swift - SwiftUI Picker triggers onChange twice after selection - Stack Overflow

Following full code example is firing onChange twice with the same values for old and new state. Similar topic solutions here on Stackoverflow are not solving the issue unfortunately. Help is much appreciated.

import SwiftUI

struct ContentView: View {
    @State var value = "v1"
    var values = ["v1","v2","v3"]
    
    var body: some View {
        if #available(iOS 17.0, *) {
            Picker("", selection: $value) {
                ForEach(values, id: \.self) { entry in
                    Text(entry)
                }.onChange(of: value) { oldValue, newValue in
                    print("currentValue: \(value), oldValue: \(oldValue), newValue: \(newValue)")
                }
            }
        }
    }
}

Following full code example is firing onChange twice with the same values for old and new state. Similar topic solutions here on Stackoverflow are not solving the issue unfortunately. Help is much appreciated.

import SwiftUI

struct ContentView: View {
    @State var value = "v1"
    var values = ["v1","v2","v3"]
    
    var body: some View {
        if #available(iOS 17.0, *) {
            Picker("", selection: $value) {
                ForEach(values, id: \.self) { entry in
                    Text(entry)
                }.onChange(of: value) { oldValue, newValue in
                    print("currentValue: \(value), oldValue: \(oldValue), newValue: \(newValue)")
                }
            }
        }
    }
}

• swift
• swiftui

Share Follow asked Jan 30 at 11:27 DuckjdDuckjd 8711 silver badge55 bronze badges

Add a comment  | 

1 Answer 1

Sorted by: Reset to default 1

Moving onChange out of the Picker fixes this.

Picker("", selection: $value) {
    ForEach(values, id: \.self) { entry in
        Text(entry)
    }
}
.onChange(of: value) { oldValue, newValue in
    print("currentValue: \(value), oldValue: \(oldValue), newValue: \(newValue)")
}