从AsyncTask返回JSON数组

Question

我有一个获取JSON数组的AsyncTask。我怎么会返回JSON数组是这样的：

JSONArray channels = new Json().execute(foo, bar);package com.example.tvrplayer; 

ECLIPS告诉我，我不能做到这一点，它应该是：

AsyncTask<Object, Integer, JSONArray> channels = new Json().execute("http://192.168.2.136:8080/rest/channel/"+ linkid +"/"+ username, "GET"); 

JSON的异步类：

public class Json extends AsyncTask<Object, Integer, JSONArray> { 

    Json(){ 
     super(); 
    } 

    @Override 
    protected JSONArray doInBackground(Object... params) { 
     // Log.i("JSON",url); 
     String url = (String) params[0]; 
     String method = (String) params[1]; 
     InputStream is = null; 
     String result = ""; 
     JSONArray jsonObject = null; 

     // HTTP 
     try { 
      HttpClient httpclient = new DefaultHttpClient(); // for port 80 requests! 
      if (method == "GET") { 
       HttpGet httpget = new HttpGet(url); 
       HttpResponse response = httpclient.execute(httpget); 
       HttpEntity entity = response.getEntity(); 
       is = entity.getContent(); 
      } else if (method == "POST") { 
       HttpPost httppost = new HttpPost(url); 
       HttpResponse response = httpclient.execute(httppost); 
       HttpEntity entity = response.getEntity(); 
       is = entity.getContent(); 
      } 


     } catch(Exception e) { 
      Log.e("JSON - 1 -", e.toString()); 
      return null; 
     } 

     // Read response to string 
     try {   
      BufferedReader reader = new BufferedReader(new InputStreamReader(is,"utf-8"),8); 
      StringBuilder sb = new StringBuilder(); 
      String line = null; 
      while ((line = reader.readLine()) != null) { 
       sb.append(line + "\n"); 
      } 
      is.close(); 
      result = sb.toString();  
//   result = result.substring(1,result.length()-1); 
//   Log.d("JSON result",result); 
     } catch(Exception e) { 
      Log.e("JSON - 2 -", e.toString()); 
      return null; 
     } 

     // Convert string to object 
     try { 
      jsonObject = new JSONArray(result);    
     } catch(JSONException e) { 
      Log.e("JSON - 3 -", e.toString()); 
      return null; 
     } 
     return jsonObject; 
    } 
    @Override 
    protected void onPostExecute(JSONArray result) 
    { 
     super.onPostExecute(result); 
     final Message msg = new Message(); 
     msg.obj = result; 
    } 
} 

这是我试图完成什么：

JSONArray channels = new Json().execute("http://192.168.2.136:8080/rest/channel/"+ linkid +"/"+ username, "GET"); 
     try { 
      for (int i=0; i < channels.length(); i++) { 
       JSONObject channel_data = channels.getJSONObject(i); 
       String channelID = channel_data.getString("ChannelID").toLowerCase(); 
       JSONArray json = new Json().execute("http://192.168.2.136:8080/rest/program/"+ linkid +"/"+ username +"/" + channelID, "GET"); 

Answer 1

您从AsyncTask开始并不是return。你指示AsyncTask在一天之前给它做些什么，但是不会有任何return给你。这就是为什么它被称为“异步”：你不会等待它，它不会等待你。

例如，借此代码与SyncTask：

result = SyncTask(); 
label.setText(result); 

这意味着setText()线将不被执行，直到SyncTask()完成并产生一result。它是同步的。相反，与异步，你做：

new AsyncTask() { 
    @Override 
    void onPostExecute(result) { 
     label.setText(result) 
    } 
}.start() 

这带来了一个全新的world of trouble。我建议你看看Loaders，它的工作方式相似，但提供了更强的抽象。

此外，我告诉你这一事实意味着有很多事情你不明白。你可能想要谷歌相关的文档，教程或文章。

Answer 2

从Asynctask执行（Runnable runnable）将返回void。

将您的结果onPostExecute（），像这样：

渠道=结果; DoSomething的（渠道）

Answer 3

你不必从的AsyncTask返回任何

@Override 
protected void onPostExecute(JSONArray result) 
{ 
    super.onPostExecute(result); 
    channels = result 
    //<here you can use channels to integrate with other code> 
} 

这里的通道将被宣布为类变量

JSONArray channels;