android - 在弹出窗口外单击时，如何关闭弹出窗口？

Question

我正在尝试制作一个弹出窗口，其中包含文本字段和信息以询问用户，但我想知道如何制作它以便用户可以通过在主要 fragment/Activity 所在的弹出窗口外部单击来关闭它是。

    @Override
    public View onCreateView(LayoutInflater inflater, ViewGroup container, Bundle savedInstanceState) {

        View rootView = inflater.inflate(R.layout.tabstudygroups, container, false);

        listview = (ListView) rootView.findViewById(R.id.clist2);
        addCourseButton = (Button) rootView.findViewById(R.id.caddcoursebutton);

        // do stuff here

        addCourseButton.setOnClickListener(this);

        return rootView;
    }

    @Override
    public void onClick(View v) {
        if(v == addCourseButton) {
            View popupView = LayoutInflater.from(getActivity()).inflate(R.layout.popup_layout, null);
            final PopupWindow popupWindow = new PopupWindow(popupView, WindowManager.LayoutParams.MATCH_PARENT, WindowManager.LayoutParams.MATCH_PARENT);

            // HERE IS WHAT I THOUGHT WOULD MAKE IT BE ABLE TO ENABLE THE OUTSIDE TOUCH
            popupWindow.setBackgroundDrawable(new BitmapDrawable());
            popupWindow.setOutsideTouchable(true);

            Button btn = (Button) popupView.findViewById(R.id.button);

            popupWindow.showAsDropDown(popupView, 0, 0);
        }
    }
}

Answer 1

将您的 PopupWindow 设置为 wrap_content 并使其可聚焦。

final PopupWindow popupWindow = new PopupWindow(popupView, WindowManager.LayoutParams.WRAP_CONTENT, WindowManager.LayoutParams.WRAP_CONTENT);

            // HERE IS WHAT I THOUGHT WOULD MAKE IT BE ABLE TO ENABLE THE OUTSIDE TOUCH
            popupWindow.setOutsideTouchable(true);
            popupWindow.setFocusable(true);
            popupWindow.setBackgroundDrawable(new ColorDrawable(Color.TRANSPARENT));


            Button btn = (Button) popupView.findViewById(R.id.button);

            popupWindow.showAsDropDown(popupView, 0, 0);