Question

我尝试使用popupWindow和自定义布局创建一种popupMenuItem。

我有一个按钮，当我点击它时会显示popupWindow。当我再次单击此按钮或单击popupWindow外部时，我想触发和事件来关闭此popupWindow。

但目前它还没有工作，我的setTouchInterceptor没有被触发，你有想法解决这个问题吗？

每次弹出窗口打开时，即使使用setOutsideTouchable（true），我也无法访问所有其他UI元素。

这是我的代码：

popupMenuImageView是我的下拉按钮（实际上是imageview），它显示了我的弹出窗口

popupWindow我的PopupWindod对象的全局变量。

popupMenuImageView = (ImageView) getView().findViewById(R.id.popup_menu_imageView);

popupMenuImageView.setOnClickListener(new View.OnClickListener() {
    @Override
    public void onClick(View v) {

        if (popupWindow == null) {
            int width = 375;
            int height = 240;

            TableLayout tableLayout = (TableLayout) getView().findViewById(R.id.custom_popup_menu_id);
            LayoutInflater layoutInflater = (LayoutInflater) getActivity().getSystemService(Context.LAYOUT_INFLATER_SERVICE);
            View layout = layoutInflater.inflate(R.layout.custom_popup_menu, tableLayout);

            popupWindow = new PopupWindow(layout, width, height, true);
            popupWindow.setOutsideTouchable(true);
            popupWindow.setFocusable(true);
            popupWindow.setTouchInterceptor(new View.OnTouchListener() {
                @Override
                public boolean onTouch(View v, MotionEvent event) {
                    if(event.getAction()==MotionEvent.ACTION_OUTSIDE){
                        popupWindow.dismiss();
                        return true;
                    }
                    return false;
                }
            });

        }
        popupWindow.showAsDropDown(popupMenuImageView);
    }
});

我也试过这个也不行的：

popupMenuImageView = (ImageView) getView().findViewById(R.id.popup_menu_imageView);
createMenuItem();
popupMenuImageView.setOnClickListener(eventOpenCloseMenuItem);

View.OnClickListener eventOpenCloseMenuItem = new View.OnClickListener(){

        @Override
        public void onClick(View v) {
            if (isShowing) {
                popupWindow.showAsDropDown(popupMenuImageView);
                isShowing = false;
            }else{
                popupWindow.dismiss();
            }
        }
    };

private void createMenuItem（）{

        if (popupWindow == null) {
            int width = 375;
            int height = 240;

            TableLayout tableLayout = (TableLayout) getView().findViewById(R.id.custom_popup_menu_id);
            LayoutInflater layoutInflater = (LayoutInflater) getActivity().getSystemService(Context.LAYOUT_INFLATER_SERVICE);
            View layout = layoutInflater.inflate(R.layout.custom_popup_menu, tableLayout);

            popupWindow = new PopupWindow(layout, width, height, true);
            popupWindow.setOutsideTouchable(false);
            popupWindow.setFocusable(true);
        }

Answer 1

如果您只是想通过点击外部来关闭弹出窗口，则不需要设置拦截器。只需将背景drawable设置为null，例如

popupWindow.setBackgroundDrawable (new BitmapDrawable());
popupWindow.setOutsideTouchable(true);

并删除：

popupWindow.setTouchInterceptor(new View.OnTouchListener() {
            @Override
            public boolean onTouch(View v, MotionEvent event) {
                if(event.getAction()==MotionEvent.ACTION_OUTSIDE){
                    popupWindow.dismiss();
                    return true;
                }
                return false;
            }
        });

请在PopupWindow - Dismiss when clicked outside

检查我对类似问题的回答

Answer 2

我认为在显示弹出窗口之前你需要setBackgroundDrawable()才能处理触摸事件：

popupWindow.setBackgroundDrawable(new ColorDrawable());

事件打开/关闭popupWindow

2 个答案: