Question

我正在创建popupwindow，如下所示

@Override
public void onBackPressed() {
    // TODO Auto-generated method stub
    super.onBackPressed();
        LinearLayout linearLayout = new LinearLayout(TimerGameActivity.this);
        TextView textView = new TextView(TimerGameActivity.this);
        textView.setText("Quit? Score will be lost....");
        linearLayout.addView(textView);
        PopupWindow popupWindow = new PopupWindow(linearLayout, 200, 100,true);
        popupWindow.showAtLocation(linearLayout, Gravity.BOTTOM, 10, 10)
}

但我面临的问题如下。当我按下后退按钮时，它不会让我显示弹出窗口并破坏活动，它会给我以下错误：

04-18 15:04:55.457: E/WindowManager(590): Activity has leaked window android.widget.LinearLayout@44f88be8 that was originally added here

帮帮我。感谢

Answer 1

您需要删除：

super.onBackPressed();

Answer 2

很明显在那里得到一个例外，因为当你按下后退按钮时super.onBackPressed();被激活，你的Activity将会完成，同时你试图显示一个PopupWindow。因此，没有用于显示PopupWindow的UI。因此，只需删除super.onBackPressed();并尝试显示PopupWindow。

Answer 3

onBackPressed()将破坏您的活动，同时您参与同一活动进行UI操作，那么它将如何运作？

参考Which actions does the back button/back key on Android trigger?

popupwindow：窗口被后退按钮泄露

3 个答案: