Swift - 推送警报操作以显示新 ViewController
Swift - push alert action to show new ViewController
当按钮处于警报状态时如何显示或link到新ViewController?
这是我的代码
let alert = UIAlertController(title: validateQRObj.responseDescription, message: validateQRObj.productName, preferredStyle: .alert)
let action = UIAlertAction(title: "OK", style: .default) { (action) -> Void in
let viewController = self.storyboard?.instantiateViewController(withIdentifier: "ProductDetialViewController")
self.present(viewController!, animated: true, completion: nil)
}
alert.addAction(action)
self.present(alert, animated: true, completion: nil)
控制从 View Controller 1(黄点)拖动到 View Controller 2 上的任意位置,然后单击 Segue。显示 Attributes inspector 和 Under Storyboard Segue identifier name the identifier VC2
如果这是您正在寻找的答案,请不要忘记排除答案。
func alert(){
let alertController = UIAlertController(title: "Open View Controller. ", message: "Press Ok to open View Controller number 2.", preferredStyle: UIAlertControllerStyle.alert)
let ok = UIAlertAction(title: "Ok", style: UIAlertActionStyle.default, handler: {(action) -> Void in
//The (withIdentifier: "VC2") is the Storyboard Segue identifier.
self.performSegue(withIdentifier: "VC2", sender: self)
})
alertController.addAction(ok)
self.present(alertController, animated: true, completion: nil)
}
对我来说它不起作用。我做了:
func alert(){
let storyboard = UIStoryboard(name: "Main", bundle: nil)
let twop = storyboard.instantiateViewController(withIdentifier: "Accueil") as! Accueil
let alertController = UIAlertController(title: "Open new View !", message: "Clic to ok", preferredStyle: UIAlertControllerStyle.alert)
let ok = UIAlertAction(title: "Ok", style: UIAlertActionStyle.default, handler: {(action) -> Void in
self.navigationController?.show(twop, sender: self)
})
alertController.addAction(ok)
self.present(alertController, animated: true, completion: nil)
}
现在可以使用了!!
当按钮处于警报状态时如何显示或link到新ViewController?
这是我的代码
let alert = UIAlertController(title: validateQRObj.responseDescription, message: validateQRObj.productName, preferredStyle: .alert)
let action = UIAlertAction(title: "OK", style: .default) { (action) -> Void in
let viewController = self.storyboard?.instantiateViewController(withIdentifier: "ProductDetialViewController")
self.present(viewController!, animated: true, completion: nil)
}
alert.addAction(action)
self.present(alert, animated: true, completion: nil)
控制从 View Controller 1(黄点)拖动到 View Controller 2 上的任意位置,然后单击 Segue。显示 Attributes inspector 和 Under Storyboard Segue identifier name the identifier VC2
如果这是您正在寻找的答案,请不要忘记排除答案。
func alert(){
let alertController = UIAlertController(title: "Open View Controller. ", message: "Press Ok to open View Controller number 2.", preferredStyle: UIAlertControllerStyle.alert)
let ok = UIAlertAction(title: "Ok", style: UIAlertActionStyle.default, handler: {(action) -> Void in
//The (withIdentifier: "VC2") is the Storyboard Segue identifier.
self.performSegue(withIdentifier: "VC2", sender: self)
})
alertController.addAction(ok)
self.present(alertController, animated: true, completion: nil)
}
对我来说它不起作用。我做了:
func alert(){
let storyboard = UIStoryboard(name: "Main", bundle: nil)
let twop = storyboard.instantiateViewController(withIdentifier: "Accueil") as! Accueil
let alertController = UIAlertController(title: "Open new View !", message: "Clic to ok", preferredStyle: UIAlertControllerStyle.alert)
let ok = UIAlertAction(title: "Ok", style: UIAlertActionStyle.default, handler: {(action) -> Void in
self.navigationController?.show(twop, sender: self)
})
alertController.addAction(ok)
self.present(alertController, animated: true, completion: nil)
}
现在可以使用了!!