Question

我在我的应用中定义了这个类来处理对后端的请求

class BackendService {      
    // Retrieves user chat channels
    class func getChatChannels(success:(response: Response)->(), failure: (response:Response)->()) {        
        let chatsURL = baseURL + "/chats"
        performRequestWith(success, failure: failure, url: chatsURL, parameters: nil)
    }

    func performRequestWith(success:(response: Response)->(), failure: (response: Response) -> (), url: String, parameters: String?) {
        var manager = Alamofire.Manager.sharedInstance

        let keychain = Keychain(service:"com.domain.app")
        let token = keychain.get("token")

        if let token = token {            
            manager.session.configuration.HTTPAdditionalHeaders = ["Authorization": "Bearer \(token)"]

            manager.request(.GET, url, parameters:nil).responseJSON { (req, res, json, err) in

                if(err != nil) {                    
                    var response = Response()
                    response.error = err
                    failure(response: response)                    
                } else {                    
                    var response = Response()

                    if let httpStatus = HTTPStatus(rawValue: res!.statusCode) {
                        response.httpStatus = httpStatus
                    }

                    response.payload = JSON(json!)
                    success(response: response)
                }
            }            
        }
    } 
}

我正在尝试将getChatChannels收到的回调/闭包参数传递给performRequestWith。在performRequestWith(success, failure: failure, url: chatsURL, parameters: nil)我得到了Extra argument 'failure' in call

我对Swift没什么经验，我显然在这里做了一些非常错误的事情。一些帮助将非常感激。

Answer 1

您的方法调用代码没有问题。问题是您是从类方法调用实例方法。

您应该将两种方法都更改为类方法，例如：

class func getChatChannels(success:(response: Response)->(), failure: (response:Response)->())
{        
    let chatsURL = baseURL + "/chats"
    performRequestWith(success, failure: failure, url: chatsURL, parameters: nil)
}

class func performRequestWith(success:(response: Response)->(), failure: (response: Response) -> (), url: String, parameters: String?)
{
    // Your code
}

或将两者都更改为实例方法，例如：

func getChatChannels(success:(response: Response)->(), failure: (response:Response)->())
{        
    let chatsURL = baseURL + "/chats"
    performRequestWith(success, failure: failure, url: chatsURL, parameters: nil)
}

func performRequestWith(success:(response: Response)->(), failure: (response: Response) -> (), url: String, parameters: String?)
{
    // Your code
}

另一种选择是在类方法中创建类的实例，并使用该调用实例方法。

Swift：如何将闭包参数从类方法传递给类中的函数

1 个答案: