“无法识别的选择器已迅速发送到实例”错误

问题描述 投票:0回答:1

我正在创建一个秒表,所以我创建了一个按钮并添加了一个用于启动秒表的动作,但是一旦调用它,我就会收到此错误:

2020-05-13 19:46:02.543555+1000 Timer Pro[27206:2613795] -[Timer_Pro.Stopwatch2ViewController updateStopwatch]: unrecognized selector sent to instance 0x7fb4d7425390

(lldb) 

我正在做的是在秒表中显示分钟,秒和小数,它从0开始。每0.01秒小数上升,到99时,小数秒加一并重置小数,当秒数重设时达到59会重置并增加一分钟。我有一个膝部表,当按下膝部按钮时,它会将时间放在膝部表上。在我的UI上,我有一个开始按钮,按下该按钮会变成一个停止按钮,而在停止/开始按钮旁边,我有一个膝部按钮,当按下时会变成一个重置按钮。我在名为stop.png,start.png,reset.png和lap.png的文件中有4个按钮。这是我的代码:

import UIKit

class Stopwatch2ViewController: UIViewController, UITableViewDelegate, UITableViewDataSource {


    var timer = Timer()
    var minutes: Int = 0
    var seconds: Int = 0
    var fractions: Int = 0

    var stopWatchString: String = ""

    var startStopWatch: Bool = true
    var addLap: Bool = false

    @IBOutlet weak var stopwatchLabel: UILabel!

    @IBOutlet weak var lapsTableView: UITableView!

    @IBOutlet weak var startstopButton: UIButton!
    @IBOutlet weak var lapresetButton: UIButton! 

    @IBAction func startStop(_ sender: Any) {

        if startStopWatch == true {

            timer = Timer.scheduledTimer(timeInterval: 0.01, target: self, selector: Selector(("updateStopwatch")), userInfo: nil, repeats: true)
            startStopWatch = false

            startstopButton.setImage(UIImage(named: "stop.pnng"), for: UIControl.State.normal)
            lapresetButton.setImage(UIImage(named: "lap.png"), for: UIControl.State.normal)

            addLap = true

        }else {

            timer.invalidate()
            startStopWatch = true

            startstopButton.setImage(UIImage(named: "start.png"), for: .normal)
            lapresetButton.setImage(UIImage(named: "reset.png"), for: .normal)

            addLap = false

        }


    }

    @IBAction func lapReset(_ sender: Any) {

    }

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

        stopwatchLabel.text = "00:00.00"
    }

    func updateStopWatch() {

        fractions += 1
        if fractions == 100 {

            seconds += 1
            fractions = 0
        }

        if seconds == 60 {

            minutes += 1
            seconds = 0
        }

        let fractionsString = fractions > 9 ? "\(fractions)" : "0\(fractions)"
        let secondsString = seconds > 9 ? "\(seconds)" : "0\(seconds)"
        let minutesString = minutes > 9 ? "\(minutes)" : "0\(minutes)"

        stopWatchString = "\(minutesString):\(secondsString).\(fractionsString)"
        stopwatchLabel.text = stopWatchString

    }



    func tableView(_ tableView: UITableView, cellForRowAt indexPath: IndexPath) -> UITableViewCell {

        let cell = UITableViewCell(style: UITableViewCell.CellStyle.value1, reuseIdentifier: "Cell")

        cell.backgroundColor = self.view.backgroundColor

        cell.textLabel?.text = "Lap"
        cell.detailTextLabel?.text = "00:00:00"


        return cell
    }

    func tableView(_ tableView: UITableView, numberOfRowsInSection section: Int) -> Int {

        return 3
    }

}

如果有人能够帮助我,将不胜感激,因为我是一个初学者,这个错误使我在深夜里熬了几个月。

swift selector stopwatch unrecognized-selector xcode11.4
1个回答
0
投票

您正在创建Selector(("updateStopwatch"),但您的方法称为func updateStopWatch()

改为使用#selector(),这只能让您在实际存在的方法上创建选择器。

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