如何在不检查应用程序状态的情况下确定是否点击了 UILocalNotification？

Question

我正在创建一个应用程序，当我点击 UILocalNotification 时应该打开特定的视图控制器。我的代码是：

func application(application:UIApplication!, didReceiveLocalNotification notification: UILocalNotification)
{
  var root = self.window!.rootViewController as ViewController
  let main: UIStoryboard = UIStoryboard(name: "Main", bundle: nil)
  var setview = main.instantiateViewControllerWithIdentifier("destination") as tapViewController

  if application.applicationState == UIApplicationState.Inactive
  {
     root.presentViewController(setview, animated:false , completion: nil)
  }
}

当我从我的应用程序下拉通知中心时，即使我没有点击通知，也会显示 TapViewController，因为当我这样做时，应用程序状态正在从非活动变为活动。

那么在这种情况下如何检测通知是否被点击？与应用程序状态一起检查的条件是什么？

Answer 1

在收到通知时尝试使用此代码检查活动/非活动状态。

func application(application: UIApplication, didReceiveLocalNotification userInfo: [NSObject : AnyObject]) {

    if application.applicationState == UIApplicationState.Inactive || application.applicationState == UIApplicationState.Background {
        //opened from a push notification when the app was on background

    }

}

Answer 2

试试这段代码（用Objective C编写，你可以轻松切换到Swift）

- (void)application:(UIApplication *)application didReceiveLocalNotification:(UILocalNotification *)notification {

    NSAssert(notification.fireDate != nil, @"Please use scheduleLocalNotification instead of presentLocalNotificationNow with fireDate.");

    // 0.2s is threshold time, you can adjust it
    if (application.applicationState != UIApplicationStateActive && [notification.fireDate timeIntervalSinceNow] < -0.2) {
        // user tapped the notification
        ...
    }
}

我不明白为什么苹果不提供检查通知是否被点击的方法。