本屏根据2值

你可以用这种方法，使用一个多元组、开关和一些NavigationLinks:

struct ContentView: View {
    @State var xy = (5.0,7.0) // <-- use a tuple
    
    var body: some View {
        NavigationView {
            QuizView(xy: $xy)
        }.navigationViewStyle(.stack)
    }
}

struct QuizView: View {
    @Binding var xy: (Double, Double)
    
    // set of ranges of your scores for each screen
    let screen1X = 1.0..<4.0
    let screen1Y = 2.0..<4.0
    
    let screen2X = 3.0..<4.0
    let screen2Y = 4.0..<8.0
    
    let screen3X = 5.0..<9.0
    let screen3Y = 6.0..<8.0
    
    @State private var action: Int? = 0
    
    var body: some View {
        VStack {
            Button(action: {
                switch xy {
                case (screen1X,screen1Y): action = 1
                case (screen2X,screen2Y): action = 2
                case (screen3X,screen3Y): action = 3
                default:
                    print("---> default")
                }
            }) {
                Text("Click me")
            }
            NavigationLink(destination: Text("screen1X"), tag: 1, selection: $action) {EmptyView()}
            NavigationLink(destination: Text("screen2X"), tag: 2, selection: $action) {EmptyView()}
            NavigationLink(destination: Text("screen3X"), tag: 3, selection: $action) {EmptyView()}
        }
    }
}

内容查看有2值：valueX和valueY. 然后范围内的需要。 然后计算所得的财产，以决定哪些屏幕应该浏览。 按钮是唯一的创建，如果计算所得的财产不返回为零。 和目的地的航链接，有一个开关的决定的屏幕显示和标签的按钮被点击。

import SwiftUI

struct MContentView: View {
    @State var valueX = 17
    @State var valueY = 15
    
    @State var isNextViewActive = false
    
    private let range8_15 = 8...15
    private let range16_23 = 16...23
    private let range8_23 = 8...23
    private let range24_40 = 24...40
    private let range8_17 = 8...17
    
    private var screenToPresent: Int? {
        if range8_15.contains(valueX) && range8_23.contains(valueY) {
            return 1
        } else if range8_15.contains(valueX) && range24_40.contains(valueY) {
            return 2
        } else if range16_23.contains(valueX) && range8_17.contains(valueY) {
            return 3
        }
        
        return nil
    }
    
    var body: some View {
        NavigationView {
            if let screen = self.screenToPresent {
                NavigationLink(isActive: self.$isNextViewActive, destination: {
                    switch screen {
                    case 1:
                        MView1()
                    case 2:
                        MView2()
                    case 3:
                        MView3()
                    default:
                        EmptyView()
                    }
                }) {
                    Button(action: {
                        self.isNextViewActive = true
                    }) {
                        Text("Click me!")
                    }
                }
            }
        }
    }
}

struct MView1: View {
    var body: some View {
        Text("My View 1")
    }
}

struct MView2: View {
    var body: some View {
        Text("My View 2")
    }
}

struct MView3: View {
    var body: some View {
        Text("My View 3")
    }
}

struct MExample_Previews: PreviewProvider {
    static var previews: some View {
        MContentView()
    }
}

希望是你在找什么!

你可能有隐藏的 导航链接 ：

NavigationLink(destination: View1(), isActive: $condition1, label: { EmptyView() })
NavigationLink(destination: View2(), isActive: $condition2, label: { EmptyView() })
NavigationLink(destination: View3(), isActive: $condition3, label: { EmptyView() })
// note that having empty views for your links will keep them hidden on your layout

为每个X和Y条件。

因此，当检查，为X和Y的价值观，可以验证他们这样的：

condition1 = Xcode >= 8 && Xcode <= 15 && Ycode >= 8 && Code <= 23
condition2 = Xcode >= 8 && Xcode <= 15 && Ycode >= 24 && Code <= 40
condition3 = Xcode >= 16 && Xcode <= 23 && Ycode >= 8 && Code <= 17

这将激活的链接你想和目屏幕上，你需要的。