从错误中启动 segue：“接收者没有带有标识符的 segue”

Question

我正在尝试通过标有toConsoleScreen 的segue 显示一个新屏幕。我不断收到错误：

receiver() 有 没有标识符'toConsoleScreen'的segue

我 100% 确定我的 segue 标记为 toConsolescreen，并且 segue 在 Interface Builder 中连接。

'rootViewController' 没有被触及，但我为第二页创建了一个新的UIViewController，名为Console Screen，类为ConsoleViewController。

我做错了什么？它告诉我它显然存在时找不到segue toConsoleScreen。

我正在使用：

self.performSegue(withIdentifier: "toConsoleScreen", sender: self)

来自ViewController。我尝试删除 segue 并重新创建它（使用“显示”设置），以及运行 Product -> Clean 功能。

我还应该补充一点，我正在尝试在 ViewController 内部的函数中调用 performSegue。这个函数是从一个单独的Swift 文件中调用的。我对 Swift 有点陌生，那么我怎样才能做到这一点呢？

这是我当前的 ViewController.Swift 文件：

import UIKit

class ViewController: UIViewController {

@IBOutlet var UserNameField: UITextField!
@IBOutlet var PasswordField: UITextField!

func successfulLogin(Username: String) {

    // Show next view \\

        self.performSegue(withIdentifier: "toConsoleScreen", sender: self)

}

override func viewDidLoad() {
    super.viewDidLoad()

    self.hideKeyboardWhenTappedAround()

    // Do any additional setup after loading the view, typically from a nib.
}

override func didReceiveMemoryWarning() {
    super.didReceiveMemoryWarning()
    // Dispose of any resources that can be recreated.
}

@IBAction func loginButton() {

            login(Username: UserNameField.text!, Password: PasswordField.text!)
}



}

还有我的 Login.swift 文件：

    import Foundation
import UIKit



func login(Username: String, Password: String) {

    var request = URLRequest(url: URL(string: "web address")!)
    request.httpMethod = "POST"
    let postString = "action=login&username=\(Username)&password=\(Password)"
    request.httpBody = postString.data(using: .utf8)
    let task = URLSession.shared.dataTask(with: request) { data, response, error in
        guard let data = data, error == nil else {                                                 // check for fundamental networking error
        print("error=\(error)")
        return
    }

        if let httpStatus = response as? HTTPURLResponse, httpStatus.statusCode != 200 {           // check for http errors
            print("statusCode should be 200, but is \(httpStatus.statusCode)")
            print("response = \(response)")
    }

        let responseString = String(data: data, encoding: .utf8)

        if responseString! == "success" {
            print("Good")

            // Send to new page \\


            ViewController().successfulLogin(Username: Username)
    }



        if responseString! == "fail" {

            print("failed")
            // Alert Error \\


        func alertPopup(title: String, message: String) {

            let alertController = UIAlertController(title: title, message: message, preferredStyle: UIAlertControllerStyle.alert)
            UIApplication.shared.keyWindow?.rootViewController?.present(alertController, animated: true, completion: nil)
            alertController.addAction(UIAlertAction(title: "Try Again", style: UIAlertActionStyle.default, handler:  nil))
            }

            // Add to main Queue \\
            OperationQueue.main.addOperation{
            alertPopup(title: "Error", message: "Invalid Login")
            }
    }


}
task.resume()



}

Answer 1

你不应该这样调用方法：

ViewController().successfulLogin(Username: Username)

上面的代码将初始化一个新的 ViewController 实例。您应该使用委托模式或完成处理程序来调用successfulLogin 方法。

使用完成处理程序的示例：

在ViewController.swift:

@IBAction func loginButton() {
        login(Username: UserNameField.text!, Password: PasswordField.text!) { username in
            self.successfulLogin(Username: username)
        }
    }

在Login.swift:

func login(Username: String, Password: String, completion: @escaping (String) -> Void) {

    var request = URLRequest(url: URL(string: "web address")!)
    request.httpMethod = "POST"
    let postString = "action=login&username=\(Username)&password=\(Password)"
    request.httpBody = postString.data(using: .utf8)
    let task = URLSession.shared.dataTask(with: request) { data, response, error in
        guard let data = data, error == nil else {                                                 // check for fundamental networking error
        print("error=\(error)")
        return
    }

        if let httpStatus = response as? HTTPURLResponse, httpStatus.statusCode != 200 {           // check for http errors
            print("statusCode should be 200, but is \(httpStatus.statusCode)")
            print("response = \(response)")
    }

        let responseString = String(data: data, encoding: .utf8)

        if responseString! == "success" {
            print("Good")

            // Send to new page \\


            completion(Username)
    }



        if responseString! == "fail" {

            print("failed")
            // Alert Error \\


        func alertPopup(title: String, message: String) {

            let alertController = UIAlertController(title: title, message: message, preferredStyle: UIAlertControllerStyle.alert)
            UIApplication.shared.keyWindow?.rootViewController?.present(alertController, animated: true, completion: nil)
            alertController.addAction(UIAlertAction(title: "Try Again", style: UIAlertActionStyle.default, handler:  nil))
            }

            // Add to main Queue \\
            OperationQueue.main.addOperation{
            alertPopup(title: "Error", message: "Invalid Login")
            }
    }


}
task.resume()



}