如何检测一个";最近的活动选项卡按钮";从服务类(Android)中按?

如何检测一个";最近的活动选项卡按钮";从服务类(Android)中按?,android,dialog,android-input-method,Android,Dialog,Android Input Method,这是我的代码: public void displayQuickSettingsDialog() { if(quickSettingsDialog == null) { quickSettingsDialog = new Dialog(mContext); quickSettingsDialog.requestWindowFeature(Window.FEATURE_NO_TITLE); Window wi

这是我的代码:

public void displayQuickSettingsDialog() {
        if(quickSettingsDialog == null) {

            quickSettingsDialog = new Dialog(mContext);
            quickSettingsDialog.requestWindowFeature(Window.FEATURE_NO_TITLE);
            Window win = quickSettingsDialog.getWindow();
            WindowManager.LayoutParams lp = quickSettingsDialog.getWindow().getAttributes();
            lp.token = mKptAdaptxtIME.mInputView.getWindowToken();
            lp.type = WindowManager.LayoutParams.TYPE_APPLICATION_ATTACHED_DIALOG;
            lp.dimAmount = 0.5f;
            lp.y = lp.y + 150;
            win.setAttributes(lp);
            win.setBackgroundDrawable(new ColorDrawable(Color.TRANSPARENT));
            win.addFlags(WindowManager.LayoutParams.FLAG_DIM_BEHIND);
            win.addFlags(WindowManager.LayoutParams.FLAG_ALT_FOCUSABLE_IM);



            quickSettingsDialog.setContentView(R.layout.quicksettings);
            quickSettingsDialog.setCanceledOnTouchOutside(true);

            //listeners for dialog options
            quickSettingsDialog.findViewById(R.id.quick_sett_lang).setOnClickListener(this);
            quickSettingsDialog.findViewById(R.id.quick_sett_keyboard_type).setOnClickListener(this);
            quickSettingsDialog.findViewById(R.id.quick_sett_sugg_control).setOnClickListener(this);
            quickSettingsDialog.findViewById(R.id.quick_sett_settings).setOnClickListener(this);

        }

        quickSettingsDialog.show();
    }
它会显示一个警报对话框,在显示该对话框后,当我按下“最近的活动”选项卡(4.0)时,一切正常,如果我选择了其他活动,则该对话框仍不会关闭,我正在从输入法服务调用此方法…在一个服务中,当我按下“最近的活动”按钮时,将调用哪个回调?需要快速帮助…提前感谢androidians..

标题为“主页按键”,但描述为“当我按下“最近的活动”按钮时”。请你纠正一下错的那一个好吗。