Receive changing data into another controller Swift - ios

Question

I have a ViewController with a variable which value is changing every second ( from a sensor ).
I made another ViewController let's call it SensorViewController with a Label on the screen in which I want to display the value from the main ViewController.
If I use override func prepare(for segue: UIStoryboardSegue, sender: Any?) the value is send but only one time ( it doesn't refresh/update every second ).
What can I do to change the value from SensorViewController every time the value from ViewController is changing?
Example:
// ViewController example code:
class ViewController: UIViewController, CBCentralManagerDelegate, CBPeripheralDelegate {
distanta1 = String(byteArray[0]) // variable which is changing every second
#IBAction func distantaSenzori(_ sender: UIButton) { //button which send me to SensorViewController
self.performSegue(withIdentifier: "goToSenzori", sender: self)
}
override func prepare(for segue: UIStoryboardSegue, sender: Any?) { // this one is sending value only when I press the button from above ( I have to exit from SensorViewController and enter again to see updated value )
if segue.identifier == "goToSenzori"{
let destinatieVC = segue.destination as! SensorViewViewController
destinatieVC.distance1 = distanta1 } }
}
// SensorViewController code:
class SensorViewViewController: UIViewController {
#IBOutlet weak var distanta1: UILabel!
var distance1: String?
override func viewDidLoad() {
super.viewDidLoad()
distanta1.text = distance2 }
}
Thank you very much, guys! You are awesome!
For Frankenstein:
In class SensorViewViewController my code looks like that:
var distance1: String?
override func viewDidLoad() {
super.viewDidLoad()
distanta1.text = distance1
print("Distance 1 is \(distance1)")
// Do any additional setup after loading the view.
}
It's called only once and the value is nil. What should I modify at the code here so the value to be refreshed?

Answer 1

I think a cleaner solution is to create a shared manager for handling the sensor. After that, you can notify your objects about the changing value. Of course in your case your "sensor" is something bluetooth but what I wrote is only a template basically, you can fill in your necessary methods and objects, delegates, so on.
class SensorManager {
static let shared: SensorManager = SensorManager()
private var sensor: Sensor
private init() {
sensor = Sensor()
}
//MARK: - Public methods
func startTheSensor() {
//This is what you call to start your sensor
}
func getSensorData() -> YourData {
//This is from where your objects can read the sensor data
}
//MARK: - Private methods -
private func didSensorUpdatedValue() {
//This is called whenever your sensor updates
.
.
.
let newSensorValue = "yourValue"
NotificationCenter.default.post(name: .init("SensorDataChanged"), object: nil)
}
}
In your viewcontroller:
deinit() {
NotificationCenter.default.removeObserver(self)
}
override func viewDidLoad() {
super.viewDidLoad()
weak var this = self
NotificationCenter.default.addObserver(this, selector: #selector(didSensorValueChanged), name: .init("SensorDataChanged"), object: nil)
}
#objc func didSensorValueChanged() {
SensorManager.shared.getSensorData()
}

Answer 2

You need to add an observer and keep reference to the destination view controller to keep passing in the new value that has been changed to the destination view controller. Here's how:
class ViewController: UIViewController, CBCentralManagerDelegate, CBPeripheralDelegate {
var distanta1 = String(byteArray[0]) {
didSet {
destinatieVC?.distance1 = distanta1
}
}
//...
var destinatieVC: SensorViewViewController?
override func prepare(for segue: UIStoryboardSegue, sender: Any?) {
if segue.identifier == "goToSenzori" {
destinatieVC = segue.destination as? SensorViewViewController
}
}
}
In SensorViewViewController:
var distance1: String? {
didSet {
print(distance1 ?? "")
}
}
Better approach: Set the destinatieVC?.distanta1.text = distanta1 directly if you're not doing anything else in the didSet block avoid the distance1 property entirely.