I have a simple view hierarchy between two views in SwiftUI.
MainView: This is the top level screen
import SwiftUI
struct MainView: View {
var body: some View {
NavigationView {
VStack {
NavigationLink(destination: SearchView()) {
Text("Search")
}
}
.navigationBarTitle("")
.navigationBarHidden(true)
}
}
}
SearchView:
import SwiftUI
struct SearchView: View {
var body: some View {
Text("Hello, World!")
}
}
I don't want a navigation bar on my main view, but I do want a "back" button on the Search view. This is how the Search view looks:
I don't want the space above the back button - where is that coming from?
I inspected the view hierarchy and it seems the view has TWO UINavigationBar objects. If I hide the navigation bar inside the Search View, the back button goes but the space still remains.
How do I get rid of this unwanted space above the Back button?
Related
I am making an app where I need to navigate to the home page when the user clicks on the Login button and when the Login button is clicked, the navigation link code is not working and shows a warning as Result of 'NavigationLink<Label, Destination>' initializer is unused. FYI, please refer to the attached screenshot and the below code:
import SwiftUI
struct LoginView: View {
var nextButton: some View {
HStack {
Button("Next") {
NavigationLink {
HomeView(user: user)
} label: {
Text("Test")
}
}
.buttonStyle(PlainButtonStyle())
.font(.system(size: 24))
}
}
var body: some View {
NavigationStack {
nextButton
}
}
}
There are two problems here:
A NavigationLink is an alternative to a Button and can be used on its own for navigation. You can use a button to navigate programatically, but for this simple case a link works. Like Buttons, you can change the appearance of NavigationLinks if needed.
NavigationStack is a more powerful replacement for NavigationView. You should not mix them both. As the HomeView is your root view, the single NavigationStack for your app should be there (not in the LoginView).
struct HomeView: View {
var body: some View {
NavigationStack {
VStack {
NavigationLink {
LoginView()
} label: {
Text("Login")
}
}
}
}
}
The warning is because you are creating a NavigationLink in the button action code block, and returned NavigationLink is not assigned or used.
Below is the code i am trying to push to another swiftUI when button in swiftUI is pressed.
But it is not navigating to another screen
let controller = DestinationHostingController(rootView: AnotherSwiftUI())
pushViewController(controller, animated: animated)
In SwiftUI, the navigation has been improved,
You have to embed your view inside a NavigationView,
and after that you can use NavigationLink to redirect wherever you want
example:
struct ContentView: View {
var body: some View {
NavigationView {
VStack(spacing: 30) {
Text("You're going to flip a coin – do you want to choose heads or tails?")
NavigationLink(destination: ResultView(choice: "Heads")) {
Text("Choose Heads")
}
NavigationLink(destination: ResultView(choice: "Tails")) {
Text("Choose Tails")
}
}
.navigationTitle("Navigation")
}
}
}
(You can replace the NavigationLink content by whatever you want, Text -> Button)
I am new to swiftUI and am having difficulty moving from one UIViewController to another UIViewController. Rights now I have a state called navigate and a button, but I am not sure how to move this screen to a new screen. Here is the code.
import SwiftUI
struct ContentView: View {
#State var navigate = false
var body: some View {
Button(action: { self.navigate.toggle() }) {
Text("Get Involved")
.font(.custom("PlayfairDisplay-Regular", size: 18))
.foregroundColor(Color("Dark Text"))
}
}
}
Here is the code for the blank screen.
import SwiftUI
struct GoalIdeasView: View {
var body: some View {
Text("Hello")
}
}
How would I go about navigating to the second screen from the ContentView Controller once the button is pressed?
There is a view called Navigation that act like UINavigationViewController. Then you can use a NavigationLink to navigate between views. So it would be like:
struct ContentView: View {
var body: some View {
NavigationView {
NavigationLink("Get Involved", destination: GoalIdeasView())
.font(.custom("PlayfairDisplay-Regular", size: 18))
}
}
}
I try to create a splitview in SwiftUI, and I have 2 issues
1. the Detail-View not showing in the navigation
2. if I go to Detail-View from master view it shows in navigation (Navigation top bar), but when it goes to next page, back button not work
struct MainView: View {
var body: some View {
NavigationView {
ListView()
DetailView()
}
}
}
struct ListView: View {
let menuItems = [MenuItem(name: "Login"),
MenuItem(name: "KS & Token"),
MenuItem(name: "Household & Domain"),
MenuItem(name: "Register")]
var body: some View {
VStack{
List{
ForEach(self.menuItems, id:\.id) { item in
NavigationLink(destination: LoginView(menuItem: item)){
Text(item.name)
}
}
}
Spacer()
}
.navigationBarTitle(Text("User Menu"))
}
}
the issue is:
1. DetailView() not showing Navigation top bar
2. in LoginView, when I go to next page with NavigationLink, the app does not show back button, and also add it by code, not help
I am creating a sign in page for my app and would like to present the home screen in a way that the user can not go back. In Swift UI how do I present it so the new view does not present in a card like style? I know this presenting style is now default for iOS 13.
This is what I already have.
import SwiftUI
struct Test : View {
var body: some View {
PresentationButton(Text("Click to show"), destination: Extra() )
}
}
I would like the view to present full screen.
Use a NavigationView with a NavigationButton and hide the destination view's navigation bar's back button.
For example:
struct ContentView : View {
let destinationView = Text("Destination")
.navigationBarItem(title: Text("Destination View"), titleDisplayMode: .automatic, hidesBackButton: true)
var body: some View {
NavigationView {
NavigationButton(destination: destinationView) {
Text("Tap Here")
}
}
}
}
You can also disable the destination view's navigation bar altogether by doing let destinationView = Text("Destination").navigationBarHidden(true).