创建嵌套的子JSON

Question

如何在C＃中创建此嵌套子json？我无法弄清楚如何嵌套添加到子级列表。我正在使用这个。

JSON格式

 {
"name": "Top Level",
"parent": "null",
"children": [
  {
    "name": "Level 2: A",
    "parent": "Top Level",
    "children": [
      {
        "name": "Son of A",
        "parent": "Level 2: A"
        ,
         "children": [
      {
        "name": "Son of A",
        "parent": "Level 2: A"

      },
      {
        "name": "Daughter of A",
        "parent": "Level 2: A"
      }
    ]

      },
      {
        "name": "Daughter of A",
        "parent": "Level 2: A"
      }
    ]
  },
  {
    "name": "Level 2: B",
    "parent": "Top Level"
  }
 ]
}

根与子的图示

类

 public class d3_mitch
{


    public string  name { get; set; }


    public string parent { get; set; }

    public List<d3_mitch> children { get; set; }
}

我在做什么

  d3_mitch d3 = new d3_mitch();
        d3.id = 1;
        d3.name = "Root";
        d3.type = "Root";
        d3.description = "abc blah blah";
        d3.children = new List<d3_mitch>()
        {

              new d3_mitch() { name = "Carnivores", type = "Type", id = 2, description = "Diet consists solely of animal materials",
            children=new List<d3_mitch>(){ new d3_mitch() { id= 3 ,name="Felidae",type="Family",description="Also known as cats"} }
           }
        };

现在问题出在

如何将孩子添加到ID = N的父母的每个列表中？

像这样

 d3.children = new List<d3_mitch>()
        {

              new d3_mitch() { name = "Carnivores", type = "Type", id = 2, description = "Diet consists solely of animal materials",
            children=new List<d3_mitch>(){ new d3_mitch() { id= 3 ,name="Felidae",type="Family",description="Also known as cats",



           //This one
            children=new List<d3_mitch>(){ new d3_mitch() { id = 4, name = "Felidae4", type = "Family", description = "Also known as cats" } }



            } }
           }
        };

Answer 1

由于您的类定义在示例中有所不同，因此出于此示例的目的，我做了自己的定义：

public class MyNode
{
    public int Id { get; set; }
    public int ParentId { get; set; }
    public string Name { get; set; }
    public List<MyNode> Nodes { get; set; }

    public MyNode()
    {
        Nodes = new List<MyNode>();
    }
}

接下来，我们可以创建一个平面列表，其节点之间的关系由其ID定义：

var rootNode = new MyNode() { ParentId = -1, Id = 1, Name = "Root" };
var nodes = new [] {
    rootNode,
    new MyNode() { ParentId = 1, Id = 2, Name = "ChildA" },
    new MyNode() { ParentId = 1, Id = 3, Name = "ChildB" },
    new MyNode() { ParentId = 2, Id = 4, Name = "ChildA.A" },
    new MyNode() { ParentId = 2, Id = 5, Name = "ChildA.B" },
    new MyNode() { ParentId = 3, Id = 6, Name = "ChildB.A" },
    new MyNode() { ParentId = 3, Id = 7, Name = "ChildB.B" }
};

然后，我们可以遍历每个项目，查找其父项并将其添加到其父项的子级列表中：

var nodeDict = nodes.ToDictionary(n => n.Id);
foreach (var item in nodes)
{
    MyNode parentNode = null;
    if (item.ParentId == -1 || !nodeDict.TryGetValue(item.ParentId, out parentNode)) // we don't want to look up the root node since it doesn't have a parent. you might want to add error handling if the parent node isn't found
    {
        continue;
    }

    parentNode.Nodes.Add(item);
}

然后我们可以序列化结果：

Console.WriteLine(JsonConvert.SerializeObject(rootNode, Formatting.Indented));

在线尝试

使用类似技术的稍微复杂一点的方法是创建一个“构建器”：

public interface IRootHierarchyBuilder 
{
    IHierarchyBuilder AddRootNode(MyNode rootNode);
}

public interface IHierarchyBuilder
{
    IHierarchyBuilder AddNode(MyNode childNode);
    MyNode Build();
}

public class HierarchyBuilder : IRootHierarchyBuilder, IHierarchyBuilder
{
    private readonly IDictionary<int, MyNode> _nodes;
    private MyNode _rootNode;

    private HierarchyBuilder()
    {
        _nodes = new Dictionary<int, MyNode>();
    }

    public static IRootHierarchyBuilder Create()
    {
        return new HierarchyBuilder();
    }

    IHierarchyBuilder IRootHierarchyBuilder.AddRootNode(MyNode rootNode)
    {
        if (_rootNode != null)
        {
            throw new InvalidOperationException("Root node already exists.");
        }
        _rootNode = rootNode;
        _nodes[rootNode.Id] = rootNode;
        return this;
    }

    IHierarchyBuilder IHierarchyBuilder.AddNode(MyNode childNode)
    {
        if (_rootNode == null)
        {
            throw new InvalidOperationException("Root node not set.");
        }

        if (_nodes.ContainsKey(childNode.Id))
        {
            throw new Exception("This child has already been added.");
        }

        MyNode parentNode;
        if (!_nodes.TryGetValue(childNode.ParentId, out parentNode))
        {
            throw new KeyNotFoundException("Parent node not found.");
        }
        parentNode.Nodes.Add(childNode);

        _nodes[childNode.Id] = childNode;
        return this;
    }

    MyNode IHierarchyBuilder.Build()
    {
        if (_rootNode == null)
        {
            throw new InvalidOperationException("Root node not set.");
        }
        return _rootNode;            
    }
}

然后可以这样使用：

var rootNode = HierarchyBuilder.Create()
    .AddRootNode(new MyNode() { Id = 1, ParentId = -1, Name = "Root" })
    .AddNode(new MyNode() { Id = 2, ParentId = 1, Name = "ChildA" })
    .AddNode(new MyNode() { Id = 3, ParentId = 1, Name = "ChildB" })
    .AddNode(new MyNode() { Id = 4, ParentId = 2, Name = "ChildA.A" })
    .AddNode(new MyNode() { Id = 5, ParentId = 2, Name = "ChildA.B" })
    .AddNode(new MyNode() { Id = 6, ParentId = 3, Name = "ChildB.A" })
    .AddNode(new MyNode() { Id = 7, ParentId = 3, Name = "ChildB.B" })
    .Build();

再一次，我们可以序列化结果：

Console.WriteLine(JsonConvert.SerializeObject(rootNode, Formatting.Indented));

在线尝试

Answer 2

@John已经展示了两种自动的方法来实现此目的，如果我们手动构建树，也许您会更清楚地看到该模式。

首先，有几点要点：

1. 有意义地命名您的班级，d3_mitch听起来就像下一部《星球大战》电影一样。
2. C＃属性按惯例以大写字母开头。

请注意，您的图形表示形式也将从更好的命名约定中受益。就目前而言，节点C1R3出现在树中的两个位置，这是不可能的，因为树不允许循环。您需要为此使用图表。无论如何，我创建了一个Node具有相同名称的新名称，只是类似于该图。

根据类表示形式以及如何馈送数据，有两个选项：

1. 通过构造函数注入

节点类

public class Node
{
    public Node(int id, int parentId, string name, List<Node> children)
    {
        Id = id;
        ParentId = parentId;
        Name = name;
        Children = children;
    }

    public int Id { get; }
    public int ParentId { get; }
    public string Name { get; }
    public List<Node> Children { get; }
}

手工组装树

var tree =
    new Node(1, -1, "R", new List<Node>
    {
        new Node(2, 1, "C1R1", new List<Node>()),
        new Node(3, 1, "C2R2", new List<Node>()),
        new Node(4, 1, "C3R3", new List<Node>
        {
            new Node(5, 4, "C1R3", new List<Node>()),
            new Node(6, 4, "C2R3", new List<Node>
            {
                new Node(7, 6, "C1R3", new List<Node>())
            })
        })
    });

2. 通过设置属性

节点类

public class Node
{
    public int Id { get; set; }
    public int ParentId { get; set; }
    public string Name { get; set; }
    public List<Node> Children { get; set; }
}

手工组装树

var tree =
    new Node
    {
        Id = 1,
        ParentId = -1,
        Name = "R",
        Children = new List<Node>
        {
            new Node
            {
                Id = 2,
                ParentId = 1,
                Name = "C1R1",
                Children = new List<Node>()
            },
            new Node
            {
                Id = 3,
                ParentId = 1,
                Name = "C2R2",
                Children = new List<Node>()
            },
            new Node
            {
                Id = 4,
                ParentId = 1,
                Name = "C3R3",
                Children = new List<Node>
                {
                    new Node
                    {
                        Id = 5,
                        ParentId = 4,
                        Name = "C1R3",
                        Children = new List<Node>()
                    },
                    new Node
                    {
                        Id = 6,
                        ParentId = 4,
                        Name = "C2R3",
                        Children = new List<Node>
                        {
                            new Node
                            {
                                Id = 7,
                                ParentId = 6,
                                Name = "C1R3",
                                Children = new List<Node>()
                            },
                        }
                    }
                }
            },
        }
    };