CODEWITHSUNDEEP
iosswiftswift3segue
Behrouz Riahi
Behrouz Riahi

Reputation: 1801

prepare for segue doesn't work as expected

I want to move to another view using performSegue as below

self.performSegue(withIdentifier: "successRegistration", sender: nil)

and I want to show an alert in my destination view so I overrides prepare function as below

override func prepare(for segue: UIStoryboardSegue, sender: Any?) {
        if segue.identifier == "successRegistration" {
            loginModel.alert(fromController: self)
            print("working")
        }
    }

but the prepare function doesn't seem to work because working is not printed to the console

I tried removing my prepare function and change my performSegue as below

self.performSegue(withIdentifier: "successRegistration", sender: self.loginModel.alert(fromController: self))

the above line did print the alert but it didn't take me to the other view.

How can I show the alert after segueing to the other view?

Upvotes: 2

Views: 754

Answers (3)

santoshi
santoshi

Reputation: 380

Check this one:

self.performSegue(withIdentifier: "successRegistration", sender: self)

Upvotes: 0

Jonathan
Jonathan

Reputation: 66

Showing an alert on your destination view controller in prepare would result in:

"Attempt to present UIAlertController on SecondViewController whose view is not in the window hierarchy!"

To achieve what you would like you could simply add a variable in your second view controller class and default it to false

//In your second view controller
var shouldPresentAlertOnOpen: Bool = false

Then you could add the logic to show your alert in that view controller's viewWillAppear method like so:

//Still in your second view controller
override func viewWillAppear(_ animated: Bool) {
    super.viewWillAppear(animated)

    if shouldPresentAlertOnOpen {
        //Present your alert here
        let alert = UIAlertController(title: "My Alert", message: "My awesome message", preferredStyle: .alert)
        alert.addAction(UIAlertAction(title: "OK", style: .default, handler: nil))
        self.present(alert, animated: true, completion: nil)
    }
}

Now back to your first view controller you could update your prepare method to:

override func prepare(for segue: UIStoryboardSegue, sender: Any?) {
    if segue.identifier == "successRegistration" {
        let destinationVC = segue.destination as! SecondViewController
        destinationVC.shouldPresentAlertOnOpen = true
    }
}

note that SecondViewController should be updated to your second view controller class.

I'm pretty sure there are other ways you could do it, this is just my approach.

Upvotes: 4

NeverHopeless
NeverHopeless

Reputation: 11233

You may try like this:

override func prepare(for segue: UIStoryboardSegue, sender: Any?) {
        if segue.identifier == "successRegistration" {
            loginModel.alert(fromController: segue.destinationViewController)
            print("working")
        }
}

and perform segue like this:

self.performSegue(withIdentifier: "successRegistration", sender: self)

It would be even more good if you pass some value to destinationViewController and show alert on next screen based on this value you are trying to display. Passed value could be like shouldDisplayLoginAlertOnLoad etc.

Upvotes: 0

Related Questions