如何使快速变量进入另一个视图

时间:2020-02-03 21:34:15

标签: swift variables view

我创建了这个脚本,想问一下如何将JSON决定的日期获取到另一个视图中?

let url = URL(string: "http://192.168.178.26/iso/loginserv.php")
                        guard let requestUrl = url else { fatalError() }

                        var request = URLRequest(url: requestUrl)
                        request.httpMethod = "POST"

                        let postString = "user=\(self.user)&pass=\(self.pass)";

                        request.httpBody = postString.data(using: String.Encoding.utf8);

                        let task = URLSession.shared.dataTask(with: request) {
                            (data, response, error) in

                            if let error = error {
                                print("Error took place \(error)")
                                return
                            }

                            guard let data = data else {return}

                            do{
                                let users = try JSONDecoder().decode(User.self, from: data)
                                print("Response data Name: \n \(users.Name)")



                                if !(data.isEmpty) {
                                    self.signedIn = true
                                }

                            }catch let jsonErr{
                                print(jsonErr)
                            }

                        }


                        task.resume()

对于JSON解码的数据,我的意思是/(users.Name)变量...

0 个答案:

没有答案