java - Android程序连接错误

Question

最终，我的目标是获得一个 Android 应用程序，它从 URL 接收 XML 文件，解析它，并显示相关信息(嗯，最终目标比这复杂得多，但这就是当前的目标)。我使用的是 Wei-Meng Lee 所著的《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(特别是错误命名的变量，如果 Eclipse 没有指出它，我就会错过它)完全)。

然而，目前该应用程序无法连接到互联网，尽管有权限、有访问权限(3g、4g 和 wifi 均已测试)并且能够访问机载中的测试 URL浏览器。

以下是相关的代码 fragment 。我做错了什么吗？ (注:我在模拟器和 Galaxy S2 上进行了测试)

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;
}

我尝试了不同的 URL，在使用它们之前检查是否可以在手机浏览器中连接到每个 URL。模拟器也没有任何运气。

更新:使用以下内容的异步代码:

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);
}

上面的方法调用:

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

您希望将 AsyncTask 用于任何需要网络连接的事情。您可以按如下方式设置异步:(这需要一个字符串作为参数并返回一个输入流)

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;

    }

}

然后你可以像这样调用/运行异步。

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