捕获在Android中启用互联网时用户正在键入的内容

时间:2014-08-07 05:49:43

标签: android service

我正在尝试创建一个密钥记录器类型的应用程序,供家长在互联网打开时在他们的孩子手机上使用。主要是聊天和上网。我知道有应用程序,但我需要自学和构建一个。我尽可能地用我的小知识准备了一个代码。但我不确定我是否可以通过在后台运行来捕获所有键盘输入事件。

接收者类

public void onReceive(Context context, Intent intent) {
        // TODO Auto-generated method stub
        ConnectivityManager connMgr = (ConnectivityManager) context
                .getSystemService(Context.CONNECTIVITY_SERVICE);
        NetworkInfo wifi = connMgr
                .getNetworkInfo(ConnectivityManager.TYPE_WIFI);
        NetworkInfo mobile = connMgr
                .getNetworkInfo(ConnectivityManager.TYPE_MOBILE);

        boolean isConnected = wifi != null && wifi.isConnectedOrConnecting()
                || mobile != null && mobile.isConnectedOrConnecting();
        if (isConnected) {
            Log.d("Network Available ", "Yes");
            Intent i = new Intent(context, services.class);
            context.startService(i);
        } else {
            Log.d("Network Available ", "NO");
        }

    }

服务类

public class services extends Service{

    @Override
    public void onCreate() {
        Log.i("worked", "A");
    }

    public IBinder onBind(Intent intent) {
        return null;
    }

    boolean doKeyDown(int keyCode, KeyEvent msg) {
        switch (keyCode) {
        case KeyEvent.KEYCODE_A:
            Log.i("Pressed : ", "A");
            break;
        case KeyEvent.KEYCODE_B:
            Log.i("Pressed : ", "B");
            break;
        case KeyEvent.KEYCODE_C:
            Log.i("Pressed : ", "C");
            break;
        case KeyEvent.KEYCODE_D:
            Log.i("Pressed : ", "D");
            break;
        case KeyEvent.KEYCODE_E:
            Log.i("Pressed : ", "E");
            break;
        case KeyEvent.KEYCODE_F:
            Log.i("Pressed : ", "F");
            break;
        case KeyEvent.KEYCODE_G:
            Log.i("Pressed : ", "G");
            break;
        case KeyEvent.KEYCODE_H:
            Log.i("Pressed : ", "H");
            break;
        }
        return false;
    }
}

0 个答案:

没有答案