Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
866 views
in Technique[技术] by (71.8m points)

swift - iOS SwiftUI: pop or dismiss view programmatically

I couldn't find any reference about any ways to make a pop or a dismiss programmatically of my presented view with SwiftUI.

Seems to me that the only way is to use the already integrated slide dow action for the modal(and what/how if I want to disable this feature?), and the back button for the navigation stack.

Does anyone know a solution? Do you know if this is a bug or it will stays like this?

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

This example uses the new environment var documented in the Beta 5 Release Notes, which was using a value property. It was changed in a later beta to use a wrappedValue property. This example is now current for the GM version. This exact same concept works to dismiss Modal views presented with the .sheet modifier.

import SwiftUI

struct DetailView: View {
    @Environment(.presentationMode) var presentationMode: Binding<PresentationMode>
    var body: some View {
        Button(
            "Here is Detail View. Tap to go back.",
            action: { self.presentationMode.wrappedValue.dismiss() }
        )
    }
}

struct RootView: View {
    var body: some View {
        VStack {
            NavigationLink(destination: DetailView())
            { Text("I am Root. Tap for Detail View.") }
        }
    }
}

struct ContentView: View {
    var body: some View {
        NavigationView {
            RootView()
        }
    }
}

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...