how to chang to value in Skscene ? i have a class (xiangcene) and the time1 need to change through the button. how can i do that?

struct xiang: View {
    @ObservedObject var skc = xiangcene()
    var xiangview:SKScene {
        let scene = xiangcene()
        scene.scaleMode = .fill
        return scene
    }
    var body: some View {
        ZStack{
            SpriteView(scene: xiangview)
            Button("timeset"){
                skc.time1 = 60 // it no change the value
            }
        }
    }
}
class xiangcene:SKScene,ObservableObject{
    @Published var time1 = 1

    override func touchesMoved(_ touches: Set<UITouch>, with event: UIEvent?) {

        let wait1 = SKAction.wait(forDuration: TimeInterval(time1))
        let seque = SKAction.sequence([wait1])
        self.run(seque)
    }

CodePudding user response：

Every time you call xiangcene(), you're creating a new instance of xiangcene. Instead, you need to create one instance that is stored on the view, passed to the SpriteView, and manipulated by the button.

Here's one solution:

struct xiang: View {
    @StateObject var skc : xiangcene
    
    init() {
        let scene = xiangcene()
        scene.scaleMode = .fill
        _skc = StateObject(wrappedValue: scene)
    }
    
    var body: some View {
        ZStack{
            SpriteView(scene: skc)
            Button("timeset"){
                skc.time1 = 60
            }
        }
    }
}

Here's another:

struct xiang: View {
    @StateObject var skc = xiangcene()
    
    var body: some View {
        ZStack{
            SpriteView(scene: skc)
            Button("timeset"){
                skc.time1 = 60
            }
        }.onAppear {
            skc.scaleMode = .fill
        }
    }
}

Note that in both places, I'm using StateObject rather than ObservedObject, which is what I assume you want, but you can change that if that's an incorrect assumption.

I'd also consider using the Swift conventions of capitalizing type names -- it'll make it easier for people to read/parse your code.