2017-01-15 40 views
0

嘿我希望在用户按下按钮而不是它后,提示推送通知一旦应用程序启动请求。我会怎么做?这是我在我的应用程序委托中所拥有的。特别提醒用户何时允许推送通知

OneSignal.initWithLaunchOptions(launchOptions, appId: "myKey") 

if application.responds(to: #selector(UIApplication.registerUserNotificationSettings(_:))) { 
     let settings = UIUserNotificationSettings(types: [.alert, .badge, .sound], categories: nil) 


     application.registerUserNotificationSettings(settings) 
     application.registerForRemoteNotifications() 

    } else { 

     let types: UIRemoteNotificationType = [UIRemoteNotificationType.badge, UIRemoteNotificationType.alert, UIRemoteNotificationType.sound] 
     application.registerForRemoteNotifications(matching: types) 
    } 

回答

0

难道你不能只使用链接到IBAction的UIButton?这样你就需要将应用程序更改为UIApplication.shared

+0

试过了,但它似乎仍显示应用程序何时启动。它有一些事情要做w/OneSignal.initWithLaunchOptions(launchOptions,appId:“myKey”),但我不知道我应该用什么来替代launchOptions –

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