import SwiftUI
struct ButtonUI: View {
var body: some View {
ZStack {
Color.white
RoundedRectangle(cornerRadius: 50)
.fill(.blue)
.frame(width: 250, height: 75, alignment: .center)
Text("Enable Location")
.font(.title3)
.fontWeight(.bold)
.foregroundColor(Color.white)
}
.offset(x: 0, y: 300)
}
}
struct ButtonUI_Previews: PreviewProvider {
static var previews: some View {
ButtonUI()
}
}
// I have used this code to create a button like shape in Xcode how to do i make it work like open the location settings in iPhone
// it should work like a button to open the location settings in iPhone
Replies
You can access your app's settings page URL from the UIApplication.openSettingsURLString
property.
Use it like this:
Button("Open Settings") {
// Get the settings URL and open it
if let url = URL(string: UIApplication.openSettingsURLString) {
UIApplication.shared.open(url)
}
}
There is currently no way to deep link directly to your app's location settings.
iOS 16 did, however, introduce a UIApplication.openNotificationSettingsURLString
property for the notifications settings section, so maybe the location settings URL is possible in the future, or through this.
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I will try it and update it to you thank you
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sorry to bother as I have just started to learn Swift I don't know how to take the url can you give me the full code containing the settings url
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That is all the code you need.
UIApplication.openSettingsURLString
is a predefined string essentially provided to you by your app – you don't need to type it out yourself.
What do you mean precisely by location settings
I need a button in my app , when i click that button it should open the iPhone settings app and automatically navigate through the location settings