安卓popupwindow浮动父视图外 [英] android popupwindow float outside of parent view

问题描述

而不是解雇一个弹出式的，我想隐藏它关闭屏幕上的位置 - 宽度，-heigt。当我尝试更新弹出关闭屏幕它被停在父视图的边界。如何更新它关闭屏幕或半屏幕外？

Instead of dismissing a popup, I would like to hide it off screen at position -width,-heigt. When I try to update the popup off screen it gets stopped at the bounds of the parent view. How do I update it off screen, or semi offscreen?

   View view = LayoutInflater.from(getBaseContext()).inflate(R.layout.mylayout,null);
    pop = new PopupWindow(this);
    pop.setTouchable(false);
    pop.setHeight(200);
    pop.setWidth(200);
    pop.setContentView(view);
    pop.showAtLocation(myparentview, 0, 50, 50);    
    pop.update(-200,-200,-1,-1);

问题快照：

Snapshot of problem:

推荐答案

我觉得你缺少的是用setClippingEnabled假。什么

I think what you are missing is using the setClippingEnabled with false.

<一个href="http://developer.android.com/reference/android/widget/PopupWindow.html#setClippingEnabled(boolean">http://developer.android.com/reference/android/widget/PopupWindow.html#setClippingEnabled(boolean)

您应该更新调用之前调用它，这应该让你绘制窗口画面bounderies之外。

You should call it before the update call and this should let you paint the window outside screen bounderies.

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