如何手动对选定的按键进行解码，以及如何使用快速的Decodable自动解码? [英] How to Decode selected keys manually and rest with the automatic decoding with swift Decodable?

问题描述

这是我正在使用的代码，

Here is the code I am using,

struct CreatePostResponseModel : Codable{
    var transcodeId:String?
    var id:String = ""
    enum TopLevelCodingKeys: String, CodingKey {
        case _transcode = "_transcode"
        case _transcoder = "_transcoder"
    }
    enum CodingKeys:String, CodingKey{
        case id = "_id"
    }
    init(from decoder: Decoder) throws {
        let container = try decoder.container(keyedBy: TopLevelCodingKeys.self)
        if let transcodeId = try container.decodeIfPresent(String.self, forKey: ._transcode) {
            self.transcodeId = transcodeId
        }else if let transcodeId = try container.decodeIfPresent(String.self, forKey: ._transcoder) {
            self.transcodeId = transcodeId
        }

    }
}

在这里，transcodeId由_transcode或_transcoder决定. 但我希望id和其余键(此处未包括)能够自动解码.我该怎么办?

Here, transcodeId is decided by either _transcode or _transcoder. But I want id and rest of the keys (not included here) to be automatically decoded. How can I do it ?

推荐答案

由编译器生成的init(from:)是全有还是全无.您不能让它解码某些密钥，而不能手动"解码其他密钥.

The compiler-generated init(from:) is all-or-nothing. You can’t have it decode some keys and "manually" decode others.

使用编译器生成的init(from:)的一种方法是给struct和两者提供可能的编码属性，并使transcodeId为计算属性:

One way to use the compiler-generated init(from:) is by giving your struct both of the possible encoded properties, and make transcodeId a computed property:

struct CreatePostResponseModel: Codable {
    var transcodeId: String? {
        get { _transcode ?? _transcoder }
        set { _transcode = newValue; _transcoder = nil }
    }

    var _transcode: String? = nil
    var _transcoder: String? = nil

    var id: String = ""
    // other properties
}

这篇关于如何手动对选定的按键进行解码，以及如何使用快速的Decodable自动解码?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！