使用辅助函数从另一个类执行 Segue答案

【问题标题】：Perform Segue from another class with helper function使用辅助函数从另一个类执行 Segue
【发布时间】：2017-02-25 16:13:34
【问题描述】：

我有两个班级，ChargeCustomer 和 TwelvethViewController。在 TwelvethViewController 中，我有一个 func goNextView()，它应该在调用时执行到下一个视图控制器的 segue。 goNextView() 在 Alamofire 完成处理程序中从 class ChargeCustomer 调用。问题是这个函数goNextView() 从来没有被调用过，我不知道为什么。我试图根据stackoverflow上的这个答案使其工作Perform Segue From Another Swift File via a class function

 class TwelvethViewController:UIViewController {

  //when this func is called, it should segue to Thiretheen
  func goNextView() {
    performSegue(withIdentifier: "twelvethToThiretheen", sender: self)
      }
 }

 class ChargeCustomer {

    //send a post request to the server
    Alamofire.request(urlString, method: .post, parameters: params, encoding: JSONEncoding.default, headers: ["Content-Type":"application/json"]).responseJSON {
        (response: DataResponse<Any>) in
                print("received reply from ALamofire") //prints string

           //perform segue to Thireteen View Controller
        func showNextView(fromViewController:TwelvethViewController) {
              print("went through")      //doesn't print
                 fromViewController.goNextView()
                    print("it has segued")      //doesn't print
           }      
      } ///end of Alamofire
   } //end of chargeUsingCustomer

根据 José Neto 的更新答案

 class ChargeCustomer {
 //create instance of the viewcontroller that has segue with identifier    twelvethToThiretheen
   let twelvethVC = TwelvethViewController()

    //send a post request to the server
    Alamofire.request(urlString, method: .post, parameters: params, encoding: JSONEncoding.default, headers: ["Content-Type":"application/json"]).responseJSON {
        (response: DataResponse<Any>) in
                print("received reply from ALamofire") //prints string

//perform segue to Thirteen View Controller
   //crashes: <CleaningApp.TwelvethViewController: 0x7fd4e16ad2a0>) has no 
     //segue with identifier 'twelvethToThiretheen''
 self.showNextView(fromViewController: self.twelvethVC )
      } //end of Alamofire

    //perform segue to Thireteen View Controller
   func showNextView(fromViewController:TwelvethViewController) {
        fromViewController.goNextView()      
       }      
} //end of chargeUsingCustomer

【问题讨论】：

您在 alamofire 闭包中声明 showNextView 函数，但您从未调用该函数。
@JoséNeto 我已更新，但应用程序崩溃并出现错误 <CleaningApp.TwelvethViewController: 0x7fd4e16ad2a0>) has no segue with identifier 'twelvethToThiretheen'' 。我已经检查了 sotryboard 中的 segue 标识符是否正确，并且是正确的。谢谢
尝试不使用 segue，实例化 TwelvethViewController 控制器并推送它。
试试这个 let storyBoard = UIStoryboard(name: "youStoryBoard", bundle: nil) let vc = storyboard?.instantiateViewController(withIdentifier: "youVCIndentifier") as！ TwelvethViewController self.navigationController?.pushViewController(vc, animated: true)
@JoséNeto 它有效。非常感谢

标签： swift segue

【解决方案1】：

你必须实例化你想要的故事板，然后是 TwelvethViewController，最后你可以推送。像这样：

let storyBoard = UIStoryboard(name: "youStoryBoard", bundle: nil) 
let vc = storyboard?.instantiateViewController(withIdentifier: "youVCIndentifier") as! TwelvethViewController 
self.navigationController?.pushViewController(vc, animated: true)

【讨论】：