使用可选初始化程序在@AppStorage 中存储自定义数据类型？

Question

我正在尝试将自定义数据类型存储到 AppStorage。为此，模型符合RawRepresentable（遵循此tutorial）。它工作正常，但是当我初始化 @AppStorage 变量时，它需要一个初始 UserModel 值。我想让变量可选，所以如果用户退出，它可以是 nil 。这可能吗？

在一个类/视图中，我可以这样初始化：

@AppStorage("user_model") private(set) var user: UserModel = UserModel(id: "", name: "", email: "")

但我想这样初始化：

@AppStorage("user_model") private(set) var user: UserModel?

型号：

struct UserModel: Codable {
        
    let id: String
    let name: String
    let email: String
    
    enum CodingKeys: String, CodingKey {
        case id
        case name
        case email
    }
    
    init(from decoder: Decoder) throws {        
        let values = try decoder.container(keyedBy: CodingKeys.self)
        
        do {
           id = try String(values.decode(Int.self, forKey: .id))
        } catch DecodingError.typeMismatch {
           id = try String(values.decode(String.self, forKey: .id))
        }
        self.name = try values.decode(String.self, forKey: .name)
        self.email = try values.decode(String.self, forKey: .email)
    }
    
    init(id: String, name: String, email: String) {
        self.id = id
        self.name = name
        self.email = email
    }
    
}

// MARK: RAW REPRESENTABLE

extension UserModel: RawRepresentable {
    
    // RawRepresentable allows a UserModel to be store in AppStorage directly.
    
    public init?(rawValue: String) {
        guard let data = rawValue.data(using: .utf8),
            let result = try? JSONDecoder().decode(UserModel.self, from: data)
        else {
            return nil
        }
        self = result
    }

    var rawValue: String {
        guard let data = try? JSONEncoder().encode(self),
            let result = String(data: data, encoding: .utf8)
        else {
            return "[]"
        }
        return result
    }
    
    func encode(to encoder: Encoder) throws {
        var container = encoder.container(keyedBy: CodingKeys.self)
        try container.encode(id, forKey: .id)
        try container.encode(name, forKey: .name)
        try container.encode(email, forKey: .email)
    }
    
}

Answer 1

以下代码有效，因为您添加了一致性UserModel: RawRepresentable：

@AppStorage("user_model") private(set) var user: UserModel = UserModel(id: "", name: "", email: "")

如果您希望以下操作起作用，您需要对 UserModel? 执行相同操作：

@AppStorage("user_model") private(set) var user: UserModel? = nil

---

这是一个可能的解决方案：

extension Optional: RawRepresentable where Wrapped == UserModel {
    public init?(rawValue: String) {
        guard let data = rawValue.data(using: .utf8),
              let result = try? JSONDecoder().decode(UserModel.self, from: data)
        else {
            return nil
        }
        self = result
    }

    public var rawValue: String {
        guard let data = try? JSONEncoder().encode(self),
              let result = String(data: data, encoding: .utf8)
        else {
            return "[]"
        }
        return result
    }

    func encode(to encoder: Encoder) throws {
        var container = encoder.container(keyedBy: UserModel.CodingKeys.self)
        try container.encode(self?.id, forKey: .id)
        try container.encode(self?.name, forKey: .name)
        try container.encode(self?.email, forKey: .email)
    }
}

注意：我重用了您对 UserModel: RawRepresentable 已有的实现 - 对于这种情况，它可能需要一些更正。

也因为你符合Optional: RawRepresentable 你需要做 UserModel public 也是如此。

Answer 2

任何可选Codable 的可能通用方法：

extension Optional: RawRepresentable where Wrapped: Codable {
    public var rawValue: String {
        guard let data = try? JSONEncoder().encode(self),
              let json = String(data: data, encoding: .utf8)
        else {
            return "{}"
        }
        return json
    }

    public init?(rawValue: String) {
        guard let data = rawValue.data(using: .utf8),
              let value = try? JSONDecoder().decode(Self.self, from: data)
        else {
            return nil
        }
        self = value
    }
}

有了这个，any Codable 现在可以保存在应用存储中：

@AppStorage("user_model") var user: UserModel? = nil