辞退PopupWindow上弹出外触控，无需使用pcated德$ P $构造 [英] Dismiss PopupWindow on touch outside popup, without using deprecated constructor

问题描述

我有一个PopupWindow，我想，当用户触摸外面解雇，所以我调查并发现，我不得不使用 popup.setBackgroundDrawable（新BitmapDrawable（））; 。问题在于构造新BitmpaDrawable（）是德precated。我想找到一个解决方案，而无需使用它。

任何人知道如何解决此问题？

谢谢！

 最后PopupWindow弹出=新PopupWindow（sortByView，
                                          ViewGroup.LayoutParams.WRAP_CONTENT，
                                          ViewGroup.LayoutParams.WRAP_CONTENT，
                                          真正）;
                popup.setBackgroundDrawable（新BitmapDrawable（））;
                popup.setOutsideTouchable（真正的）;
                popup.showAsDropDown（五）;
 

解决方案

我不得不这样做，以得到它的工作：

  popup.setBackgroundDrawable（新ColorDrawable（android.R.color.transparent））;
popup.setOutsideTouchable（真正的）;
 

I have a PopupWindow and I wanted it to dismiss when the user touches outside, so I looked into and found out that I had to use popup.setBackgroundDrawable(new BitmapDrawable());. The problem is that the constructor new BitmpaDrawable() is deprecated. I Would like to find a solution without using it.

Anybody knows how to solve this?

Thanks!

                final PopupWindow popup = new PopupWindow(sortByView,
                                          ViewGroup.LayoutParams.WRAP_CONTENT,
                                          ViewGroup.LayoutParams.WRAP_CONTENT,            
                                          true);
                popup.setBackgroundDrawable(new BitmapDrawable());
                popup.setOutsideTouchable(true);
                popup.showAsDropDown(v);

解决方案

What I had to do to get it to work:

popup.setBackgroundDrawable(new ColorDrawable(android.R.color.transparent));
popup.setOutsideTouchable(true);

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