UIRemoteNotificationType types = [[UIApplication sharedApplication] enabledRemoteNotificationTypes];
if (types & UIRemoteNotificationTypeAlert)
    // blah blah blah
{
    NSLog(@"Notification Enabled");
}
else
{
    NSLog(@"Notification not enabled");
}

这里我们从UIApplication获得UIRemoteNotificationType。它代表了这个应用程序在设置中的推送通知的状态，然后你可以很容易地检查它的类型

尽管Zac的答案在iOS 7之前是完全正确的，但自从iOS 8到来后，它就发生了变化。因为enabledRemoteNotificationTypes从iOS 8开始就已经弃用了。对于iOS 8和更高版本，你需要使用isregisteredforremotenotifizations。

—>使用enabledRemoteNotificationTypes >使用isregisteredforremotenotifizations。

在尝试同时支持iOS8和更低版本时，我并没有像Kevin建议的那样使用isregisteredforremotenotifations。相反，我使用了currentUserNotificationSettings，它在我的测试中工作得很好。

+ (BOOL)notificationServicesEnabled {
    BOOL isEnabled = NO;

    if ([[UIApplication sharedApplication] respondsToSelector:@selector(currentUserNotificationSettings)]){
        UIUserNotificationSettings *notificationSettings = [[UIApplication sharedApplication] currentUserNotificationSettings];

        if (!notificationSettings || (notificationSettings.types == UIUserNotificationTypeNone)) {
            isEnabled = NO;
        } else {
            isEnabled = YES;
        }
    } else {
        UIRemoteNotificationType types = [[UIApplication sharedApplication] enabledRemoteNotificationTypes];
        if (types & UIRemoteNotificationTypeAlert) {
            isEnabled = YES;
        } else{
            isEnabled = NO;
        }
    }

    return isEnabled;
}

调用enabledremotenotificationsttypes并检查掩码。

例如:

UIRemoteNotificationType types = [[UIApplication sharedApplication] enabledRemoteNotificationTypes];
if (types == UIRemoteNotificationTypeNone) 
   // blah blah blah

iOS8及以上:

[[UIApplication sharedApplication] isRegisteredForRemoteNotifications]

为了完成这个答案，它可以这样工作……

UIRemoteNotificationType types = [[UIApplication sharedApplication] enabledRemoteNotificationTypes];
switch (types) {
   case UIRemoteNotificationTypeAlert:
   case UIRemoteNotificationTypeBadge:
       // For enabled code
       break;
   case UIRemoteNotificationTypeSound:
   case UIRemoteNotificationTypeNone:
   default:
       // For disabled code
       break;
}

编辑:这是不对的。因为这些是位智慧的东西，它不会与开关工作，所以我结束使用这个:

UIRemoteNotificationType types = [[UIApplication sharedApplication] enabledRemoteNotificationTypes];
UIRemoteNotificationType typesset = (UIRemoteNotificationTypeAlert | UIRemoteNotificationTypeBadge);
if((types & typesset) == typesset)
{
    CeldaSwitch.chkSwitch.on = true;
}
else
{
    CeldaSwitch.chkSwitch.on = false;
}

这个Swifty解决方案对我来说效果很好(iOS8+)，

方法:

func isNotificationEnabled(completion:@escaping (_ enabled:Bool)->()){
    if #available(iOS 10.0, *) {
        UNUserNotificationCenter.current().getNotificationSettings(completionHandler: { (settings: UNNotificationSettings) in
            let status =  (settings.authorizationStatus == .authorized)
            completion(status)
        })
    } else {
        if let status = UIApplication.shared.currentUserNotificationSettings?.types{
            let status = status.rawValue != UIUserNotificationType(rawValue: 0).rawValue
            completion(status)
        }else{
            completion(false)
        }
    }
}

用法:

isNotificationEnabled { (isEnabled) in
            if isEnabled{
                print("Push notification enabled")
            }else{
                print("Push notification not enabled")
            }
        }

Ref