Question

最终，我的目标是获得一个Android应用程序，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，但这是当前的目标）。我正在使用Wei-Meng Lee的“开始Android应用程序开发”中的示例，我已经注意到本书示例代码中的一些错误（特别是一个错误名称变量，如果Eclipse没有指出它我会错过它完全地）。

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都经过测试），并且能够到达机载测试URL浏览器。

以下是相关的代码段。我做错了吗？（注意：我在模拟器和Galaxy S2上测试过）

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;
}

我尝试过不同的网址，在使用之前检查我是否可以在手机浏览器中连接到每个网址。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用于需要网络连接的任何内容。您可以按如下方式设置Async :(这会将String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;

    }

}

然后你可以像这样调用/运行Async。

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

Android程序错误连接

1 个答案: