Not able to pass returned value from viewB to viewA in SwiftUI

Question

So I have a ViewA which returns a value called idd after defining a function addNote() and I wish to use that idd value in ViewB. I'm able to define idd in ViewB as viewA.addNote() however I'm not able to get the value of idd passed to viewB. And even Xcode says that my result from addNote() us unused. I'm new to swiftUI and learning everyday so I'd appreciate if you could let me know how to approach this problem. Thanks!

View A:

class viewA: ObservableObject{
@Published var idd: String = ""

func addNote()-> String{
  
   // some Code

         return idd

}

}

View B:

   struct ViewB: View {

    @StateObject var viewTwo = viewTwo()

    @State var idd = String()

     var body: some View {

        NavigationView {

            ZStack{
   .navigationTitle("Notes")
                
            .navigationBarItems(
                trailing: Button (action: {},
                                  
 label: { NavigationLink(destination: NoteView(newNote: "", idd: viewA.addNote() ) )
                   
                       
          {Image(systemName: "plus")} }
                     ))
                    .listStyle(.plain)
                    .buttonStyle(PlainButtonStyle())
          
                
                
                }
}
        

Answer 1

1. viewA is your viewModel for viewB. So the name of viewA should be ViewBViewModel.

2. class/struct names should start with Capital letters.

3. In your viewModel (in this case in viewA) just set a value to your Published property , it will be auto updated throughout all the Views where it is being used. In this case:

   class viewA: ObservableObject{
       @Published var idd: String = ""
   
       func addNote()-> String {
   
      // some Code
   
           self.idd = idd // setting idd here, which will be auto updated (since its a published variable) in the 
   //views where @Stateobject property wrapper is being used to instantiate viewA
   // (which is a viewModel actually).
   
       }
   }