Android程序错误连接

Question

最终，我的目标是获得一个Android应用程序，该应用程序从URL接收XML文件，对其进行解析并显示相关信息（当然，最终目标比这更复杂，但是是当前的目标）。我正在使用Wei-Meng Lee的“Beginning Android Application Development”中的示例，并且我已经注意到本书示例代码中的一些错误（尤其是错误变量，如果Eclipse没有指出它，我会错过它完全）。

然而，目前，尽管有权限访问（3g，4g和wifi已全部测试过），但该应用程序仍无法连接到互联网，并且能够通过网络访问测试网址，板载浏览器。

以下是相关的代码片段。我做错了什么？

private InputStream OpenHttpConnection(String urlString) 
throws IOException 
{ 
    InputStream in = null; 
    int response = 01; 

    URL url = new URL(urlString); 
    URLConnection conn = url.openConnection(); 

    if (!(conn instanceof HttpURLConnection)) 
     throw new IOException("Not an HTTP connection"); 
    try{ 
     HttpURLConnection httpConn = (HttpURLConnection) conn; 
     httpConn.setAllowUserInteraction(false); 
     httpConn.setInstanceFollowRedirects(true); 
     httpConn.setRequestMethod("Get"); 
     httpConn.connect(); 
     response = httpConn.getResponseCode(); 
     if (response == HttpURLConnection.HTTP_OK){ 
      in = httpConn.getInputStream(); 
     } 
    } 
    catch (Exception ex) 
    { 
     throw new IOException("Error connecting"); 
    } 
    return in; 
} 

我曾尝试不同的URL：（请注意我都在模拟器和我的Galaxy S2测试），检查我可以连接到他们每个人在手机的浏览器中使用它们。模拟器也没有任何运气。

UPDATE：利用下面的非同步代码：

private class BackgroundTask extends AsyncTask 
<String, Void, Bitmap> { 
    protected Bitmap doInBackground(String... url){ 
     // download an image 
     Bitmap bitmap = DownloadImage(url[0]); 
     return bitmap; 
    } 
} 

protected void onPostExecute(Bitmap bitmap) { 
    ImageView img = (ImageView) findViewById(R.id.img); 
    img.setImageBitmap(bitmap); 
} 

方法调用上面：

protected void onCreate(Bundle savedInstanceState) { 
    super.onCreate(savedInstanceState); 
    setContentView(R.layout.activity_main); 
    new BackgroundTask().execute("http://www.google.com/intl/en_ALL/images/logos/images_logo_lg.gif"); 

} 

Answer 1

您想使用的AsyncTask用于需要网络连接的任何东西。你可以设置你的异步如下：（这需要一个字符串作为参数，并返回一个InputStream）

public class OpenHttpConnection extends AsyncTask<String, Void, InputStream> { 

    @Override 
    protected String doInBackground(String... params) { 
     String urlstring = params[0]; 
     InputStream in = null; 
     int response = 01; 

     URL url = new URL(urlString); 
     URLConnection conn = url.openConnection(); 

     if (!(conn instanceof HttpURLConnection)) 
     throw new IOException("Not an HTTP connection"); 
     try{ 
     HttpURLConnection httpConn = (HttpURLConnection) conn; 
     httpConn.setAllowUserInteraction(false); 
     httpConn.setInstanceFollowRedirects(true); 
     httpConn.setRequestMethod("Get"); 
     httpConn.connect(); 
     response = httpConn.getResponseCode(); 
     if (response == HttpURLConnection.HTTP_OK){ 
      in = httpConn.getInputStream(); 
     } 
     } 
     catch (Exception ex) 
     { 
     throw new IOException("Error connecting"); 
     } 
     return in; 

    } 

} 

然后就可以调用/运行你异步这样。

OpenHttpConnection connection = new OpenHttpConnection().execute("http://YourURL.com"); 
InputStream is = connection.get();