警报对话框冻结了我的应用 [英] Alert dialog freeze my app

问题描述

这里有代码检查Internet连接，然后如果isconnected为false，则在allert对话框中显示一条消息.问题是，如果isconnected为true，并且(我尝试放置if(isconnected)子句，而不是if(！isconnected))，每个方法都可以工作.但是，如果我在VM执行show()每次冻结时都关闭了手机上的每个网络.为什么?感谢所有人:

Here there is the code that check the internet connection and then if the isconnected is false show a message in a allert dialog. The problem is that if isconnected is true and (i tried to put if(isconnected) clause instead if(!isconnected)) every works. But if i put off every network on my phone when VM execute show() every freeze. Why? Thanks to all:

final AlertDialog.Builder dialog= new AlertDialog.Builder(this);


    ((Button)findViewById(R.id.listabutton)).setOnClickListener(new View.OnClickListener() {
        @Override
        public void onClick(View v) {

            AsyncTask asyncTask= new AsyncTask() {
                boolean isconnected=true;
                @Override
                protected Object doInBackground(Object[] params) {


                        ConnectivityManager conMgr = (ConnectivityManager)getApplicationContext().getSystemService(Context.CONNECTIVITY_SERVICE);

                        NetworkInfo activeNetwork = conMgr.getActiveNetworkInfo();
                        isconnected = activeNetwork != null &&
                                activeNetwork.isConnectedOrConnecting();
                    Log.i("StateNet",isconnected+"");


                    return null;
                }

                @Override
                protected void onPostExecute(Object o) {
                    if(!isconnected){

                        dialog.setMessage("Controlla la tua conessione a internet")
                                .setTitle("Ops problemino con internet")
                                .setPositiveButton("Ok", ok)
                                .show();

                    }

                    super.onPostExecute(o);
                }

            };

推荐答案

您不需要AsyncTask来检查Internet连接，因此请保持简单.

You do not need AsyncTask to check the internet connection, so please keep it simple.

我认为您的onClick方法应如下所示:

I think that your onClick method should looks like this:

@Override
public void onClick(View v) {
      if (isConnectionAvailable(context)) {
          // connected
      } else {
          // not connected
      }
};

public static boolean isConnectionAvailable(Context context) {
    ConnectivityManager manager = (ConnectivityManager) context.getSystemService(Context.CONNECTIVITY_SERVICE);
    NetworkInfo networkInfo = manager.getActiveNetworkInfo();
    return networkInfo != null && networkInfo.isAvailable() && networkInfo.isConnected();
}

不要忘记清单中的ACCESS_NETWORK_STATE和INTERNET权限.

Do not forget the ACCESS_NETWORK_STATE and INTERNET permissions in your Manifest.

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