Is it possible to make a modal non-dismissible in SwiftUI?

Question

I am creating an App where the login / register part is inside a modal, which is shown if the user is not logged in.

The problem is, that the user can dismiss the modal by swiping it down...

Is it possible to prevent this?

var body: some View {
    TabView(selection: $selection) {
        App()
    }.sheet(isPresented: self.$showSheet) { // This needs to be non-dismissible
        LoginRegister()
    }
}

Second example:

I am using a modal to ask for information. The user should not be able to quit this process except by dismissing the modal with save button. The user has to input information before the button works. Unfortunately the modal can be dismissed by swiping it down.

Is it possible to prevent this?

Answer 1

iOS 15 and later:

Use .interactiveDismissDisabled(true) on the sheet, that's all.

Prev iOS 15:

You can try to do this by using a highPriorityGesture. Of course the blue Rectangle is only for demonstration but you would have to use a view which is covering the whole screen.

struct ModalViewNoClose : View {
    @Environment(\.presentationMode) var presentationMode
    
    let gesture = DragGesture()
    
    var body: some View {
        
        Rectangle()
            .fill(Color.blue)
            .frame(width: 300, height: 600)
            .highPriorityGesture(gesture)
            
            .overlay(
                VStack{
                    Button("Close") {
                        self.presentationMode.value.dismiss()
                    }.accentColor(.white)
                    Text("Modal")
                        .highPriorityGesture(gesture)
                    TextField("as", text: .constant("sdf"))
                        .highPriorityGesture(gesture)
                } .highPriorityGesture(gesture)
        )
            .border(Color.green)
    }
}

Answer 2

This is a common problem and a "code smell"... well not really code but a "design pattern smell" anyway.

The problem is that you are making your login process part of the rest of the app.

Instead of presenting the LoginRegister over the App you should really be showing either App or LoginRegister.

i.e. you should have some state object like userLoggedIn: Bool or something and depending on that value you should show either App or LoginRegister.

Just don't have both in the view hierarchy at the same time. That way your user won't be able to dismiss the view.

Answer 3

If you dont mind using Introspect:

import Introspect

@available(iOS 13, *)
extension View {
    /// A Boolean value indicating whether the view controller enforces a modal behavior.
    ///
    /// The default value of this property is `false`. When you set it to `true`, UIKit ignores events
    /// outside the view controller's bounds and prevents the interactive dismissal of the
    /// view controller while it is onscreen.
    public func isModalInPresentation(_ value: Bool) -> some View {
        introspectViewController {
            $0.isModalInPresentation = value
        }
    }
}

Usage:

.sheet {
    VStack {
        ...
    }.isModalInPresentation(true)
}

Answer 4

iOS 15+

Starting from iOS 15 you can use interactiveDismissDisabled.

You just need to attach it to the sheet:

var body: some View {
    TabView(selection: $selection) {
        App()
    }.sheet(isPresented: self.$showSheet) {
        LoginRegister()
            .interactiveDismissDisabled(true)
    }
}

Regarding your second example, you can pass a variable to control when the sheet is disabled:

.interactiveDismissDisabled(!isAllInformationProvided)

You can find more information in the documentation.

Answer 5

theoretically this may help you (I didn't tryed it)

private var isDisplayedBind: Binding<Bool>{ Binding(get: { true }, set: { _ = $0 }) }

and usage:

content
    .sheet(isPresented: isDisplayedBind) { some sheet }