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swift - 无法快速在 View Controller 和 TableView Controller 之间发送 int 值

转载 作者:可可西里 更新时间:2023-11-01 01:36:59 26 4
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PreferencesViewController 类通过 prepare for segue 发送 nil,无论是否采用 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 值发送到类 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)


}
}
}

最佳答案

Ok segue 直接连接到按钮。您必须在您的 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

关于swift - 无法快速在 View Controller 和 TableView Controller 之间发送 int 值,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/36039488/

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