Push View programmatically in callback, SwiftUI

Question

It seems to me that Apple is encouraging us to give up using UIViewController in SwiftUI, but without using view controllers, I feel a little bit powerless. What I would like is to be able to implement some sort of ViewModel which will emit events to View.

ViewModel:

public protocol LoginViewModel: ViewModel {
  var onError: PassthroughSubject<Error, Never> { get }
  var onSuccessLogin: PassthroughSubject<Void, Never> { get }
}

View:

public struct LoginView: View {
  fileprivate let viewModel: LoginViewModel
  
  public init(viewModel: LoginViewModel) {
    self.viewModel = viewModel
  }
  
  public var body: some View {
    NavigationView {
      MasterView()
        .onReceive(self.viewModel.onError, perform: self.handleError)
        .onReceive(self.viewModel.onSuccessLogin, perform: self.handleSuccessfullLogin)
    }
  }

  func handleSuccessfullLogin() {
    //push next screen
  }
  
  func handleError(_ error: Error) {
    //show alert
  }
}

Using SwiftUI, I don't know how to push another controller if login is successful

Also, I would appreciate any advice about how to implement what I want in a better way. Thanks.

You are correct, if you have a SwiftUI project, IMHO you should only use a UIViewController if you need to use something like delegates or something not (yet) available in native SwiftUI. Let's start at square one; your model needs to drive the views, not the reverse. it likely should be a class instance that conforms to the ObservableObject protocol. (Beware, this is still early beta, and has changed in beta 5.) As for the next two things? Ask one at a time. Show code. Be aware that #2 (show alert) has undergone THREE changes over these 5 betas. — user7014451
– user7014451, Commented Aug 1, 2019 at 19:23

LinusG. · Accepted Answer · 2021-05-12 14:07:40Z

90

I've found the answer. If you want to show another view on callback you should

Create state @State var pushActive = false

When ViewModel notifies that login is successful set pushActive to true

func handleSuccessfullLogin() {
    self.pushActive = true
    print("handleSuccessfullLogin")
}

Create hidden NavigationLink and bind to that state

NavigationLink(destination: 
   ProfileView(viewModel: ProfileViewModelImpl()),
   isActive: self.$pushActive) {
     EmptyView()
}.hidden()

edited May 12, 2021 at 14:07

LinusG.

29.4k32 gold badges124 silver badges183 bronze badges

answered Aug 2, 2019 at 7:31

Bohdan Savych

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8 Comments

sbdchd Over a year ago

Text("") added extra space in layout for me, replacing it with EmptyView() did the trick

Leena Over a year ago

I was looking for the similar answer where I had to login the user and navigate to detail screen. Thanks...

m0bl Over a year ago

In step 3, need to add a $ for the binding... i.e. self.$pushActive

Anthony. Over a year ago

Thanks! Although this works fine, I think Apple should change SwiftUI to make this possible without needing to resort to such an unintuitive hack to open a new view programmatically.

Abdullah Over a year ago

isActive depreciated in IOS 16

|

Logan · Accepted Answer · 2020-11-24 01:49:04Z

I'm adding some snippets here because I think it simplifies some things and makes reusing navigation links easier:

1. Add View Navigation Extensions

extension View {
    func navigatePush(whenTrue toggle: Binding<Bool>) -> some View {
        NavigationLink(
            destination: self,
            isActive: toggle
        ) { EmptyView() }
    }

    func navigatePush<H: Hashable>(when binding: Binding<H>,
                                   matches: H) -> some View {
        NavigationLink(
            destination: self,
            tag: matches,
            selection: Binding<H?>(binding)
        ) { EmptyView() }
    }

    func navigatePush<H: Hashable>(when binding: Binding<H?>,
                                   matches: H) -> some View {
        NavigationLink(
            destination: self,
            tag: matches,
            selection: binding
        ) { EmptyView() }
    }
}

Now, you can call on any view (make sure they (or a parent) are in a navigation view)

2. Use at leisure

struct Example: View {
    @State var toggle = false
    @State var tag = 0

    var body: some View {
        NavigationView {
            VStack(alignment: .center, spacing: 24) {
                Text("toggle pushed me")
                    .navigatePush(whenTrue: $toggle)
                Text("tag pushed me (2)")
                    .navigatePush(when: $tag, matches: 2)
                Text("tag pushed me (4)")
                    .navigatePush(when: $tag, matches: 4)

                Button("toggle") {
                    self.toggle = true
                }

                Button("set tag 2") {
                    self.tag = 2
                }

                Button("set tag 4") {
                    self.tag = 4
                }
            }
        }
    }
}

Malcolm · Accepted Answer · 2020-05-23 22:54:16Z

as @Bhodan mentioned you can do it by changing state

Using EnvironmentObject with SwiftUI

Add UserData ObservableObject :

class UserData: ObservableObject, Identifiable {

    let id = UUID()
    @Published var firebase_uid: String = ""
    @Published var name: String = ""
    @Published var email: String = ""
    @Published var loggedIn: Bool = false
}

the loggedIn property will be used to monitor when a change in user logs in or out

Now add it as an @EnvironmentObject in your SceneDelegate.swift file in Xcode this just makes it so its accessible everywhere in your app

class SceneDelegate: UIResponder, UIWindowSceneDelegate {

    var window: UIWindow?


    func scene(_ scene: UIScene, willConnectTo session: UISceneSession, options connectionOptions: UIScene.ConnectionOptions) {
        // Use this method to optionally configure and attach the UIWindow `window` to the provided UIWindowScene `scene`.
        // If using a storyboard, the `window` property will automatically be initialized and attached to the scene.
        // This delegate does not imply the connecting scene or session are new (see `application:configurationForConnectingSceneSession` instead).

        // Create the SwiftUI view that provides the window contents.
        let userData = UserData()
        let contentView = ContentView().environmentObject(userData)

        // Use a UIHostingController as window root view controller.
        if let windowScene = scene as? UIWindowScene {
            let window = UIWindow(windowScene: windowScene)
            window.rootViewController = UIHostingController(rootView: contentView)
            self.window = window
            window.makeKeyAndVisible()
        }
    }

Once you make any change to the loggedIn property any UI that is Binded to it will respond to the true/false value change

the as @Bhodan mentioned just add this to your view and it will respond to that change


struct LoginView: View {
@EnvironmentObject var userData: UserData

var body: some View {
NavigationView {
VStack {
NavigationLink(destination: ProfileView(), isActive: self.$userData.loggedin) {
    EmptyView()
    }.hidden()
   }
  }
 }
}

Bryan Bartow · Accepted Answer · 2019-08-01 20:04:10Z

3

As of beta 5, NavigationLink is the mechanism used to programmatically push views. You can see an example of it here.

answered Aug 1, 2019 at 20:04

Bryan Bartow

413 bronze badges

Comments

user8354447 · Accepted Answer · 2021-10-25 13:32:15Z

2

CleanUI makes it extremely easy.

import SwiftUI
import CleanUI

struct ContentView: View {
    var body: some View {
        NavigationView {
            Button(action: {
                CUNavigation.pushToSwiftUiView(YOUR_VIEW_HERE)
            }){
                Text("Push To SwiftUI View")
            }
        }
    }
}

answered Oct 25, 2021 at 13:32

user8354447

3 Comments

smat88dd Over a year ago

The github link throws a 404. Unfortunate.

user8354447 Over a year ago

It's available again

dknchris Over a year ago

Amazing. So clean and CleanUI has a lot of other good features as well.

Igor Kasuan · Accepted Answer · 2019-12-10 00:27:32Z

0

Workaround without creating additional empty views.

You can use .disabled(true) or .allowsHitTesting(false) modifiers to disable taps on NavigationLink.

Disadvantage: You loose default button tap highlighting.

NavigationLink(destination: EnterVerificationCodeScreen(), isActive: self.$viewModel.verifyPinIsShowing) {
    Text("Create an account")
}
.allowsHitTesting(false) // or .disabled(true) 
.buttonStyle(ShadowRadiusButtonStyle(type: .dark, height: 38))

answered Dec 10, 2019 at 0:27

Igor Kasuan

8623 gold badges11 silver badges27 bronze badges

1 Comment

Paresh. P Over a year ago

verifyPinIsShowing is variable with @State or simple variable

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