swift - 无法在 View Controller 和 TableView Controller 之间快速发送 int 值

Question

PreferencesViewController 类通过 prepare for segue 发送nil，无论 IBAction 是否被执行。我的准备工作似乎出了问题，但我不确定。我不想分配给任何标签或任何东西，我只是想将 int 值发送到 Pick2_1 类中的另一个变量。提前致谢 导入 UIKit

class PreferencesViewController: UIViewController {
    var preference = 0


@IBAction func nImportant(sender: UIButton) {
    preference = -1
    print(preference)
}

@IBAction func N(sender: UIButton) {
    preference = 0
    print(preference)

}

@IBAction func mImportant(sender: UIButton) {
    preference = 1
    print(preference)

}


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

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



override func prepareForSegue(segue: UIStoryboardSegue, sender: AnyObject?) { 

    if segue.identifier == "preferenceSegue" {
        let preferenceVC = segue.destinationViewController as! Pick2_1
        preferenceVC.preferenceSent = preference
        print(preference)


    }
}
}

Answer 1

Ok segue 直接连接到按钮。您必须在 2 个 Controller 之间设置 segue 设置标识符，并且在 @IBAction 中您应该调用 PerformSegueWithIdentifier func