ios - Swift:我的便利 init 看不到正常的 init

Question

我正在尝试为我的类创建一个方便的初始化:User。我之前为另一个类(class)做过这个，并且 - 再次创建它 - 我使用了相同的代码，只是我的用户类(class)有所不同。

这是我的用户类:

import Foundation

class User {
    //Database Variables
    let userID: String?
    let firstName: String?
    let lastName: String?
    let password: String?
    let emailID: String?
    let dob: String? //timestamp
    let picture: String? //URL?
    let location: Location?
    let sex: String?

    convenience init(data: [[String: AnyObject]]) {
        self.init(userID: String(data["user_id"]!),
            firstName: String(data["first_name"]!),
            lastName: String(data["last_name"]!),
            password: String(data["password"]!),
            emailID: String(data["email"]!),
            dob: String(data["dob"]!),
            picture: String(data["picture"]!),
            location: Location(
                String(data["street"]!),
                String(data["city"]!),
                String(data["state"]!),
                String(data["zip"]!),
                String(data["country"]!)),
            sex: String(data["sex"]!))
    }

    init (userID: String, firstName: String, lastName: String,
          password: String, emailID: String, dob: String,
          picture: String, location: Location, sex: String) {
        self.userID = userID
        self.firstName = firstName
        self.lastName = lastName
        self.password = password
        self.emailID = emailID
        self.dob = dob
        self.picture = picture
        self.location = location
        self.sex = sex
    }

但是，Swift 看不到 self.init 方法。我得到一个 Could not find an overload for init that accepts the supplied arguments

怎么了？

Answer 1

除了Peter所说的字典数组，Location 对象的创建

Location(String(data["street"]!), String(data["city"]!), String(data["state"]!), String(data["zip"]!)

是错误的，因为缺少参数名称。根据您的其他代码假设这是 Swift 2，它应该是

Location(streetAddress: String(data["street"]!), city: String(data["city"]!), state: String(data["state"]!)