rav*_*024 2 android android-service android-layout android-view android-windowmanager
我正在尝试创建一个覆盖窗口,但是当我尝试将视图添加到 WindowManager 时,它给了我一个异常。我添加了“SYSTEM_ALERT_WINDOW”权限,并在应用信息中启用了“绘制其他应用”。我从服务的 onCreate 函数中调用它。
代码:
WindowManager manager = (WindowManager)getSystemService(WINDOW_SERVICE);
LayoutInflater inflater = (LayoutInflater) getSystemService(LAYOUT_INFLATER_SERVICE);
RelativeLayout overlay = (RelativeLayout) inflater.inflate(R.layout.button_main, null);
final WindowManager.LayoutParams params =
new WindowManager.LayoutParams(
WindowManager.LayoutParams.MATCH_PARENT,
WindowManager.LayoutParams.WRAP_CONTENT,
WindowManager.LayoutParams.TYPE_APPLICATION_PANEL,
0,
PixelFormat.TRANSLUCENT);
params.gravity = Gravity.TOP | Gravity.START;
params.x = 0;
params.y = 0;
manager.addView(overlay, params);
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异常堆栈跟踪:
Caused by: android.view.WindowManager$BadTokenException: Unable to add window -- token null is not valid; is your activity running?
at android.view.ViewRootImpl.setView(ViewRootImpl.java:764
at android.view.WindowManagerGlobal.addView(WindowManagerGlobal.java:356)
at android.view.WindowManagerImpl.addView(WindowManagerImpl.java:92)
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无论我为 LayoutParams 使用什么类型,我总是会遇到这种崩溃。
小智 6
使用 TYPE_APPLICATION_OVERLAY。它是 Android O 允许在其他应用上显示的唯一窗口类型。
看看这个解释:https : //developer.android.com/preview/behavior-changes.html#cwt