从IME: BadTokenException创建PopupWindow

Question

我正在尝试从IME服务(屏幕键盘)显示一个没有活动的弹出窗口。当我调用popUp.showAtLocation(layout，Gravity.CENTER，0，-100)时，我得到一个"Windowmanager$BadTokenException: Unable to add window -- token null is is“。我知道从没有关联活动的服务中打开弹出窗口有点不寻常--这可能吗？

下面是我的代码：

public void initiatePopupWindow()
{
    try {

        LayoutInflater inflater = (LayoutInflater) this.getSystemService(Context.LAYOUT_INFLATER_SERVICE);
        View layout = inflater.inflate(R.layout.popup_layout,null);

        // create a 300px width and 470px height PopupWindow
        popUp = new PopupWindow(layout, 300, 470, true);
        popUp.showAtLocation(layout, Gravity.CENTER, 0, -100);


        Button cancelButton = (Button) layout.findViewById(R.id.popup_cancel_button);
        cancelButton.setOnClickListener(inputView.cancel_button_click_listener);

    } catch (Exception e) {
        e.printStackTrace();
    }
}

任何帮助都将不胜感激。谢谢

Answer 1

我发现了问题--我试图使用布局绘制弹出窗口，并使用与父级相同的布局。解决方案是将父视图设置为另一个视图。我发现在创建弹出窗口之前，确保视图已经创建(例如，使用处理程序/runnable)也很重要。

public void initiatePopupWindow()
{
    try {
        Log.i("dotdashkeyboard","initiatePopupWindow (from IME service)");
        LayoutInflater inflater = (LayoutInflater) this.getBaseContext().getSystemService(Context.LAYOUT_INFLATER_SERVICE);
        View layout = inflater.inflate(R.layout.popup_layout,null);

        // create a 300px width and 470px height PopupWindow
        popUp = new PopupWindow(layout, 300, 470, false);
        popUp.showAtLocation(inputView, Gravity.CENTER, 0, -100);

        Button cancelButton = (Button) layout.findViewById(R.id.popup_cancel_button);
        cancelButton.setOnClickListener(inputView.cancel_button_click_listener);

    } catch (Exception e) {
        e.printStackTrace();
    }
}