嘿,我在我的应用程序中使用HttpUrlConnection。在我看来,每次我打.getInputStream()或urlConnection.getResponseCode()等电话时,都会提出另一个请求,所以当我发出帖子请求时,这对我没有好处。是否有一种方法可以获得某种类型的响应对象,该对象封装响应数据并可以从UI线程访问,如下所示:
private class RegisterAsync extends AsyncTask<String, Void, HttpResponse> {
protected String doInBackground(String... strings) {
HttpURLConnection urlConnection = null;
String message = null;
try {
URL url = new URL(REGISTER_URL);
urlConnection = (HttpURLConnection) url.openConnection();
urlConnection.setDoOutput(true);
urlConnection.setChunkedStreamingMode(0);
OutputStream out = new BufferedOutputStream(urlConnection.getOutputStream());
out.write(strings[0].getBytes("UTF-8"));
out.flush();
out.close();
HttpResponse response = urlConnection.getResponse();
} catch (MalformedURLException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
} finally {
urlConnection.disconnect();
}
return Response;
}
protected void onPostExecute(HttpResponse response) {
//Do some with response object: get status, headers, content etc?
}
}发布于 2013-06-25 15:13:32
您可以执行以下步骤:
onReceive()方法BroadcastReceiver运行在UI线程上,因此您可以在那里快速更新UI。在您实现之后,您可以简单地从您的活动调用单例webservice调用方法,就在您注册到将从单例发送的意图之后.
https://stackoverflow.com/questions/17293463
复制相似问题