Push View Controller in SwiftUI

Question

Created Jun ’20

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I need to push to other view controller when I get error from API call. All SwiftUI examples push are using navigation view and navigation link. I can't seem to make it work to push without action button.

Answered by mtsrodrigues in 614994022

You can use the isActive parameter of NavigationLink to bind a property and trigger the push changing the property value to true:

Code Block swiftstruct ContentView: View {
    @State private var isActive = false
    var body: some View {
        NavigationView {
            VStack {
                Button("Present") {
                    isActive = true
                }
                NavigationLink(destination: Color.red, isActive: $isActive) { }
            }
        }
    }
}

I'm using a Button action to change the value but you can do it in the error handling for example.

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Answer 1

mtsrodrigues OP

Jun ’20

Accepted Answer

You can use the isActive parameter of NavigationLink to bind a property and trigger the push changing the property value to true:

Code Block swiftstruct ContentView: View {
    @State private var isActive = false
    var body: some View {
        NavigationView {
            VStack {
                Button("Present") {
                    isActive = true
                }
                NavigationLink(destination: Color.red, isActive: $isActive) { }
            }
        }
    }
}

I'm using a Button action to change the value but you can do it in the error handling for example.

0

Answer 2

anthonius OP

Jun ’20

thank you @mtsrodrigues, that will be fine to show error message in that particular View. But what if we have a bunch of API calls from different source and we would want to centralized the "push to error view" from that global network call?

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Answer 3

mjfigueroa OP

Jun ’20

You can implement your error view in your root view with an if statement and show/hide the error view using either an environment object or a shared model/viewmodel property. This way you can control it directly from any view and/or from the model/viewmodel itself.

1

Answer 4

anthonius OP

Jul ’20

Yeah, I guess currently the only approach is to centralised all the logic in the root view. Thank you mjfigueroa.

0