Question

我想用我的代码将数组转换为通用数组。关于泛型，我是新手。

这是我的代码：

class APIServices {
    static let youtubeBaseURL = "https://www.googleapis.com/youtube/v3/search"

    static let shared = APIServices()

    func fetchVideo(name: String, completionHandler: @escaping ([Item]) -> ()) {
        let apiKey = "mykey"
        let params = ["part": "snippet", "q": "tausiyah \(name)", "key": apiKey]

        Alamofire.request(APIServices.youtubeBaseURL, method: .get, parameters: params, encoding: URLEncoding.default, headers: nil).responseData { (dataResponse) in
            if let err = dataResponse.error {
                print("Failed to get data:", err)
                return
            }
            guard let data = dataResponse.data else { return }

            do {
                let youtubeData = try JSONDecoder().decode(YoutubeModel.self, from: data)

                completionHandler(youtubeData.items)
            } catch let decodeErr {
                print("Failed to decode youtube data:", decodeErr)
            }
        }
    }
}

这里是我的YoutubeModel

struct YoutubeModel: Decodable {
var items: [Item]
}

struct Item: Decodable {
   var id: VideoId
   var snippet: Snippet
}

struct VideoId: Decodable {

var videoId: String

init(dictionary: [String: Any]) {
    self.videoId = dictionary["videoId"] as? String ?? ""
  }
}

struct Snippet: Decodable {

   func encode(with aCoder: NSCoder) {
      aCoder.encode(title, forKey: "titleStringKey")
   }

   var title: String
}

在我想变成通用的completionHandler: @escaping ([item])中。如何更改JSONDecoder下的completionHandler？

Answer 1

您可以尝试以下通用实现，

public class YoutubeModel<T: Decodable>: Decodable {

    public var items: [T] = []
}

class APIServices {
    static let youtubeBaseURL = "https://www.googleapis.com/youtube/v3/search"

    static let shared = APIServices()

    func fetchVideo<T: Decodable>(name: String, of type: T.Type, completionHandler: @escaping ([T]) -> ()) {
        let apiKey = "mykey"
        let params = ["part": "snippet", "q": "tausiyah \(name)", "key": apiKey]

        Alamofire.request(APIServices.youtubeBaseURL, method: .get, parameters: params, encoding: URLEncoding.default, headers: nil).responseData { (dataResponse) in
            if let err = dataResponse.error {
                print("Failed to get data:", err)
                return
            }
            guard let data = dataResponse.data else { return }

            do {
                let youtubeData = try JSONDecoder().decode(YoutubeModel<T>.self, from: data)

                completionHandler(youtubeData.items)
            } catch let decodeErr {
                print("Failed to decode youtube data:", decodeErr)
            }
        }
    }
}

用法

APIServices.shared.fetchVideo(name: "name", of: Item.self) { list in
    print(list)
}

从单例制作通用数组

1 个答案: