如何使用异步任务返回对象？

Question

我正在尝试从异步任务中解析后返回推文列表，但是我无法从任务中获取到ArrayList。 请问有没有人能提供一个解决方案？

public class Main extends ListActivity {
    String MY_APP_TAG = "com.list";

    /** Called when the activity is first created. */
    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        ArrayList listItems = new ArrayList();
        new myAsyncTask().execute(listItems);

        setListAdapter(new ArrayAdapter(this, R.layout.tweet, R.id.tweet,listItems));
        ListView lv = getListView();
        lv.setTextFilterEnabled(true);
        lv.setOnItemClickListener(new OnItemClickListener() {
            @Override
            public void onItemClick(AdapterView<?> parent, View view, int position, long id) {
                // When clicked, show a Toaster
                Toast.makeText(getApplicationContext(), ((TextView) view).getText(), Toast.LENGTH_SHORT).show();
            }
        });
    }

    private class myAsyncTask extends AsyncTask<ArrayList<Object>, Void, Void>
    {
        ArrayList<Object> listItems;
        ProgressDialog dialog;
        @Override
        protected void onPreExecute() {
            dialog = ProgressDialog.show(Main.this, "", "Loading....");
        }
        @Override
        protected Void doInBackground(ArrayList<Object>... params) {

            String host = "api.twitter.com";
            String twitterURL = "http://"+host+"/1/statuses/user_timeline.json?screen_name=i1990jain&amp;count;=10";
            try {
                HttpClient client = new DefaultHttpClient();
                BasicHttpContext localContext = new BasicHttpContext();
                HttpHost targetHost = new HttpHost(host, 80, "http");
                HttpGet httpget = new HttpGet(twitterURL);
                httpget.setHeader("Content-Type", "application/json");
                HttpResponse response = client.execute(targetHost, httpget, localContext);
                HttpEntity entity = response.getEntity();
                Object content = EntityUtils.toString(entity);
                Log.d(MY_APP_TAG, "OK: " + content.toString());

                JSONArray ja = new JSONArray(content.toString());

                for(int i = 0; i < ja.length(); i++){
                    JSONObject jo = ja.getJSONObject(i);
                    listItems.add(jo.getString("text"));
                }
            } catch(Exception e) {
                e.printStackTrace();
            }
            return null;
        }
        @Override
        protected void onPostExecute(Void result) {
            dialog.dismiss();
        }
    }
}

- Rishabh

---

要么让两个线程异步运行并使用AsyncTask.onPostExecute()处理结果，要么显式地使用AsyncTask.get()阻塞UI线程并等待工作线程返回结果。请注意，后者实际上牺牲了AsyncTask的好处，并可能导致ANR异常。 - yorkw

Answer 1

这是一个 AsyncTask<Params, Progress, Result>。因此，您应该将其声明为 AsyncTask<Void, Void, ArrayList<Object>>。

从 doInBackground 方法中返回列表。

- Jong

---

无法从doInBackground返回列表。 - Rishabh

不是吗？但这正是你在发布的答案中所做的...就是我说的。 - Jong

Answer 2

如果您想从异步线程更新活动中的新数据，可以使用onProgressUpdate()/onPostExecute()方法，或者如果您正在使用线程，则应该使用Handler。

//aciticty class:
final Handler mHandler = new Handler();

//inside an async thread
handler.post(new Runnable() {
    @Override
    public void run() {
        //update GUI thread here
    }
});

- Kirill Kulakov

Answer 3

我是如何让它工作的

public class Main extends ListActivity {

    String MY_APP_TAG = "com.vidyut";
    ArrayList<Object> list = new ArrayList<Object>();
        /** Called when the activity is first created. */
    @SuppressWarnings("unchecked")
    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        myAsyncTask task = new myAsyncTask();
        task.execute();

}

    private class myAsyncTask extends AsyncTask<ArrayList<Object>, ArrayList<Object>, ArrayList<Object>>
    {
        ProgressDialog dialog;
        @Override
        protected void onPreExecute() {
            dialog = ProgressDialog.show(Main.this, "", "Loading....");
        }

        @Override
        protected ArrayList<Object> doInBackground(ArrayList<Object>... params) {

            String host = "api.twitter.com";
            String twitterURL = "http://"+host+"/1/statuses/user_timeline.json?screen_name=i1990jain&amp;count;=10";
         try {
          HttpClient client = new DefaultHttpClient();
             BasicHttpContext localContext = new BasicHttpContext();
             HttpHost targetHost = new HttpHost(host, 80, "http");
             HttpGet httpget = new HttpGet(twitterURL);
             httpget.setHeader("Content-Type", "application/json");
             HttpResponse response = client.execute(targetHost, httpget, localContext);
             HttpEntity entity = response.getEntity();
             Object content = EntityUtils.toString(entity);
             Log.d(MY_APP_TAG, "OK: " + content.toString());

             JSONArray ja = new JSONArray(content.toString());

          for(int i = 0; i < ja.length(); i++){
           JSONObject jo = ja.getJSONObject(i);
           list.add(jo.getString("text"));
          }
         } catch(Exception e) {
          e.printStackTrace();
         }
            return list;
        }
        protected void onPostExecute(ArrayList<Object> list) {
            dialog.dismiss();

             Log.d(MY_APP_TAG, "The returned list contains " +list.size()+ "elements");
             setListAdapter(new ArrayAdapter<Object>(Main.this, R.layout.tweet, R.id.tweet,list));
                ListView lv = getListView();
                lv.setTextFilterEnabled(true);
                lv.setOnItemClickListener(new OnItemClickListener() {
                 @Override
                 public void onItemClick(AdapterView<?> parent, View view, int position, long id) {
                  // When clicked, show a Toaster
                      Toast.makeText(getApplicationContext(), ((TextView) view).getText(), Toast.LENGTH_SHORT).show();
             }
            });

        }

    }
}

- Rishabh

Answer 4

您想将结果返回到哪里？您可以在以下位置访问listItems：

protected void onPostExecute(Void result) {

这就是你应该使用listItems的地方，例如填充列表。此方法在UI线程上运行，因此在其中执行是安全的。

在你的doInBackground中，我没有看到listItems被初始化为任何值，所以它可能为空。你应该创建一个单独的listItems实例，在doInBackground中使用它，然后在onPostExecute中将其分配给listView。

- marcinj