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社区首页 >问答首页 >获取推送通知令牌- iOS 10、Swift 3

获取推送通知令牌- iOS 10、Swift 3
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Stack Overflow用户
提问于 2016-09-19 12:02:42
回答 10查看 26.6K关注 0票数 20

如何从新的xCode 8iOS 10中的Swift 3获取设备令牌

以下是注册通知的代码:

代码语言:javascript
复制
func application(_ application: UIApplication, didFinishLaunchingWithOptions launchOptions: [UIApplicationLaunchOptionsKey: Any]?) -> Bool {    

    DispatchQueue.main.async {
        let settings = UIUserNotificationSettings(types: [.badge, .sound, .alert], categories: nil)
        UIApplication.shared.registerUserNotificationSettings(settings)
    }
    return true
}

func application(_ application: UIApplication, didRegister notificationSettings: UIUserNotificationSettings) {

    if notificationSettings.types != UIUserNotificationType() {
        application.registerForRemoteNotifications()
    }
}

在这里,我得到了令牌,但我得到了以下错误

代码语言:javascript
复制
Cannot invoke initializer for type 'UnsafePointer<CChar>' with an argument list of type '(UnsafeRawPointer)':

代码语言:javascript
复制
func application(_ application: UIApplication, didRegisterForRemoteNotificationsWithDeviceToken deviceToken: Data) {

    let chars = UnsafePointer<CChar>((deviceToken as NSData).bytes)
    var token = ""

    for i in 0..<deviceToken.count {
        token += String(format: "%02.2hhx", arguments: [chars[i]])
    }

    print("Registration succeeded!")
    print("Token: ", token)
}

有人能帮我解决这个问题吗?

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页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/39565172

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