json - 带可配置键的Swift 4 JSON解码

我刚接触Swift，需要用一些可配置的键解析JSON。
与我在这里看到的许多例子相反，密钥在解码操作开始之前就已经知道了，它们仅仅依赖于传递给端点的一些参数。
例子：

https://some.provider.com/endpoint/?param=XXX

和

https://some.provider.com/endpoint/?param=YYY

将分别回答：

[
    {
        "fixed_key1": "value1",
        "fixed_key2": "value2",
        "variable_key_1_XXX": "some value",
        "variable_key_2_XXX": "some other value"
    },
    ...
]

和

[
    {
        "fixed_key1": "value1",
        "fixed_key2": "value2",
        "variable_key_1_YYY": "some value",
        "variable_key_2_YYY": "some other value"
    },
    ...
]

考虑到这些密钥在解码之前就已经知道了，我希望能够巧妙地声明一个可解码的结构和/或编码密钥，而无需编写

init(from decoder: Decoder)

不幸的是，我没能提出这样的声明。
当然，我不想为每个可能的参数值编写一个可解码/编码键结构：—）
有什么建议吗？

最佳答案

除非所有的JSON键都是编译时常量，否则编译器无法合成解码方法。但是，您可以做一些事情来减少手动解码的麻烦。
首先，一些助手结构和扩展：

/*
Allow us to initialize a `CodingUserInfoKey` with a `String` so that we can write:
    decoder.userInfo = ["param": "XXX"]

Instead of:
    decoder.userInfo = [CodingUserInfoKey(rawValue:"param")!: "XXX"]
*/
extension CodingUserInfoKey: ExpressibleByStringLiteral {
    public typealias StringLiteralType = String

    public init(stringLiteral value: StringLiteralType) {
        self.rawValue = value
    }
}

/*
This struct is a plain-vanilla implementation of the `CodingKey` protocol. Adding
`ExpressibleByStringLiteral` allows us to initialize a new instance of
`GenericCodingKeys` with a `String` literal, for example:
    try container.decode(String.self, forKey: "fixed_key1")

Instead of:
    try container.decode(String.self, forKey: GenericCodingKeys(stringValue: "fixed_key1")!)
*/
struct GenericCodingKeys: CodingKey, ExpressibleByStringLiteral {
    // MARK: CodingKey
    var stringValue: String
    var intValue: Int?

    init?(stringValue: String) { self.stringValue = stringValue }
    init?(intValue: Int) { return nil }

    // MARK: ExpressibleByStringLiteral
    typealias StringLiteralType = String
    init(stringLiteral: StringLiteralType) { self.stringValue = stringLiteral }
}

然后手动解码：

struct MyDataModel: Decodable {
    var fixedKey1: String
    var fixedKey2: String
    var variableKey1: String
    var variableKey2: String

    enum DecodingError: Error {
        case missingParamKey
        case unrecognizedParamValue(String)
    }

    init(from decoder: Decoder) throws {
        let container = try decoder.container(keyedBy: GenericCodingKeys.self)

        // Decode the fixed keys
        self.fixedKey1 = try container.decode(String.self, forKey: "fixed_key1")
        self.fixedKey2 = try container.decode(String.self, forKey: "fixed_key2")

        // Now decode the variable keys
        guard let paramValue = decoder.userInfo["param"] as? String else {
            throw DecodingError.missingParamKey
        }

        switch paramValue {
        case "XXX":
            self.variableKey1 = try container.decode(String.self, forKey: "variable_key_1_XXX")
            self.variableKey2 = try container.decode(String.self, forKey: "variable_key_2_XXX")
        case "YYY":
            self.variableKey1 = try container.decode(String.self, forKey: "variable_key_1_YYY")
            self.variableKey2 = try container.decode(String.self, forKey: "variable_key_2_YYY")
        default:
            throw DecodingError.unrecognizedParamValue(paramValue)
        }
    }
}

最后，下面是您如何使用它：

let jsonData = """
[
    {
        "fixed_key1": "value1",
        "fixed_key2": "value2",
        "variable_key_1_XXX": "some value",
        "variable_key_2_XXX": "some other value"
    }
]
""".data(using: .utf8)!

// Supplying the `userInfo` dictionary is how you "configure" the JSON-decoding
let decoder = JSONDecoder()
decoder.userInfo = ["param": "XXX"]
let model = try decoder.decode([MyDataModel].self, from: jsonData)

print(model)

关于json - 带可配置键的Swift 4 JSON解码，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/48832709/