Question

你好我有这个代码来显示弹出窗口......

final WindowManager.LayoutParams parameters = new WindowManager.LayoutParams(
            (width/100)*50, height, WindowManager.LayoutParams.TYPE_PHONE,
            WindowManager.LayoutParams.FLAG_NOT_FOCUSABLE,
            PixelFormat.TRANSLUCENT);
parameters.gravity = Gravity.CENTER | Gravity.CENTER;
parameters.x = (width/100)*50;
parameters.y = 0;

但我需要在屏幕外显示它... 当我将x位置设置为更大，然后将窗口大小设置在角落时，它不会越过屏幕......

感谢。

编辑：因为有些人不知道我的意思..我的意思是一半外屏幕不满..

Answer 1

您需要再添加一个flag_layout_no_limits

标记

final WindowManager.LayoutParams parameters = new WindowManager.LayoutParams(
            (width/100)*50, height, 
            WindowManager.LayoutParams.TYPE_PHONE,
            WindowManager.LayoutParams.FLAG_LAYOUT_NO_LIMITS | WindowManager.LayoutParams.FLAG_NOT_FOCUSABLE,
            PixelFormat.TRANSLUCENT);  

parameters.gravity = Gravity.CENTER | Gravity.CENTER;
parameters.x = (width/2);
parameters.y = 0;

将弹出窗口移出屏幕

1 个答案: