Question

我是新手，请向我建议如何为此服务功能创建完成处理程序

func postRequest(router: Router, postData: [String: String], completion: @escaping (Result<Int, Error>) -> ()) { ......

我被困在这里

ServiceLayer.postRequest(router: Router.register, postData: postData) { (<#Result<Int, Error>#>) in
            return ....
        }

Answer 1

只是要求您给闭包参数起一个名字。您可以这样做：

ServiceLayer.postRequest(router: Router.register, postData: postData) { result in
}

Answer 2

Result有.success和.failure的情况，因此您可以像使用->

一样使用Result

postRequest(router: Router.register, postData: postData) { result in
    switch result {
    case .success(let value): 
        // Do something

    case .failure(let error): 
        // Do something
    }
}

奖励

Understanding Result Type in Swift 5

Answer 3

ServiceLayer.postRequest(router: Router.register, postData: postData) { (response) in
            switch response{
                 case .success(let id):
                     print(id)
                     //do something with return id
                 case .failure(let error):
                    //error handle
            }
        }

更多信息Result type tutorial

Answer 4

已经有几个正确答案。如果您有2个具有不同回调的类似发布函数。 F.e

func postRequest(router: Router, postData: [String: String], completion: @escaping (Result<Int, Error>) -> Void) {}
func postRequest(router: Router, postData: [String: String], completion: @escaping (Error) -> Void) {}

然后您可以像这样指定通话类型：

ServiceLayer.postRequest(router: Router.register, postData: postData) { (result: Result<Int, Error>) in
}

ServiceLayer.postRequest(router: Router.register, postData: postData) { (error: Error) in
}

迅速如何为此功能创建完成处理程序

4 个答案: