无法使用类型为（（T，from：Data）'的参数列表调用'decode' [英] Cannot invoke 'decode' with an argument list of type '(T, from: Data)'

问题描述

我正在尝试创建一个函数，该函数根据要传递给它的自定义JSON模型采用类型为 Codable的参数。错误：

I'm trying to create a function that takes in a parameter of type 'Codable' based on the custom JSON models being passed to it. The error :

 Cannot invoke 'decode' with an argument list of type '(T, from: Data)'

在解码行上出现，这是函数：

happens on the decode line, here is the function:

static func updateDataModels <T : Codable> (url: serverUrl, type: T, completionHandler:@escaping (_ details: Codable?) -> Void) {

guard let url = URL(string: url.rawValue) else { return }

URLSession.shared.dataTask(with: url) { (data, response, err) in

    guard let data = data else { return }

    do {
        let dataFamilies = try JSONDecoder().decode(type, from: data)// error takes place here

        completionHandler(colorFamilies)

    } catch let jsonErr {
        print("Error serializing json:", jsonErr)
        return
    }
    }.resume()
}

这是用于函数参数中类型值的示例模型（减小尺寸以节省空间） ：

This is what a sample model to be used for the 'type' value in the function's parameters (made much smaller to save space):

struct MainDataFamily: Codable {

    let families: [Family]

    enum CodingKeys: String, CodingKey {

        case families = "families"
    }
}

推荐答案

T 类型的类型是其元类型 T.Type ，因此必须将
函数参数声明为 type：T.Type 。

The type of the type T is its metatype T.Type, therefore the function parameter must be declared as type: T.Type.

您可能还想使完成句柄采用 T 类型的参数，而不是 Codable ：

You may also want to make the completion handle take a parameter of type T instead of Codable:

static func updateDataModels <T : Codable> (url: serverUrl, type: T.Type,
         completionHandler:@escaping (_ details: T) -> Void) 

调用函数时，使用 .self 将类型作为
参数传递：

When calling the function, use .self to pass the type as an argument:

updateDataModels(url: ..., type: MainDataFamily.self) { ... }

这篇关于无法使用类型为（（T，from：Data）'的参数列表调用'decode'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！