使用三星手机的未知问题

时间:2014-03-06 06:21:20

标签: android performance web-services android-wifi galaxy

我在android中开发了一个应用程序。该应用程序包括通过webservice进行的服务器和客户端交互。我检查了wifi连接状态,如果连接状态为false,我已经显示了注意框,如果连接状态为true,我将执行一些操作。但我遇到的问题是LG,索尼和其他智能手机上的注意框都显示在除SA​​MSUNG之外的所有型号,包括Galaxy S4。

任何人都可以知道这个......

我将发布下面的代码,检查wifi,4g和状态。

boolean connected = false;
     ConnectivityManager connectivityManager = (ConnectivityManager)getSystemService(Context.CONNECTIVITY_SERVICE);
     if(connectivityManager.getNetworkInfo(ConnectivityManager.TYPE_MOBILE).getState() == NetworkInfo.State.CONNECTED || 
             connectivityManager.getNetworkInfo(ConnectivityManager.TYPE_WIFI).getState() == NetworkInfo.State.CONNECTED||
             connectivityManager.getNetworkInfo(ConnectivityManager.TYPE_WIMAX).getState() == NetworkInfo.State.CONNECTED)
        {
     // some action
            //we are connected to a network
            connected = true;
        }
    if(connected == false){
    //      System.out.println("The Network error");
            attentionBox("There is no network connection right now. Please try again later.","Internet Connection Error"); 
        } 

和注意框编码是:

 public void attentionBox(final String msg, final String title)
    {
        AlertDialog.Builder builder = new AlertDialog.Builder(ViewYourPlanList.this);
        builder.setMessage(msg)
                .setTitle(title)
                .setPositiveButton("OK", new android.content.DialogInterface.OnClickListener() {
                    public void onClick(DialogInterface dialog, int which) {
                        ((AlertDialog)dialog).getButton(which).setVisibility(View.INVISIBLE);
                         //finish();  
                         System.exit(0);
                    }
                });
        AlertDialog alert = builder.create();
        alert.show();
    }

2 个答案:

答案 0 :(得分:0)

尝试这样

    public boolean isConnectingToInternet() {
       ConnectivityManager connectivity = (ConnectivityManager) _context.getSystemService(Context.CONNECTIVITY_SERVICE);
    if (connectivity != null) {
        NetworkInfo[] info = connectivity.getAllNetworkInfo();
        if (info != null)
            for (int i = 0; i < info.length; i++)
                if (info[i].getState() == NetworkInfo.State.CONNECTED) {
                    return true;
                }

    }
    return false;
}

答案 1 :(得分:0)

试试这个,它为我工作

public void showAlertDialog(Context context, String title, String message) {
    AlertDialog alertDialog = new AlertDialog.Builder(context).create();

    // Setting Dialog Title
    alertDialog.setTitle(title);

    // Setting Dialog Message
    alertDialog.setMessage(message);


    // Setting OK Button
    alertDialog.setButton("OK", new DialogInterface.OnClickListener() {
        public void onClick(DialogInterface dialog, int which) {
        }
    });

    // Showing Alert Message
    alertDialog.show();
}

如果它对您有用。投票给我。