试图在Swift中传递Generic类型作为参数

Question

我正在使用AlamofireObjectMapper我需要创建一个带有这样的Generic参数的函数：

func doSomething < T : BaseMappable > (myCustomClass : T) 
{
    Alamofire.request("url", method: .get, parameters: nil, encoding: JSONEncoding.default, headers: APIKeys().AuthorizedHeader).responseObject(completionHandler: { ( response :DataResponse<T>) in

             let data = response.result.value

            if let array = data?.objects
            {
                for ar in array
                {
                   self.allPromotions.append(ar)
                }
            }

        })


}

但我得到错误：

使用未声明的类型'myCustomClass' 编辑，因为你们在评论中回答我我修复了错误但是当我试图调用这个方法时我得到了另一个错误

我这样称呼方法

doSomething(myCustomClass: Promotions)

但我有另一个错误

参数类型'Promotions.Type'不符合预期类型'BaseMappable'

这是我的促销课程

import ObjectMapper


class Promotions : Mappable  {





        var id:Int?
        var workshop_id:Int?
        var title:String?
        var desc:String?
        var start_date:String?
        var expire_date:String?
        var type:String?

        var objects  = [Promotions]()




        required init?(map: Map){

        }

        func mapping(map: Map) {
            id <- map["id"]
            workshop_id <- map["workshop_id"]
            title <- map["title"]
            desc <- map["desc"]
            start_date <- map["start_date"]
            expire_date <- map["expire_date"]
            type <- map["type"]
            objects <- map["promotions"]
        }




}

我该如何解决这个问题

Answer 1

只需将Promotions.self作为参数传递即可。

doSomething(myCustomClass: Promotions.self)

这将克服您在函数调用中遇到的错误。

Answer 2

您需要将Generic Type T的类型作为参数传递。尝试更改

myCustomClass : T

通过

myCustomClass : T.Type

结果如下：

斯威夫特4：

    func doSomething<T>(myCustomClass : T.Type) where T : BaseMappable
    {
        Alamofire.request("url", method: .get, parameters: nil, encoding: JSONEncoding.default, headers: APIKeys().AuthorizedHeader).responseObject(completionHandler: { ( response :DataResponse<T>) in

                 let data = response.result.value

                if let array = data?.objects
                {
                    for ar in array
                    {
                       self.allPromotions.append(ar)
                    }
                }

            })
    }

然后，您应该调用该方法：

doSomething(myCustomClass: Promotions.self)

Answer 3

myCustomClass只是doSomething的输入参数的名称。 泛型类型的名称是T ，因此DataResponse应该是DataResponse<T> 。

Answer 4

这是使用Alamofire ， ObjectMapper和PromiseKit的解决方案，

• 用户可Mappable对象。

   class User: Mappable { var id: Int! var userName: String! var firstName: String? var lastName: String? required init?(map: Map) { } // Mappable func mapping(map: Map) { id <- map["id"] userName <- map["userName"]\\ firstName <- map["firstName"] lastName <- map["lastName"] } } 

• 基于类模板的功能

   func sendRequest<T: Mappable>(_ url: URLConvertible, method: HTTPMethod = .get, parameters: Parameters? = nil, encoding: ParameterEncoding = URLEncoding.default, headers: HTTPHeaders? = nil, responseObject: T.Type) -> Promise<T> { return Promise { resolve, reject in Alamofire.request(url, method: method, parameters: parameters, encoding: encoding, headers: headers) .responseJSON() { response in switch response.result { case .success(let data): let json = JSON(data as Any) let dataObj = json["data"].object let resObj = Mapper<T>().map(JSONObject: dataObj) resolve(resObj!) case .failure(_): reject(self.generateError(message: nil)) } } } } 

• 用法

在实用程序类中定义

    func loginWith(userName: String, and password: String) -> Promise<User> {

        let apiPath = "localhost:3000"

        let parameters: Parameters = [
            "email": userName,
            "password": password
        ]

        return sendRequest(apiPath, method: .post, parameters: parameters, encoding: JSONEncoding.default, headers: nil, responseObject: User.self)
    }

在视图控制器中使用

    loginWith(userName: name, and: password)
        .then { response -> Void in
            // response is `User` type object!!!
        }
        .catch { error in
            // show error message
        }
        .always {
            // do something like hide activity indicator
        }