swift - 参数类型不符合 decodable

Question

不确定在哪里做什么；我相信我的类型符合 decodable!

let dm = DataManager(networkManagers: [mockHTTPManager])
let ep = Endpoint(scheme: .http, host: "api.nytimes.com",path: "/search/repositories")


dm.object(from: ep, with: DisplayContent.self) {result in
    print (result)
}

其中 DisplayContent 是一个结构

struct DisplayContent:Decodable {
    var title: String?
    var abstract: String?
    var thumbnailImageString: String?
    var date: String?
    var image: String?
}

我正在尝试创建一种方法来将数据一般地转换为对象，但我认为这里只有签名是相关的

func object<T : Decodable>(from endpoint: Endpoint, with object: T, completion: @escaping (Result<T, Error>) -> Void) {
    let error = NSError(domain:"", code:-1009, userInfo:[ NSLocalizedDescriptionKey: "Internet Offline"]) as Error
    let url = endpoint.url!
    networkManagers.first!.get(url: url) { result in
        switch result {
        case .failure: print ("failure")
        case .success(let success) :
            do {
                let decoder = JSONDecoder()
                let content = try decoder.decode(T.self, from: success)
                print ("content")
            } catch {

            }
        }
    }
}

错误是“参数类型‘DisplayContent.Type’不符合预期类型‘Decodable’”但 DisplayContent 符合 decodatble!

Answer 1

您的函数需要一个类型为 T 的对象(with object: T)，但您正试图传递一个类型( >DisplayContent.self).

您可以:

1. 传递正确类型的对象(例如 DisplayContent())，或者
2. 告诉函数期望传递一个类型(例如 with object: T.type)。