I am looking at the Apple Reminders app and want to build the same pop over like view on iPhone. This the screen I am referring to:
So, I can present a popover like UI on an iPad using the popover modifier.
struct ContentView: View {
// 1.
#State var isPopoverPresented = false
var body: some View {
// 2.
Button(action: {
self.isPopoverPresented = true
}) {
Text("Some content")
}
// 3.
.popover(isPresented: $isPopoverPresented) {
Text("Popover is Presented")
.font(.largeTitle)
.frame(width: 200, height: 300)
}
}
}
However when that code runs on an iPhone, the popover turns in a fullscreen modal coming up from the bottom.
I want to know if there is a native way to build a screen like the one shown in the Reminders app or, is that screen a custom View with custom layout logic on an iPhone?
You're looking for a Menu (UIMenu in UIKit). Note that it's iOS 14+ only.
Related
I'm learning SwiftUI and having trouble with closing Each Tab View Element.
My App shows photos from user's album by TabView with pageViewStyle one by one.
And What I want to make is user can click save button in each view, and when button is clicked, save that photo and close only that view while other photos are still displayed. So unless all photos are saved or discarded, if user clicks save button, TabView should automatically move to another one.
However, I don't know how to close only one Tab Element. I've tried to use dismiss() and dynamically changing vm.images element. Latter one actually works, but it displays awkward movement and it also requires quite messy code. How could I solve this issue?
Here is my code.
TabView {
ForEach(vm.images, id: \.self) { image in
TestView(image: image)
}
}
.tabViewStyle(.page(indexDisplayMode: .never))
struct TestView: View {
#ObservedObject var vm: TestviewModel
...
var body: some View {
VStack(spacing: 10) {
Image(...)
Spacer()
Button {
...
} label: {
Text("Save")
}
}
You need actually to remove saved image from the viewModel container, and UI will be updated automatically
literally
Button {
vm.images.removeAll { $0.id == image.id } // << here !!
} label: {
Text("Save")
}
You need to use the selection initializer of TabView in order to control what it displays. So replace TabView with:
TabView(selection: $selection)
Than add a new property: #State var selection: YourIdType = someDefaultValue, and in the Button action you set selection to whatever you want to display.
Also add .tag(TheIdTheViewWillUse) remember that whatever Id you use must be the same as your selection variable. I recommend you use Int for the simple use.
I am trying to create a situation in a SwiftUI Mac app where when I click a Button inside a parent view, only the Button's action is trigger—not any of the tap gestures attached to its parent.
Here is a simple example that reproduces the problem:
import SwiftUI
struct ContentView: View {
var body: some View {
VStack(spacing: 30){
Button("Button"){
print("button clicked")
}
.padding(5)
.background(Color.blue)
.buttonStyle(PlainButtonStyle())
}
.frame(width: 500, height: 500)
.background(Color.gray)
.padding(100)
.gesture(TapGesture(count: 2).onEnded {
//Double click (open message in popup)
print("double click")
})
.simultaneousGesture(TapGesture().onEnded {
if NSEvent.modifierFlags.contains(.command){
print("command click")
}else{
print("single click")
}
})
}
}
Clicking the button triggers both button clicked and single click.
If you comment out the buttonStyle...
//.buttonStyle(PlainButtonStyle())
It works how I want it to. Only button clicked is fired.
It doesn't seem to matter which button style I use, the behavior persists. I really need a custom button style on the child button in my situation, so how do I get around this?
If you replace your .simultaneousGesture with a regular .gesture it works for me – and still recognizes the outer single and double taps.
I have set up a TabView in my application, so that I can swipe horizontally between multiple pages, but I also have an unwanted vertical scroll that may appear, with a bounce effect so. How can I disable this vertical scroll?
My code:
struct ContentView: View {
#State private var currentTabIndex: Double = 0
var body: some View {
VStack {
TabView(selection: $currentTabIndex) {
Text("Text n°1")
.tag(0)
Text("Text n°2")
.tag(1)
}
.border(Color.black)
.tabViewStyle(PageTabViewStyle(indexDisplayMode: .never))
}
}
}
I had this same problem. It's not an exact solution, but you can turn off bouncing on scrollviews (which is used within a TabView). And as long as the items within the TabView are not larger than the TabView frame, it should act as if you disabled vertical scrolling.
I would call it either .onAppear or in your init function:
.onAppear(perform: {
UIScrollView.appearance().bounces = false
})
Note: this disables the bouncing on ALL scrollviews across your app... So you may want to re-enable it .onDisappear.
Still an issue with Xcode 12.4.
I managed to workaround that by wrapping the TabView within a ScrollView and using the alwaysBounceVertical property set to false, as follow:
ScrollView(.horizontal) {
TabView {
///your content goes here
}
.tabViewStyle(PageTabViewStyle())
}
.onAppear(perform: {
UIScrollView.appearance().alwaysBounceVertical = false
})
.onDisappear(perform: {
UIScrollView.appearance().alwaysBounceVertical = true
})
I actually came across this because I saw this effect in a tutorial but couldn’t replicate it on iOS 15.2. However, I managed to replicate it on iOS 14.4 on another simulator side by side. So I guess this behaviour is disabled or fundamentally changed in the newer iOS.
Demonstration
How to create such a view floating view with a custom boarder as shown in the picture? And such that it disappears as soon as the user clicks outside of the view.
Normally you would do that with a Popover like this:
#State var isPresented = false
var body: some View {
Button(action: {
self.isPresented = true
}) {
Text("Press me")
}.popover(isPresented: $isPresented, arrowEdge: .top) {
Text("Pop!") // You can put you own custom view here for the popover
}
}
Although it works as intended on the iPad (and I believe tvOS too, but I haven't tested it), it does not work properly with the current version of SwiftUI (as of 10/12/2019) on iPhones. Currently, the above code will just result in a somewhat glitchy modal on an iPhone, which I don't think is the intended function of it on iPhones. Apple's documentation for popover isn't very helpful right now, but here it is anyway.
For you information .popover is unabailable in tvOS.
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?
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)
}
}
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.
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)
}
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.
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 }