Android 为什么可以'；将数据从internet加载到listview中

我使用自定义适配器和AsyncTask读取数据并将其放入listview。但它不起作用

Thread t;
int res;
ArrayList<mLink> values;
AdapterLink adapter;
ListView lv;
String title;
InputStream is;
private ReadXML readXML;
TextView status;
@Override
public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);

    status = (TextView)findViewById(R.id.textView1);
    lv = (ListView)findViewById(R.id.lView);
    values = new ArrayList<mLink>();       
    readXML = new ReadXML(getApplicationContext()); 
    load(); 
}

此代码不是错误，但它不会在列表视图中显示数据

new DownloadWebpageText().execute(stringUrl) 

这是你的问题。

---

函数

doInBackground

返回您的输入流。将此输入流保存到变量中并将其传递到适配器中

我们需要阅读大量代码。您是否遇到任何异常。如果是，那么在这里发布logcat报告。我没有得到任何例外。但它不会在我的布局中加载任何内容。我不保存它，我只是从internet上读取它并将其加载到ListView中。

public void load(){
    String stringUrl = "http://vnexpress.net/rss/gl/trang-chu.rss";
    ConnectivityManager connMgr = (ConnectivityManager) 
        getSystemService(Context.CONNECTIVITY_SERVICE);
    NetworkInfo networkInfo = connMgr.getActiveNetworkInfo();
    if (networkInfo != null && networkInfo.isConnected()) {
        new DownloadWebpageText().execute(stringUrl);
    } else {
        status.setText("No network connection available.");
    }
    adapter = new AdapterLink(getApplicationContext(), R.layout.item_layout, values);
    lv.setAdapter(adapter);
}


public InputStream connectT3H(String mURL){

    int responseCode = 0;

    InputStream in = null;
    try {
        URL url = new URL(mURL);

        URLConnection connection = url.openConnection();

        HttpURLConnection httpURLConnection = (HttpURLConnection)connection;

        responseCode = httpURLConnection.getResponseCode();

        if (responseCode == HttpURLConnection.HTTP_OK){
            in = httpURLConnection.getInputStream();
        }   
    } catch (MalformedURLException e) {
        e.printStackTrace();
    } catch (IOException e) {
        e.printStackTrace();
    }   
    return in;
}   

new DownloadWebpageText().execute(stringUrl)