swift - 如何以通用格式编写可编码

Question

这个问题已经有答案了:

What Is Preventing My Conversion From String to Int When Decoding Using Swift 4’s Codable? (2 个回答)

已关闭 3 年前。

我已经了解如何为服务响应结构制作可编码的包装类。但有时在服务器端属性值会有所不同，可能是 Int 或 String。

示例

struct ResponseDataModel : Codable{
    let data : DataClass?
    enum  CodingKey: String, CodingKey{
        case data = "data"

    }
    init(from decoder: Decoder) throw {
        let values = try decoder.container(keyedBy: CodingKeys.self)
        data = try values.decodeIfPresent(DataClass.self, forKey:.data)
    }
}

struct DataClass : Codable{
    let id : Int
    let name : String?
    let age : Int?

    enum  CodingKey: String, CodingKey{
        case id = "id"
        case name = "name"
        case age = "age"

    }
    init(from decoder: Decoder) throw {
        let values = try decoder.container(keyedBy: CodingKeys.self)
        id = try values.decodeIfPresent(Int.self, forKey:.it)
        name = try values.decodeIfPresent(String.self, forKey:.name)
        age = try values.decodeIfPresent(Int.self, forKey:.age)
    }
}

我想使用通用方式，如果 id int string 无论它是什么，它都应该使用 id 值数据绑定(bind)到我的 Controller 。

let id : <T>

如何以通用格式编写可编码代码。

Answer 1

您可以使用以下模型来做到这一点:

struct ResponseDataModel<T: Codable>: Codable{
    let data : DataClass<T>?
    enum  CodingKeys: String, CodingKey{
        case data = "data"

    }
    init(from decoder: Decoder) throws {
        let values = try decoder.container(keyedBy: CodingKeys.self)
        data = try values.decodeIfPresent(DataClass<T>.self, forKey:.data)
    }
}

struct DataClass<T: Codable>: Codable {
    let id: T?
    let name: String?
    let age: Int?

    enum  CodingKeys: String, CodingKey{
        case id = "id"
        case name = "name"
        case age = "age"

    }
    init(from decoder: Decoder) throws {
        let values = try decoder.container(keyedBy: CodingKeys.self)
        id = try values.decodeIfPresent(T.self, forKey:.id)
        name = try values.decodeIfPresent(String.self, forKey:.name)
        age = try values.decodeIfPresent(Int.self, forKey:.age)
    }
}

但是，当您调用 JSONDecoder 的 decode(_:from:) 函数时，您应该始终知道 id 属性的类型，例如这个:

let decoder = JSONDecoder()

do {
    let decoded = try decoder.decode(ResponseDataModel<Int>.self, from: data)
    print(decoded)
} catch {
    print(error)
}

或者，您可以使用以下模型始终将 id 映射为 Int，即使您的服务器将其作为 String 发送:

struct ResponseDataModel: Codable{
    let data : DataClass?
    enum  CodingKeys: String, CodingKey{
        case data = "data"

    }
    init(from decoder: Decoder) throws {
        let values = try decoder.container(keyedBy: CodingKeys.self)
        data = try values.decodeIfPresent(DataClass.self, forKey:.data)
    }
}

struct DataClass: Codable {
    let id: Int?
    let name: String?
    let age: Int?

    enum  CodingKeys: String, CodingKey{
        case id = "id"
        case name = "name"
        case age = "age"

    }
    init(from decoder: Decoder) throws {
        let values = try decoder.container(keyedBy: CodingKeys.self)
        do {
            id = try values.decodeIfPresent(Int.self, forKey:.id)
        } catch DecodingError.typeMismatch {
            if let idString = try values.decodeIfPresent(String.self, forKey:.id) {
                id = Int(idString)
            } else {
                id = nil
            }
        }
        name = try values.decodeIfPresent(String.self, forKey:.name)
        age = try values.decodeIfPresent(Int.self, forKey:.age)
    }
}