Question

我有一个popupWindow，我用它来显示谷歌地图，但一旦它打开我似乎无法关闭它，我试图设置后退按钮关闭它但不能调用onBackButtonClick覆盖，因为我在片段内。似乎没有做任何事情

   @Override
    public View onCreateView(LayoutInflater inflater, ViewGroup container,
                             Bundle savedInstanceState) {
    View rootView = inflater.inflate(R.layout.fragment_go, container, false);
    mRouteBtn = (Button) rootView.findViewById(R.id.routeBtn);

    mRouteBtn.setOnClickListener(new View.OnClickListener() {
        @Override
        public void onClick(View view) {
            View popupView = getActivity().getLayoutInflater().inflate(R.layout.fragment_map, null);

            mPopupWindow = new PopupWindow(popupView,
                    ViewGroup.LayoutParams.WRAP_CONTENT, ViewGroup.LayoutParams.WRAP_CONTENT);

            // If the PopupWindow should be focusable
            mPopupWindow.setFocusable(true);
            mPopupWindow.showAtLocation(getActivity().findViewById(R.id.routeBtn), Gravity.CENTER, 0, 0);
            mPopupWindow.update(0, 0, 800, 800); //don't hardcode

后退按钮代码部分：

            mPopupWindow.getContentView().setOnKeyListener( new View.OnKeyListener(){
                @Override
                public boolean onKey( View v, int keyCode, KeyEvent event ){
                    if( keyCode == KeyEvent.KEYCODE_BACK ){
                        mPopupWindow.dismiss();
                        return true;
                    }
                    return false;
                }
            } );
        }
    });

好的排序，在按照 Deepak Singh的建议后我开始工作了，虽然当我重新打开它时，由于尝试重新创建仍然存在的视图，我遇到了崩溃的问题。这是我的工作代码供将来参考：

固定版本：

@Override
public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
}

@Override
public View onCreateView(LayoutInflater inflater, ViewGroup container,
                         Bundle savedInstanceState) {
    View rootView = inflater.inflate(R.layout.fragment_go, container, false);

    mRouteBtn = (Button) rootView.findViewById(R.id.routeBtn);

    mRouteBtn.setOnClickListener(new View.OnClickListener() {
        @Override
        public void onClick(View view) {
            if (mPopupView != null) {
                ViewGroup parent = (ViewGroup) mPopupView.getParent();
                if (parent != null)
                    parent.removeView(mPopupView);
            }
            try {
                mPopupView = getActivity().getLayoutInflater().inflate(R.layout.fragment_map, null);
            } catch (InflateException e) {
                /* map is already there, just return view as it is */
            }
            mPopupWindow = new PopupWindow(mPopupView,
                    ViewGroup.LayoutParams.WRAP_CONTENT, ViewGroup.LayoutParams.WRAP_CONTENT);

            mPopupWindow.setBackgroundDrawable(new BitmapDrawable());
            mPopupWindow.setOutsideTouchable(true);
            mPopupWindow.showAtLocation(getActivity().findViewById(R.id.routeBtn), Gravity.CENTER, 0, 0);
            mPopupWindow.update(0, 0, 800, 800); //don't hardcode

            mPopupView.findViewById(R.id.doneBtn).setOnClickListener(new View.OnClickListener() {
                @Override
                public void onClick(View view) {
                    mPopupWindow.dismiss();
                }
            });
        }
    });

Answer 1

在弹出式窗口布局中添加一个按钮，然后使用popupView.findviewbyid(id)找到按钮的ID，然后在此按钮上设置 clicklisener 并致电mPopupWindow.dismiss();。

你也可以试试这个：

 popupView.setOutsideTouchable(true);
 popupView.setFocusable(true);

Answer 2

添加到您的代码popupWindow.setOutsideTouchable(true);

Answer 3

根据this answer，你需要在弹出窗口中设置一个可绘制的背景。

我测试了它，你甚至不需要自己处理OnKeyListener。

这足以使'后退按钮'工作：

popupWindow.setBackgroundDrawable(new ColorDrawable());
popupWindow.setFocusable(true);

用后退按钮关闭PopupWindow？

3 个答案: