Question

我有一个获取JSON数组的AsyncTask。我将如何返回JSON数组：

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

Eclips告诉我，我不能那样做，应该是：

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

Json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;
    }
}

这就是我想要完成的事情：

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

你不是来自return的{{1}}。你指示AsyncTask在每天打电话之前做一些事情，但它不会给你AsyncTask任何事情。这就是为什么它被称为异步＆＃34;：你不等待它，它不会等你。

例如，将此代码与return：

一起使用

SyncTask

这意味着result = SyncTask(); label.setText(result);行不会被执行，直到setText()完成并产生SyncTask()。它是同步的。相反，使用异步，您可以：

result

这带来了全新的world of trouble。我建议你看看new AsyncTask() { @Override void onPostExecute(result) { label.setText(result) } }.start()，它的工作方式相似但提供了更强大的抽象。

此外，我告诉你这一事实意味着你不了解很多事情。您可能想要了解相关文档，教程或文章。

Answer 2

您不必从ASyncTask返回任何内容

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

这里的频道将被声明为类变量

 JSONArray channels;

Answer 3

来自Asynctask的

execute（Runnable runnable）返回void。

在onPostExecute（）中分配结果，如下所示：

渠道=结果; doSomething的（信道）

从AsyncTask返回JSON数组

3 个答案: