Swift 4 JSON 仅在运行时知道类型时解码

Swift 4 JSON decoding when type is only known at runtime

当要解码的类型是 only 时,Swift 4 中的 Decodable 协议是否可以解码 JSON 对象在运行时已知?

我有一个注册表,可以将 String 标识符映射到我们要解码的类型,如下所示:

import Foundation

struct Person: Decodable {
    let forename: String
    let surname: String
}

struct Company: Decodable {
    let officeCount: Int
    let people: [Person]
}

let registry: [String:Decodable.Type] = [
    "Person": Person.self,
    "Company": Company.self
]

let exampleJSON = """
{
    "forename": "Bob",
    "surname": "Jones"
}
""".data(using: .utf8)!

let t = registry["Person"]!

try! JSONDecoder().decode(t, from: exampleJSON) // doesn't work :-(

我的思路是否正确,或者有更好的方法吗?

您的设计确实独一无二,但不幸的是,我认为您遇到了 Swift 类型系统的边缘情况。基本上,协议不符合自身,因此,您的通用 Decodable.Type 在这里是不够的(即,您 really 需要一个具体类型来满足类型系统要求)。这可能解释了您遇到的错误:

Cannot invoke decode with an argument list of type (Decodable.Type, from: Data). Expected an argument list of type (T.Type, from: Data).

但是,话虽如此,这确实有一个(肮脏的!)破解方法。首先,创建一个虚拟 DecodableWrapper 来保存你的运行时 Decodable 类型:

struct DecodableWrapper: Decodable {
    static var baseType: Decodable.Type!
    var base: Decodable

    init(from decoder: Decoder) throws {
        self.base = try DecodableWrapper.baseType.init(from: decoder)
    }
}

然后像这样使用它:

DecodableWrapper.baseType = registry["Person"]!
let person = try! JSONDecoder().decode(DecodableWrapper.self, from: exampleJSON).base
print("person: \(person)")

打印预期结果:

person: Person(forename: "Bob", surname: "Jones")

Paulo 的 缺点是它不是线程安全的。这是一个更简单的解决方案示例,它允许您在没有可用具体类型的情况下解码值:

struct DecodingHelper: Decodable {
    private let decoder: Decoder

    init(from decoder: Decoder) throws {
        self.decoder = decoder
    }

    func decode(to type: Decodable.Type) throws -> Decodable {
        let decodable = try type.init(from: decoder)
        return decodable
    }
}

func decodeFrom(_ data: Data, to type: Decodable.Type) throws -> Decodable {
    let decodingHelper = try JSONDecoder().decode(DecodingHelper.self, from: data)
    let decodable = try decodingHelper.decode(to: type)
    return decodable
}