Java onClickListener用于在弹出窗口外进行触摸

从这个问题中，我学会了如何在这之前隐藏popupWindow，popupWindow；在下面的示例中，代码不起作用

private void initFAB(){
        fab = (FloatingActionButton) findViewById(R.id.fab);
        fab.setOnClickListener(new View.OnClickListener() {

        boolean showPopupWindow = false;

        public void onClick(View view) {

            LayoutInflater layoutInflater = (LayoutInflater) getBaseContext()
                    .getSystemService(LAYOUT_INFLATER_SERVICE);
            View popupWindowContent = layoutInflater.inflate(R.layout.popup_window_content, null);
            final PopupWindow popupWindow = new PopupWindow(popupWindowContent,
                    ViewGroup.LayoutParams.WRAP_CONTENT, ViewGroup.LayoutParams.WRAP_CONTENT);
            popupWindow.setOutsideTouchable(true);

            if (!showPopupWindow){
                rotateFabForward();

                View.OnClickListener onClickListener = new View.OnClickListener() {
                    public void onClick(View view) {
                        //...
                    }
                };

                // ...

                popupWindow.showAtLocation(fab, Gravity.END | Gravity.BOTTOM, 50, 400);
                showPopupWindow = true;
            }
            else{
                rotateFabBackward();
                popupWindow.dismiss();
                showPopupWindow = false;
            }
        }
    });
}

现在，当我在弹出窗口外单击时，它消失了，FloatingActionButton不向后旋转，当弹出窗口隐藏时，它有+图标，旋转45度，当弹出窗口可见时，图标变为×

有允许在弹出窗口外注册触摸并执行RotateFabbBackward的侦听器吗

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另外，谢谢你回答关于晶圆厂轮换的问题。

你应该像那样添加onDismiss监听器

popupWindow.setOnDismissListener(new PopupWindow.OnDismissListener() {
    @Override
    public void onDismiss() {
        rotateFabBackward();
        showPopupWindow = false;
        // end may TODO anything else                   
    }
});