Question

是否可以在Swift中创建可选的初始化参数，这样我就可以使用API调用返回的值从JSON创建一个对象，但是当我稍后保存该对象时，我也可以保存下载的UIImage我之前收到的一个网址。

示例：

  class Story: NSObject, NSCoding {
        var id: Int?
        var title, coverImageURL: String?
        var coverImage: UIImage?

    required init?(anId: Int?, aTitle: String?, aCoverImageURL: String?) {
            self.id = anId
            self.title = aTitle
            self.coverImageURL = aCoverImageURL
    }
    convenience init?(json: [String: Any]) {
            let id = json["id"] as? Int
            let title = json["title"] as? String
            let coverImageURL = json["cover_image"] as? String

            self.init(
                anId: id,
                aTitle: title,
                aCoverImageURL: coverImageURL,
            )
      }

然后我想将对象保存到内存

    //MARK: Types
    struct PropertyKey {
        static let id = "id"
        static let title = "title"
        static let coverImageURL = "coverImageURL"
        static let coverImage = "coverImage"
    }

    //MARK: NSCoding
    func encode(with aCoder: NSCoder) {
        aCoder.encode(id, forKey: PropertyKey.id)
        aCoder.encode(title, forKey: PropertyKey.title)
        aCoder.encode(coverImageURL, forKey: PropertyKey.coverImageURL)
        aCoder.encode(coverImage, forKey: PropertyKey.coverImage)
    }

    required convenience init?(coder aDecoder: NSCoder) {
        guard let id = aDecoder.decodeObject(forKey: PropertyKey.id) as? Int else {
            os_log("Unable to decode the id for a Story object.", log: OSLog.default, type: .debug)
            return nil
        }
        guard let title = aDecoder.decodeObject(forKey: PropertyKey.title) as? String else {
            os_log("Unable to decode the title for a Story object.", log: OSLog.default, type: .debug)
            return nil
        }

        let coverImageURL = aDecoder.decodeObject(forKey: PropertyKey.coverImageURL) as? String
        let coverImage = aDecoder.decodeObject(forKey: PropertyKey.coverImage) as? UIImage

        self.init(
            anId: id,
            aTitle: title,
            aCoverImageURL: coverImageURL,
            coverImage: coverImage,
        )
    }

这有意义吗？我希望能够在从API获得响应后立即保存Story对象，但稍后当我将故事保存到内存时，我希望能够为coverImage保存已获取的UIImage。

我该怎么做？

Answer 1

我不确定为什么没有人对这个答案采取简单的观点，但答案是简单地使你的属性选项，然后你可以设置一个值，或者为零。您还可以创建便利初始值设定项，如果需要，可以自动将某些值设置为nil。因此，以我的应用程序为例，我有一个从API调用构建的模型。该模型具有id，created_at等值，在将记录保存到服务器之前不存在，但我在本地创建对象，存储它们，最后将它们发送到服务器，所以我需要能够仅在从JSON创建对象时设置上述值，所以这就是我所做的：

class Story: NSObject, NSCoding {
        var id: Int?
        var title, coverImageURL: String?
        var coverImage: UIImage?

    required init?(anId: Int?, aTitle: String?, aCoverImageURL: String?) {
            self.id = anId
            self.title = aTitle
            self.coverImageURL = aCoverImageURL
    }
    convenience init?(json: [String: Any]) {
            let id = json["id"] as? Int
            let title = json["title"] as? String
            let coverImageURL = json["cover_image"] as? String

            self.init(
                anId: id,
                aTitle: title,
                aCoverImageURL: coverImageURL,
            )
      }

      convenience init?(aTitle: String, aCoverImage: UIImage?) {
            let title = aTitle
            let subtitle = aSubtitle
            let coverImage = aCoverImage
            let isActive = activeStatus

            self.init(
                anId: nil,
                aTitle: title,
                aCoverImageURL: nil,
                aCoverImage: coverImage,
            )
        }

如您所见，我在本地创建对象时只设置了两个值，而其他值只设置为nil。要允许将值设置为nil，请在设置时将其设置为可选。简单！

Swift 3可选参数

1 个答案: