:android="http://schemas.android.com/apk/res/android"
    android:layout_width="match_parent" android:layout_height="match_parent">
    android:id="@+id/shwevnt"

</LinearLayout>

这里调用者mapactivity类

 public class MapActivity extends AppCompatActivity  implements GoogleApiClient.ConnectionCallbacks,
        GoogleApiClient.OnConnectionFailedListener {
    private int strokeColor = 0xffff0000; //red outline
    private int shadeColor = 0x44ff0000; //opaque red fill
    private int count=0;
    private GoogleMap googleMap;
    private GoogleApiClient mGoogleApiClient;
    private Location mLastLocation;
    private LocationRequest mLocationRequest;
    private Context mContext;
    private String TAG = "Chic";
    private final static int CONNECTION_FAILURE_RESOLUTION_REQUEST = 9000;
    private int radius;//in meters
    @TargetApi(Build.VERSION_CODES.HONEYCOMB)
    protected void onCreate(Bundle savedInstanceState) {
        Log.d(TAG, "inside map oncreaste");
        mContext =  getApplicationContext();
        View myFragmentView = inflater.inflate(R.layout.show_events, container, false);
        super.onCreate(savedInstanceState);
        setUpMapIfNeeded();
        Log.d(TAG, "buildgoogleapi called");
        buildGoogleApiClient();
        Log.d(TAG, "after buildgoogleapi called");
     }//end of oncreate

这里有错误

View myFragmentView = inflater.inflate(R.layout.show_events, container, false);

因为我没有打气筒

我不想从执行后返回到主类或另一个包装类。我想我可以在 postexecute 中做，不是吗？

LayoutInflater inflater = (LayoutInflater)context.getSystemService
  (Context.LAYOUT_INFLATER_SERVICE);

如果我这样做，我将采用当前 View ，而不是我想要创建的 View ？

我试过了

private Inflater inflater;
     View myFragmentView = inflater.inflate(R.layout.show_events, container, false);

Answer 1

是的，只需将 Activity 传递给 Asynctask:

   AsyncTask myAsyncTask = new MyTask(this);

然后您将在 Activity 的 layout 中找到该元素:

public class MyTask extends AsyncTask<String, String, String>{
    public MyActivity activity;

    public MyTask(MyActivity a){
        this.activity = a;
    }
    protected void onPostExecute(String result){
...
...
       LinearLayout shw_evnt = (LinearLayout) activity.findViewById(R.id.shwevnt);
...
...
    }
}

正如 Daniel 所描述的那样这将是一种糟糕的做法。

我向你推荐一个 Interface .

例如:

a) 创建接口(interface)类。

public interface AsyncResponse {
    void processFinish(String output);
}

b) 转到您的 AsyncTask 类，并将接口(interface) AsyncResponse 声明为一个字段:

public class MyAsyncTask extends AsyncTask{
    public AsyncResponse delegate = null;
    @Override
    protected void onPostExecute(String result) {
         delegate.processFinish(result);
    }
}

c) 在您的主 Activity 中，您需要实现接口(interface) AsyncResponse。

public class MainActivity implements AsyncResponse{    
MyAsyncTask asyncTask =new MyAsyncTask();
      @Override
      public void onCreate(Bundle savedInstanceState) {
         //this to set delegate/listener back to this class
         asyncTask.delegate = this;
    
         //execute the async task 
         asyncTask.execute();
      }
      //this override the implemented method from asyncTask
      void processFinish(String output){
         //Here you will receive the result fired from async class 
         //of onPostExecute(result) method.
           LinearLayout shw_evnt = (LinearLayout) findViewById(R.id.shwevnt);
       }
     }