'SwiftUI can not add onChange event
I have a view in which I display a list of my messages. Whenever a new message is added I want to scroll to the bottom. I am trying to add an onChange
event to the ForEach
but this breaks my code with some weird errors:
Referencing instance method 'onChange(of:perform:)' on 'Array' requires that 'Message' conform to 'Equatable'
and sometimes
The compiler is unable to type-check this expression in reasonable time; try breaking up the expression into distinct sub-expressions
When I remove onChange
everything complies.
Here is the Message
model:
struct Message: Identifiable {
var id = UUID()
var text: String
var createdAt: Date = Date()
var senderID: String
var seen: Bool
init(dictionary: [String: Any]) {
self.text = dictionary["text"] as? String ?? ""
self.senderID = dictionary["senderID"] as? String ?? ""
self.seen = dictionary["seen"] as? Bool ?? false
}
}
What am I doing wrong here?
`ForEach
struct MessagesView: View {
@ObservedObject var messagesViewModel: MessagesViewModel
var body: some View {
VStack {
ScrollView(.vertical, showsIndicators: false, content: {
ScrollViewReader { reader in
VStack(spacing: 20) {
ForEach(messagesViewModel.request.messages) { message in
Text(message.text)
.id(message.id)
}
.onChange(of: messagesViewModel.request.messages) { (value) in
reader.scrollTo(value.last?.id)
}
}
}
})
}
}
}
Solution 1:[1]
If your messages
is a @Published
the following should work (as I tested in Make a list scroll to bottom with SwiftUI)
struct MessagesView: View {
@ObservedObject var messagesViewModel: MessagesViewModel
var body: some View {
VStack {
ScrollView(.vertical, showsIndicators: false, content: {
ScrollViewReader { reader in
VStack(spacing: 20) {
ForEach(messagesViewModel.request.messages) { message in
Text(message.text)
.id(message.id)
}
.onReceive(messagesViewModel.request.$messages) { (value) in
guard !value.isEmpty else { return }
reader.scrollTo(value.last!.id)
}
}
}
})
}
}
}
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
Solution | Source |
---|---|
Solution 1 |