Question

请帮助，我真的很为此苦苦挣扎，已经阅读了多个线程并对此进行了讨论，但似乎找不到问题。我在我的应用程序中采用了与另一段代码相同的方法。

当我从源var实例化dictionary时，我为ViewController ViewController设置了一个值。

但是控制台继续输出包装为Optional的字符串，因此也在UI上输出，为什么？

sender.name: Liam, 01/13/1990, Optional("Actor")

为dictionary设置值：

var dictionary: [String : Any]!{

    didSet{
        print("inside InviteVC")


        inviteType = dictionary["type"] as? String
        inviteId = dictionary["complimentId"] as? String
        status = dictionary["status"] as? Bool
        timeStamp = dictionary["timeStamp"] as? Int

        sender = dictionary["fromUser"] as? User

        print("sender.name: \(sender!.firstName), \(sender!.birthday), \(String(describing: sender?.occupation))")

    }//end didSet

}//end var

这是我的用户模型：

struct User {

    let uid: String
    let name: String
    let email: String
    let profilePictureURL: String

    var occupation: String?

    let birthday: String
    let firstName: String
    let lastName: String
    let gender: String
    let discoverable: Bool
    let online: Bool

    let discoveryPrefs: [String : Any]

    var profileImages = [String]()

    init(uid: String, dictionary: [String: Any]) {

        self.uid = uid
        self.name = dictionary["name"] as? String ?? ""
        self.email = dictionary["email"] as? String ?? ""
        self.profilePictureURL = dictionary["profilePictureURL"] as? String ?? ""

        self.occupation = dictionary["occupation"] as? String ?? ""

        self.birthday = dictionary["birthday"] as? String ?? ""
        self.firstName = dictionary["firstName"] as? String ?? ""
        self.lastName = dictionary["lastName"] as? String ?? ""
        self.gender = dictionary["gender"] as? String ?? ""
        self.discoverable = dictionary["discoverable"] as? Bool ?? false
        self.online = dictionary["online"] as? Bool ?? false

        self.discoveryPrefs = dictionary["discoveryPrefs"] as? [String : Any] ?? [String : Any]()
        self.profileImages = dictionary["profileImages"] as! [String]


    }//end init

}//end class

这是构建用户对象的地方：

func getUserInfo(forUserId forId: String, handler: @escaping (User) -> ()) {

        REF_USERS.child(forId).observeSingleEvent(of: .value, with: { (snapshot) in

            //handle snapshot code here...

            var occupa: String?

            if let occupation = snapshot.childSnapshot(forPath: "occupation").value as? String {
                occupa = occupation
            } else {
                occupa = ""
            }



            let dictionary: [String : Any] = ["uid": uid, "name": name, "email": email, "profilePictureURL": profilePictureURL, "birthday": birthday, "firstName": firstName, "lastName": lastName, "gender": gender, "discoverable": discoverable, "online": online, "discoveryPrefs": discoveryPrefs, "profileImages": profileImages!, "occupation": occupa!]

            let user = User(uid: uid, dictionary: dictionary)

            handler(user)
        }, withCancel: { (error) in
            print(error)
        })


    }//end func

Answer 1

如果您查看要打印值的行，则会注意到您正在尝试打印可选值。

print("sender.name: \(sender!.firstName), \(sender!.birthday), \(String(describing: sender?.occupation))")

在最后一部分中，您尚未解包sender，它仍然是可选的，而属性变量occupation也是可选的。用

替换最后一部分

print(sender?.occupation ?? "")

或

if let occupation = sender?.occupation {
    print(occupation)
}

Answer 2

因为您的var occupation: String?是可选的。 String(describing:object)与"\(object)"相同，如果对象是Optional类型，则文本将由Optional()换行。您必须打开包装才能摆脱它。

控制台继续输出在Swift中包装为Optional的字符串，为什么？

2 个答案: