Question

如何以编程方式检查“设置”中“显示通知”复选框的状态。该应用程序支持的最低API为16。

建议的重复问题没有回答我的问题，我找到了一个并在下面发布。

谢谢，肖恩

Answer 1

我从这个网址找到了我想要的内容：https://code.google.com/p/android/issues/detail?id=38482。我在这里复制了答案：

public class NotificationsUtils {

    private static final String CHECK_OP_NO_THROW = "checkOpNoThrow";
    private static final String OP_POST_NOTIFICATION = "OP_POST_NOTIFICATION";

    public static boolean isNotificationEnabled() {

        AppOpsManager mAppOps = (AppOpsManager) GlobalContext.getContext().getSystemService(Context.APP_OPS_SERVICE);

        ApplicationInfo appInfo = GlobalContext.getContext().getApplicationInfo();

        String pkg = GlobalContext.getContext().getApplicationContext().getPackageName();

        int uid = appInfo.uid;

        Class appOpsClass = null; /* Context.APP_OPS_MANAGER */

        try {

            appOpsClass = Class.forName(AppOpsManager.class.getName());

            Method checkOpNoThrowMethod = appOpsClass.getMethod(CHECK_OP_NO_THROW, Integer.TYPE, Integer.TYPE, String.class);

            Field opPostNotificationValue = appOpsClass.getDeclaredField(OP_POST_NOTIFICATION);
            int value = (int)opPostNotificationValue.get(Integer.class);

            return ((int)checkOpNoThrowMethod.invoke(mAppOps,value, uid, pkg) == AppOpsManager.MODE_ALLOWED);

        } catch (ClassNotFoundException e) {
            e.printStackTrace();
        } catch (NoSuchMethodException e) {
            e.printStackTrace();
        } catch (NoSuchFieldException e) {
            e.printStackTrace();
        } catch (InvocationTargetException e) {
            e.printStackTrace();
        } catch (IllegalAccessException e) {
            e.printStackTrace();
        }
        return false;
    }
    }

如何按“设置”

1 个答案: