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Android Studio中窗口类型2002的权限被拒绝_Android_Android Permissions - Fatal编程技术网

Android Studio中窗口类型2002的权限被拒绝

Android Studio中窗口类型2002的权限被拒绝,android,android-permissions,Android,Android Permissions,今天我安装了最新版本的Android Studio 我正在学习Android中的浮动小部件 我从应用这个例子开始 它编译的很好 但是当我在模拟器中运行它时,它崩溃了 告诉我这个错误 08-28 22:52:02.932 7400-7400/com.asmgx.MyApp.MyApp E/AndroidRuntime: FATAL EXCEPTION: main Process: com.asmgx.MyApp.MyApp, PID: 7400 java.lang.Runtime

今天我安装了最新版本的Android Studio

我正在学习Android中的浮动小部件

我从应用这个例子开始

它编译的很好

但是当我在模拟器中运行它时,它崩溃了

告诉我这个错误

08-28 22:52:02.932 7400-7400/com.asmgx.MyApp.MyApp E/AndroidRuntime: FATAL EXCEPTION: main
    Process: com.asmgx.MyApp.MyApp, PID: 7400
    java.lang.RuntimeException: Unable to create service com.asmgx.MyApp.MyApp.FloatWidgetService: android.view.WindowManager$BadTokenException: Unable to add window android.view.ViewRootImpl$W@7c93828 -- permission denied for window type 2002
        at android.app.ActivityThread.handleCreateService(ActivityThread.java:3544)
        at android.app.ActivityThread.access$1300(ActivityThread.java:199)
        at android.app.ActivityThread$H.handleMessage(ActivityThread.java:1666)
        at android.os.Handler.dispatchMessage(Handler.java:106)
        at android.os.Looper.loop(Looper.java:193)
        at android.app.ActivityThread.main(ActivityThread.java:6669)
        at java.lang.reflect.Method.invoke(Native Method)
        at com.android.internal.os.RuntimeInit$MethodAndArgsCaller.run(RuntimeInit.java:493)
        at com.android.internal.os.ZygoteInit.main(ZygoteInit.java:858)
     Caused by: android.view.WindowManager$BadTokenException: Unable to add window android.view.ViewRootImpl$W@7c93828 -- permission denied for window type 2002
        at android.view.ViewRootImpl.setView(ViewRootImpl.java:822)
        at android.view.WindowManagerGlobal.addView(WindowManagerGlobal.java:356)
        at android.view.WindowManagerImpl.addView(WindowManagerImpl.java:93)
        at com.asmgx.MyApp.MyApp.FloatWidgetService.onCreate(FloatWidgetService.java:36)
        at android.app.ActivityThread.handleCreateService(ActivityThread.java:3532)
        at android.app.ActivityThread.access$1300(ActivityThread.java:199) 
        at android.app.ActivityThread$H.handleMessage(ActivityThread.java:1666) 
        at android.os.Handler.dispatchMessage(Handler.java:106) 
        at android.os.Looper.loop(Looper.java:193) 
        at android.app.ActivityThread.main(ActivityThread.java:6669) 
        at java.lang.reflect.Method.invoke(Native Method) 
        at com.android.internal.os.RuntimeInit$MethodAndArgsCaller.run(RuntimeInit.java:493) 
        at com.android.internal.os.ZygoteInit.main(ZygoteInit.java:858) 
我尝试解决该问题并找到了此链接

他们建议将这一行添加到清单中,清单中已经添加了

<uses-permission android:name="android.permission.SYSTEM_ALERT_WINDOW"/>

有人知道我为什么会这样吗


另外,我在Android 28上使用emulator,在Android 27上使用emulator,这是因为示例中的targetSdkVersion与您的targetSdkVersion不同。在WindowManager中使用标记类型\应用程序\覆盖,而不是类型\电话。布局参数:

WindowManager.LayoutParams.TYPE_APPLICATION_OVERLAY 

通过将targetSdkVersion更改为22,该问题将得到部分解决。对于22以上的设备,您将得到相同的错误。 我在stack overflow中找到了以下答案,但不记得链接。您可以尝试以下代码:

private void createFloatView() {
    if (Build.VERSION.SDK_INT >= Build.VERSION_CODES.M) {
        checkDrawOverlayPermission();
    } else {
        createView();
    }
}

public void checkDrawOverlayPermission() {
    if (Build.VERSION.SDK_INT >= Build.VERSION_CODES.M) {
        if (!Settings.canDrawOverlays(getContext())) {
            Intent intent = new Intent(Settings.ACTION_MANAGE_OVERLAY_PERMISSION, Uri.parse("package:" + getActivity().getPackageName()));
            startActivityForResult(intent, CommonVariables.REQUEST_CODE);
        } else {
            createView();
        }
    }
}

@Override
public void onActivityResult(int requestCode, int resultCode, Intent data) {
    if (requestCode == CommonVariables.REQUEST_CODE) {
        if (Build.VERSION.SDK_INT >= Build.VERSION_CODES.M) {
            if (Settings.canDrawOverlays(getContext())) {
                createView();
            }
        }
    }
}

 private int CommonVariables.REQUEST_CODE = 5463 & 0xffffff00

这对我很有用。。我相信它可以简化,但如果不是这样,它就可以完美地工作。
如果(且仅当)用户等于或高于Oreo:

final WindowManager.LayoutParams params;
    if (android.os.Build.VERSION.SDK_INT >= android.os.Build.VERSION_CODES.O) {
        params = new WindowManager.LayoutParams(
                WindowManager.LayoutParams.WRAP_CONTENT,
                WindowManager.LayoutParams.WRAP_CONTENT,
                WindowManager.LayoutParams.TYPE_APPLICATION_OVERLAY,
                WindowManager.LayoutParams.FLAG_NOT_FOCUSABLE,
                PixelFormat.TRANSLUCENT);
    } else {
        params = new WindowManager.LayoutParams(
                WindowManager.LayoutParams.WRAP_CONTENT,
                WindowManager.LayoutParams.WRAP_CONTENT,
                WindowManager.LayoutParams.TYPE_PHONE,
                WindowManager.LayoutParams.FLAG_NOT_FOCUSABLE,
                PixelFormat.TRANSLUCENT);
    }
或者用简单的方法来做

    int LAYOUT_FLAG;
        if (Build.VERSION.SDK_INT >= Build.VERSION_CODES.O) {
            LAYOUT_FLAG = WindowManager.LayoutParams.TYPE_APPLICATION_OVERLAY;
        } else {
            LAYOUT_FLAG = WindowManager.LayoutParams.TYPE_PHONE;
        }

        final WindowManager.LayoutParams params = new WindowManager.LayoutParams(
                WindowManager.LayoutParams.WRAP_CONTENT,
                WindowManager.LayoutParams.WRAP_CONTENT,
                LAYOUT_FLAG,
                WindowManager.LayoutParams.FLAG_NOT_FOCUSABLE,
                PixelFormat.TRANSLUCENT);
简单地替换

TYPE_PHONE,
在代码中使用

android.os.Build.VERSION.SDK_INT >= android.os.Build.VERSION_CODES.O ? TYPE_APPLICATION_OVERLAY : TYPE_PHONE,

如何尝试使用不同的标志?请尝试使用不同的标志,如覆盖以下标志:WindowManager.LayoutParams.TYPE_SYSTEM_ALERT,TYPE_SYSTEM_OVERLAY,TYPE_ACCESSIBILITY_OVERLAY,它会在所有其他类型上崩溃。我还有别的办法让它工作吗?你误解了答案。将targetSdkVersion降级为22或将其保持在26,并将标志更改为键入应用程序覆盖。有道理吗?这是你丢失的链接。。感谢您对本作品的详细回答