将通用可编码类型作为参数传递给保存到领域的方法 [英] Pass A Generic Codable Type as Parameter to a Method for Saving to Realm

问题描述

我正在尝试创建用于解码某些JSON数据的通用方法。我需要将JSON数据转换为对象，以便以后保存在Realm中。

I am trying to create a generic method for decoding some JSON data. I need to cast the JSON data into Objects for later saving in Realm.

例如：获取车辆的品牌，型号，颜色和车身类型。

For example: Getting the makes, models, colors, and body types of vehicles.

在我的每个JSON调用中，结果的格式都完全相同（请参阅问题末尾的结构）。为简便起见，我仅向您显示品牌和型号，但是颜色和主体类型完全相同，只是名称有所更改。

In each of my JSON calls to the results are formatted exactly the same (See my structs at the end of the question). For brevity, I am only showing you Makes and Models, but Colors and Body Types are exactly the same, just with the name changes.

我目前有四种方法调用，其中 Makes.self 被其他结构之一替换。

I currently have four methods that I call where Makes.self is replaced with one of the other structs.

if let results = try? JSONDecoder().decode(Makes.self, from: jsonData)) {
        DispatchQueue.main.async {
        //save data to realm
        self?.save(objects: results.data )
    }
}

一旦获得JSON数据，我想将数据发送到通用方法进行处理。

Once I get my JSON Data, I would like to send my data to a generic method for processing.

所以我可以调用类似这样的东西：

So I could call something like :

process(jsonData, modelType: Makes.self)
process(jsonData, modelType: Models.self)
process(jsonData, modelType: Colors.self)
process(jsonData, modelType: Bodies.self)

我已经尝试过泛型类型的变体，但是我不能

I have tried variations on generic types but I can't seem to get it right.

func process<T:Codable>(_ jsonData: Data, modelType: T.Type = T.self) {

        if let results = try? JSONDecoder().decode(modelType.self, from: jsonData) {
            DispatchQueue.main.async {
                //save data to realm
                self?.save(objects:results.data)
            }
        }
}

我如何通过通用的可解码协议类型？

做出

import RealmSwift

struct Makes: Codable {
    let result: String
    let data: [Make]

    enum CodingKeys: String, CodingKey {
        case result = "Result"
        case data = "Data"
    }
}

class Make: Object, Codable {
    @objc dynamic var key: String
    @objc dynamic var value: String
    @objc dynamic var shortCode: String
    @objc dynamic var active: Bool

    enum CodingKeys: String, CodingKey {
        case key = "Key"
        case value = "Value"
        case shortCode = "ShortCode"
        case active = "Active"
    }
}

模型

import RealmSwift

struct Makes: Codable {
    let result: String
    let data: [Make]

    enum CodingKeys: String, CodingKey {
        case result = "Result"
        case data = "Data"
    }
}

class Make: Object, Codable {
    @objc dynamic var key: String
    @objc dynamic var value: String
    @objc dynamic var shortCode: String
    @objc dynamic var active: Bool

    enum CodingKeys: String, CodingKey {
        case key = "Key"
        case value = "Value"
        case shortCode = "ShortCode"
        case active = "Active"
    }
}

推荐答案

为什么您希望任何 Codable 类型具有 data 属性？ results.data 无法工作...您需要使 T 成为 Object 可以将其保存到 Realm 以及 Decodable 以便将其传递给 decode 方法。

Why would you expect any Codable type to have a data property? results.data cannot work... You need to make T a subclass of Object to be able to save it to Realm and also Decodable to be able to pass it to the decode method.

func process<T:Object>(_ jsonData: Data, modelType: T.Type) where T:Decodable {
    if let results = try? JSONDecoder().decode(modelType.self, from: jsonData) {
        DispatchQueue.main.async {
            //save data to realm
            self?.save(objects:results)
        }
    }
}

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