swift - 方法无法标记为@objc，因为参数 3 的类型无法在 Objective-C 中表示 - @objc func 中无法识别错误类型？

Question

我试图从 Objective-C 类调用此函数，但遇到了问题。我收到错误消息“方法无法标记为@objc，因为参数 3 的类型无法在 Objective-C 中表示”

我假设问题出在“错误”参数上，但不明白问题到底是什么。我读过的所有内容都是关于 Swift 1 或 2 的，但这是在 Swift 5 中。

class NetworkManager {
    
    struct DataModel: Decodable {
        let id: String
        let carID: String
        let ownerId: String
        let locationId: String?
        let status: String
    }
    
    static let shared = NetworkManager()
    
    @objc static func fetchCarWarranty(for carID: String, ownerID: String, completed: @escaping (Result<[DataModel], Error>) -> Void) {
        
        let urlString = APIURL
        let session = URLSession.shared
        let url = URL(string: urlString)!
        var request = URLRequest(url: url)
        
        let task = session.dataTask(with: request) { data, response, error in
            
            if let _ = error {
                completed(.failure(error!))
                return
            }
            
            guard let response = response as? HTTPURLResponse, response.statusCode == 200 else {
                completed(.failure(error!))
                return
            }
            
            guard let data = data else {
                return
            }
            
            do {
                let decoder = JSONDecoder()
                let dataModel = try decoder.decode([DataModel].self, from: data)
                completed(Result.success(dataModel))
            } catch {
                completed(.failure(error))
            }
        }
        task.resume()
    }
}

我考虑过从第二个 Swift 函数调用 fetchCarWarranty，但不确定这是否是一个值得遵循的好习惯。

此外，如果我从 objective-c 文件/类中调用它，我如何从该函数中获取值？

Answer 1

您不能在 Objective-C 中使用 Result，您需要一个像 dataTask(with:completion:) 那样的闭包，带有可选参数，例如 error 表示它不在“错误情况”中:

//Call that one when using Objective-C
@objc static func fetchCarWarranty(for carID: String, ownerID: String, completed: @escaping (([DataModel]?, Error?) -> Void)) {
    fetchCarWarranty(for: carID, ownerID: ownerID) { result in
        switch result {
        case .success(let models):
            completed(models, nil)
        case .failure(let error):
            completed(nil, error)
        }
    }
}

//Call that one when using Swift
static func fetchCarWarranty(for carID: String, ownerID: String, completed: @escaping (Result<[DataModel], Error>) -> Void) {
    // Your method "Swifty" code
}

我为一个带有遗留 Objective-C 代码的应用程序做了这个。最后，您在内部使用 Result 调用 Swifty 方法。所以代码也是一样的。

您还可以为 Objective-C 兼容的闭包:

@objc static func fetchCarWarranty(for carID: String, ownerID: String, success: @escaping (([DataModel]) -> Void), failure: @escaping ((Error) -> Void)) {
    fetchCarWarranty(for: carID, ownerID: ownerID { result in 
        switch result {
            case .success(let models): 
                success(models)
            case .failure(let error):
                failure(error)
        }
    }
}

现在，问题是 DataModel 是一个 struct。如果您将其替换为:

// With temp [String] parameter instead of [DataModel]
@objc static func fetchCarWarranty(for carID: String, ownerID: String, completed: @escaping (([String]?, Error?) -> Void)) {
    fetchCarWarranty(for: carID, ownerID: ownerID) { result in
        switch result {
        case .success(let models):
            completed(["models"], nil) //Fake output
        case .failure(let error):
            completed(nil, error)
        }
    }
}

它有效，因为 String 对 Objective-C 可见。但是你看不到来自 Objective-C 的 Swift 结构。请参阅 documentation 或相关问题之一:

Is there a way to use Swift structs in Objective-C without making them Classes?
How to pass a Swift struct as a parameter to an Objective-C method
How to use Swift struct in Objective-C