ios - 如何在 SwiftUI 中将 View() 作为列表的一部分传递？

Question

我已经定义了一些(最多很多)不同的 SwiftUI View :

ViewA()、ViewB()、ViewC() ..ViewN()

每个 View 都包含我要根据用户所做的选择呈现的特定信息..

我创建了一个可散列的、可识别的项目列表，我想将每个项目链接到一个特定的 View 。但我似乎无法克服这些错误..

这是我正在尝试做的具体示例:

//Country Definitions:

import Foundation
import Combine
import SwiftUI

let Countries = [
    CountryModel(countryID: "US", countryName: "United States", countryView: USAView() )
    CountryModel(countryID: "UK", countryName: "United Kingdom", countryView: UKView() )
]

struct CountryModel: Identifiable, Hashable {
    
    let id:UUID
    let countryID:String
    let countryName:String
    let countryView:View   //I have tried AnyView here, different issues, still no go...
    
    init(countryID:String, countryName:String, countryView:View) {
        self.id = UUID();
        self.countryID = countryID
        self.countryName = countryName
        self.countryView = countryView
    }
}

理论上计划如何在 NavigationView/NavigationLink 中使用:

struct ContentView: View {
    
    var body: some View {
        
        NavigationView {
            
            ForEach(Countries) { thisCoutry in
                
                NavigationLink(destination: thisCoutry.countryView) { //Also tried forcing as! View, no good
                    MyItemCell(itemName: thisCoutry.countryName, itemDesc: thisCoutry.countryID)
                    
                }
            }
        }
    }
}

(MyItemCell 只是一个用于格式化项目的名称和 ID 的 View )..

到目前为止，尝试使用“查看”时，出现错误:

Protocol 'View' can only be used as a generic constraint because ithas Self or associated type requirements

还有一些关于不符合 Equatable 的内容。当我添加 Equatable 并让它添加返回 true 的 stub 时，没有帮助...

struct CountryModel: Hashable, Identifiable, Equatable {
    static func == (lhs: CountryModel, rhs: CountryModel) -> Bool {
        return true
    }

我也收到一个错误，声明这不符合 Hashable，所以我尝试将其删除，但仍然不行。

非常感谢任何想法。试图避免一长串导航链接，因为 SwiftUI 似乎在 10 点左右/之后变得很时髦，我必须开始分组，或者以其他方式更有创意......

struct ContentView: View {
    
    var body: some View {
        
        VStack{
            NavigationView {
                
                NavigationLink(destination: ViewA()) {
                    MyItemCell(itemName: "United States", itemDesc: "USA")
                    
                }
                
                NavigationLink(destination: ViewB()) {
                    MyItemCell(itemName: "United Kingdom", itemDesc: "UK")
                    
                }
            }
        }
    }
}

Answer 1

您可以使用 AnyView 类型的删除包装器(但它不仅应该在模型中声明，而且应该在创建时明确使用 - 转换在这里不起作用，我认为这是错误的来源):

let Countries = [
    CountryModel(countryID: "US", countryName: "United States", 
       countryView: AnyView(USAView()) )    // << here !!
    CountryModel(countryID: "UK", countryName: "United Kingdom", 
       countryView: AnyView(UKView()) )
]

struct CountryModel: Identifiable, Hashable {
    
    let id:UUID
    let countryID:String
    let countryName:String
    let countryView:AnyView
    
    init(countryID:String, countryName:String, countryView:View) {
        self.id = UUID();
        self.countryID = countryID
        self.countryName = countryName
        self.countryView = countryView
    }
}