如何使代码更简单，更易于在 swiftui 中使用多维字典导航链接重用？

Question

我有一个多级词典，我需要用它来建立导航链接。现在我有一个 2 级深度字典：

let multiDimDict: [String: [[String: [String]]]]
= ["A": [["A1": ["A11", "A12"]], ["A2": ["A21", "A22"]]],
   "B": [["B1": ["B11", "B12"]], ["B2": ["B21", "B22"]]]
    ]

我可以构建包含所有导航链接的导航视图。但是，我觉得我在构建导航链接时以某种方式重复了一些代码。问题是，如果我有一个多级字典，比如说 10 级，我认为构建 10 个不同的子视图生成器来构建完整的导航视图是不明智的。知道如何减少代码吗？

struct PlayingWithMultiDimNavLink: View {

    let multiDimDict: [String: [[String: [String]]]]
    = ["A": [["A1": ["A11", "A12"]], ["A2": ["A21", "A22"]]],
       "B": [["B1": ["B11", "B12"]], ["B2": ["B21", "B22"]]]
        ]




    var body: some View {

        NavigationView{
            List {
                ForEach(multiDimDict.keys.sorted(), id: \.self) { key in

                    NavigationLink(destination: GenerateChildView(key: key, dict: self.multiDimDict)){

                        Text(key)
                    }

                }
            }
        }

    }
}

这是：第一个子视图生成器：

struct GenerateChildView: View {

    var key: String
    var dict: [String: [[String: [String]]]]
    var infoList: [[String: [String]]]

    init(key: String, dict: [String: [[String: [String]]]]  ){
        self.key = key
        self.dict = dict
        self.infoList = dict[key]!
    }


    var body: some View {
        List{
            ForEach(infoList, id: \.self){ info in
                self.getSomeView(infoList: self.infoList)

            }
        }

}


    func getSomeView(infoList: [[String: [String]]] )-> AnyView{

        let dictToUse = infoList[0]
        return AnyView(
            ForEach(dictToUse.keys.sorted(), id: \.self){ key2 in

                    NavigationLink(destination: GenerateChildView2(infoList: dictToUse[key2]!)){
                        Text(key2)
                    }
                }


        )

    }

最后一个子视图生成器：

struct GenerateChildView2: View {

    var infoList: [String]

    init(infoList: [String]){
        self.infoList = infoList
    }

    var body: some View {

        List{

            ForEach(infoList, id:\.self){ content in

                Text(content)
            }

        }


    }
}

Answer 1

我将从放弃使用字典开始，使用提供更有用模型的简单递归结构：

struct Item: Identifiable {
    let id = UUID()
    var title: String
    var children: [Item] = []
}

然后你可以定义一个递归View来显示这个模型：

struct NavView: View { 
    var item: Item

    var body: some View {
        NavigationView {
            List {
                ForEach(item.children) { child in
                    if child.children.isEmpty {
                        Text(child.title)
                    } else {  
                        NavigationLink(destination: NavView(item:child)) {
                            Text(child.title)
                        }
                    }
                }
            }.navigationBarTitle(item.title)
        }
    }
}

然后您可以根据需要创建层次结构并将根传递给您的视图：

let root = Item(title:"",children:
    [Item(title:"A",children:
        [Item(title:"A1", children:
            [Item(title:"A11"),Item(title:"A12")]
            )]),
          Item(title:"B",children:
            [Item(title:"B1"),Item(title:"B2")]
        )]
)

NavView(item:root))

Answer 2

从提供的快照中不清楚应该如何呈现字典列表，实际上，为什么它使用字典作为基本模型，但与不确定性无关，这里是可能的方向（不是最终的，但经过测试 &可以考虑)

struct PlayingWithMultiDimNavLink2: View {

    let multiDimDict: [String: [[String: [String]]]]
    = ["A": [["A1": ["A11", "A12"]], ["A2": ["A21", "A22"]]],
       "B": [["B1": ["B11", "B12"]], ["B2": ["B21", "B22"]]]
        ]

    var body: some View {
       NavigationView{
            GenericChildView(model: multiDimDict)
        }
    }
}

struct GenericChildView: View {
    let model: [String: Any]

    var body: some View {
        List{
            ForEach(model.keys.sorted(), id: \.self) { key in
                self.nextLevelView(for: key)
            }
        }
    }

    private func nextLevelView(for key: String) -> some View {
        let next = model[key]
        return Group {
            if next as? String != nil {
                Text(next as! String)
            } else if next as? [Any] != nil {
                NavigationLink(destination: GenerateChildListView(infoList: next as! [Any]))
                    { Text(key) }
            } else if next as? [String: Any] != nil {
                NavigationLink(destination: GenericChildView(model: next as! [String: Any]))
                    { Text(key) }
            }
        }
    }
}

struct GenerateChildListView: View {

    var infoList: [Any]

    init(infoList: [Any]){
        self.infoList = infoList
    }

    var body: some View {
        Group {
            if infoList.count == 1 {
                self.singleContentView()
            }
            if infoList.count > 1 {
                List{
                    self.contentView()
                }
            }
        }
    }

    private func singleContentView() -> some View {
        return Group {
            if infoList as? [String] != nil {
                Text(infoList[0] as! String)
            } else if infoList as? [[String: Any]] != nil {
                GenericChildView(model: self.infoList[0] as! [String: Any])
            }
        }
    }

    private func contentView() -> some View {
        return Group {
            if infoList as? [String] != nil {
                ForEach(infoList as! [String], id:\.self) { content in
                    Text(content)
                }
            } else if infoList as? [[String: Any]] != nil {
                ForEach((infoList as! [[String: Any]]).indices, id: \.self) { i in
                    GenericChildView(model: self.infoList[i] as! [String: Any])
                }
            }
        }
    }
}

如何使代码更简单，更易于在 swiftui 中使用多维字典导航链接重用？

how to make the code simpler and more reusable for navigation links with multi dimension dictionary in swiftui?

ios

swiftui

swiftui-navigationlink