如何在 Swift 4 中引用通用的可解码结构 [英] How to reference a generic Decodable struct in Swift 4

问题描述

我有一个函数，我想重用它并让它接受一个 Decocable 结构的参数.例如，这是我当前代码的简化(假设MyDecodableStruct"是在应用程序其他地方声明的可解码结构):

I have a function that I would like to reuse and have it accept a parameter of a Decocable struct. For example, this is a simplification of my current code (assume "MyDecodableStruct" is a Decodable struct declared elsewhere in the app):

 static func getResults(url: String, parameters: Parameters) {
    // On success REST response
     if response.result.isSuccess {
        struct Results: Decodable {
          let items: [MyDecodableStruct]
         }

      if let jsonResults = try? JSONDecoder().decode(Results.self, from: response.data!) {
        //success
    }
}

而不是说MyDecodableStruct"，我希望它是我作为参数传入的任何可解码结构.像这样:

and instead of saying "MyDecodableStruct", I would like it to be any Decodable struct I pass in as a parameter. Something like this:

 static func getResults(url: String, parameters: Parameters, myStruct: Decodable) {
    // On success REST response
     if response.result.isSuccess {
        struct Results: Decodable {
          let items: [myStruct]
         }

      if let jsonResults = try? JSONDecoder().decode(Results.self, from: response.data!) {
        //success
    }
}

我会这样称呼它

 getResults(url: "url", parameters: nil, myStruct: MyDecodableStruct)

我无法弄清楚如何让它工作的语法.我得到的错误是

I cannot figure out the syntax on how to get this to work though. The errors I get are

Type 'Results' does not conform to protocol 'Decodable'
Expected element type

关于处理这个问题的最佳方法有什么想法吗?

Any ideas on the best way to handle this?

推荐答案

当你想传递一个类型作为参数时，你需要将参数的类型声明为metatype.在您的情况下，它是一个需要符合 Decodable 的泛型类型.

When you want to pass a type as a parameter, you need to declare the type of the parameter as metatype. In your case, it's a generic type which needs to conform to Decodable.

所以你可能需要这样写:

So you may need to write something like this:

struct Results<Element: Decodable>: Decodable {
    let items: [Element]
}
static func getResults<Element: Decodable>(url: String, parameters: Parameters?, myStruct: Element.Type) {
    //...
        // On success REST response
        if response.result.isSuccess {

            do {
                let jsonResults = try JSONDecoder().decode(Results<Element>.self, from: response.data!)
                //success
                print(jsonResults)
            } catch {
                //Better not dispose errors silently
                print(error)
            }
        }
    //...
}

Swift 说类型不能嵌套在泛型上下文中，所以我把它移到外部的非泛型上下文中.

Swift says types cannot be nested in generic context, so I moved it out to the outer non-generic context.

称之为:

getResults(url: "url", parameters: nil, myStruct: MyDecodableStruct.self)

这篇关于如何在 Swift 4 中引用通用的可解码结构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！