Captured previous value in onChange is equal to the new value not the old value

Question

Created Jun ’20

Replies 6

Boosts 2

Views 8.5k

Participants 6

When I try to replicate the example from SwiftUI documentation on onChange modifier and capture the previous value, inside the closure the previous value is equal to the new value, not the old one.

Code Block swiftstruct ContentView: View {
  @State private var isSubscribed = false
  var body: some View {
    Toggle("Subscribe", isOn: $isSubscribed)
       
    .onChange(of: isSubscribed) { [isSubscribed] newState in
      print(isSubscribed)
      print(newState)
    }
  }
}

In the documentation example, the onChange modifier is attached to the body property, not to a view inside it. Is this how it should be used? Attaching it to the body property in Xcode 12.0 beta (12A6159) doesn't compile.

Boost

Answer 1

tarasis OP

Jun ’20

I am having the same issue, onChange is not capturing the previous value.

Note, I am able to attach .onChange to the body view and have it compile.

1

Answer 2

natpan OP

Jun ’20

I submitted feedback FB7824176

1

Answer 3

Thomas994y OP

Aug ’21

A good alternative to get old values is to use 'willSet'

@State private var isSubscribed = false {
    willSet {
       print("oldValue: \(isSubscribed)")
    }
  }

1

Answer 4

nmecld OP

Oct ’21

In my case I was missing the part in square brackets which captures the old value.

Wrong

.onChange(of: isSubscribed) { newState in      
  print("debug", isSubscribed, newState) // will print the same value twice      
}

Correct

.onChange(of: isSubscribed) { [isSubscribed] newState in      
  print("debug", isSubscribed, newState) // will print two values - old, then new
}

15

Answer 5

Claude31 OP

May ’23

That's an old thread, but this doesn't compile:

.onChange(of: isSubscribed) { [isSubscribed] newState in      
  print("debug", isSubscribed, newState) &#x2F;&#x2F; will print two values - old, then new
}

I needed to write:

.onChange(of: isSubscribed) { [oldValue = isSubscribed] newState in      
  print("debug", oldValue, newState) &#x2F;&#x2F; will print two values - old, then new
}

2

Answer 6

cht1995 OP

Apr ’24

To anyone who find this question after iOS 17, here is a new modifier that solves exactly this problem: onChange(of:initial:_:)

0