在 Swift 中无需键即可解码 JSON

Question

我正在使用一个返回这个非常可怕的 JSON 结构的 API：

[
  "A string",
  [
    "A string",
    "A string",
    "A string",
    "A string",
    …
  ]
]

我正在尝试使用 JSONDecoder 解码嵌套数组，但它没有单个键，我真的不知道从哪里开始......

Answer 1

如果结构保持不变，您可以使用这种 Decodable 方法。

首先创建一个可解码的模型，如下所示：

struct MyModel: Decodable {
    let firstString: String
    let stringArray: [String]

    init(from decoder: Decoder) throws {
        var container = try decoder.unkeyedContainer()
        firstString = try container.decode(String.self)
        stringArray = try container.decode([String].self)
    }
}

或者如果您确实想保留 JSON 的结构，如下所示：

struct MyModel: Decodable {
    let array: [Any]

    init(from decoder: Decoder) throws {
        var container = try decoder.unkeyedContainer()
        let firstString = try container.decode(String.self)
        let stringArray = try container.decode([String].self)
        array = [firstString, stringArray]
    }
}

然后像这样使用它

let jsonString = """
["A string1", ["A string2", "A string3", "A string4", "A string5"]]
"""
if let jsonData = jsonString.data(using: .utf8) {
    let myModel = try? JSONDecoder().decode(MyModel.self, from: jsonData)
}

Answer 2

这对于解码来说有点有趣。

你没有任何

key

。所以它消除了包装的需要

struct

。

但是看看内部类型。您会得到

String

和

[String]

类型的混合。所以你需要一些能够处理这种混合物类型的东西。准确地说，您需要一个

enum

。

// I've provided the Encodable & Decodable both with Codable for clarity. You obviously can omit the implementation for Encodable
enum StringOrArrayType: Codable {
    case string(String)
    case array([String])

    init(from decoder: Decoder) throws {
        let container = try decoder.singleValueContainer()
        do {
            self = try .string(container.decode(String.self))
        } catch DecodingError.typeMismatch {
            do {
                self = try .array(container.decode([String].self))
            } catch DecodingError.typeMismatch {
                throw DecodingError.typeMismatch(StringOrArrayType.self, DecodingError.Context(codingPath: decoder.codingPath, debugDescription: "Encoded payload conflicts with expected type"))
            }
        }
    }

    func encode(to encoder: Encoder) throws {
        var container = encoder.singleValueContainer()
        switch self {
        case .string(let string):
            try container.encode(string)
        case .array(let array):
            try container.encode(array)
        }
    }
}

解码过程：

let json = """
[
  "A string",
  [
    "A string",
    "A string",
    "A string",
    "A string"
  ]
]
""".data(using: .utf8)!

do {
    let response = try JSONDecoder().decode([StringOrArrayType].self, from: json)
    // Here, you have your Array
    print(response) // ["A string", ["A string", "A string", "A string", "A string"]]

    // If you want to get elements from this Array, you might do something like below
    response.forEach({ (element) in
        if case .string(let string) = element {
            print(string) // "A string"
        }
        if case .array(let array) = element {
            print(array) // ["A string", "A string", "A string", "A string"]
        }
    })
} catch {
    print(error)
}

Answer 3

一个可能的解决方案是使用

JSONSerialization

，然后你可以简单地挖掘这样的 json，这样做：

import Foundation

let jsonString = "[\"A string\",[\"A string\",\"A string\", \"A string\", \"A string\"]]"
if let jsonData = jsonString.data(using: .utf8) {
    if let jsonArray = try JSONSerialization.jsonObject(with: jsonData, options: []) as? [Any] {
        jsonArray.forEach {
            if let innerArray = $0 as? [Any] {
                print(innerArray) // this is the stuff you need
            }
        }
    }
}