无法在 swift 中的视图控制器和 table 视图控制器之间发送 int 值

Question

class PreferencesViewController 通过 prepare for segue 发送 nil 无论是否采取 IBActions。似乎我为 segue 做的准备有问题，但我不确定。我不是要分配给任何标签或任何东西，我只是想将 int 值发送到 class Pick2_1 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。您必须在您的 2 个控制器之间设置 segue 设置标识符，并且在 @IBAction 中您应该调用 performSegueWithIdentifier func

无法在 swift 中的视图控制器和 table 视图控制器之间发送 int 值

Unable to send an int value between view controller and table view controller in swift

null

for-loop

prepare

segue

swift