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Swift 比较两个变量时出错_Swift_Xcode_Xcode11 - Fatal编程技术网

Swift 比较两个变量时出错

Swift 比较两个变量时出错,swift,xcode,xcode11,Swift,Xcode,Xcode11,这是我的登录代码。如您所见,我已经预先定义了用户名和密码 import UIKit class ViewController: UIViewController { @IBOutlet var _username: UITextField! @IBOutlet var _password: UITextField! @IBOutlet var _login_button: UIButton! @IBOutlet var label: UILabel!

这是我的登录代码。如您所见,我已经预先定义了用户名和密码

import UIKit

class ViewController: UIViewController {

    @IBOutlet var _username: UITextField!
    @IBOutlet var _password: UITextField!
    @IBOutlet var _login_button: UIButton!
    @IBOutlet var label: UILabel!
    let username = "Kiarash"
    let password = "Test"

    override func viewDidLoad() {
        super.viewDidLoad()
        // Do any additional setup after loading the view.
    }

    @IBAction func LoginButton(_ sender: Any) {

         if(_username.text == username && _password.text == password)
        {
           label.text = "You are Logged in!"

    }
    }
}
但当我输入正确的用户名和密码时,它会抛出下面的错误

unrecognized selector sent to instance 0x7f8b9450ac60
2020-05-03 19:44:02.481826-0700 KCMLogin[9156:552975] *** Terminating app due to uncaught exception 
'NSInvalidArgumentException', reason: '-[UIView setText:]: unrecognized selector sent to instance 0x7f8b9450ac60'

我不知道为什么会发生这种情况,很抱歉我是初学者。

从标签中删除当前引用插座,然后将标签重新连接到视图控制器。

这意味着标签插座已经弄乱了。@matt我如何修复它?如果值的类型为UITextField!,当它为零时,它变得非常棘手。