Newtonsoft Json - 根据给定的选定属性序列化深度嵌套对象

Question

Newtonsoft Json 是否有任何方法可以序列化仅具有选定属性的嵌套对象？ 模型如下：

class Node
{
int id{ get; set; };
string name{ get; set; };
IList<Node>children{ get; set; };
}

之后我正在创建一个看起来像这样的树结构：

{
    "id": 1,
    "name": "A",
    "children": [
        {
            "id": 2,
            "name": "B",
            "children": null
        },
        {
            "id": 3,
            "name": "C",
            "children": [
                {
                    "id": 10,
                    "name": "D",
                    "children": null
                }
            ]
        }
    ]
}

编写了我自己的序列化程序。我只通过了 Children 属性。它显示为：

{
    "children": [
        {
            "children": null
        },
        {
            "children": [
                {
                    "children": null
                }
            ]
        }
    ]
}

我想显示所有属性，而不仅仅是 1st/root 的子项下的子项。与下面类似。

{
    "children": [
        {
            "id": 2,
            "name": "B",
            "children": null
        },
        {
            "id": 3,
            "name": "C",
            "children": [
                {
                    "id": 10,
                    "name": "D",
                    "children": null
                }
            ]
        }
    ] }

还有我要如何对children.Id做同样的事情，我想序列化所有children.Id的根节点，类似于下面

{
"children": [
    {
        "id": 2
    },
    {
        "id": 3
    }
]

}

Answer 1

他们已经完成了您希望在此页面上实现的相同结果。

https://weblog.west-wind.com/posts/2012/aug/30/using-jsonnet-for-dynamic-json-parsing

另外，请看这里，

https://entityframeworkcore.com/knowledge-base/58283745/self-reference-when-serializing-nested-object

希望对你有用。

Answer 2

您需要自己的序列化程序来做什么？试试标准的

var jsonDeserialized = JsonConvert.DeserializeObject<Child>(json);
json = JsonConvert.SerializeObject(jsonDeserialized);

输出

{
  "id": 1,
  "name": "A",
  "children": [
    {
      "id": 2,
      "name": "B",
      "children": null
    },
    {
      "id": 3,
      "name": "C",
      "children": [
        {
          "id": 10,
          "name": "D",
          "children": null
        }
      ]
    }
  ]
}

类

public partial class Child
{
    [JsonProperty("id")]
    public long Id { get; set; }

    [JsonProperty("name")]
    public string Name { get; set; }

    [JsonProperty("children")]
    public List<Child> Children { get; set; }
}

为此

{
    "children": [
        {
            "id": 2,
            "name": "B",
            "children": null
        },
        {
            "id": 3,
            "name": "C",
            "children": [
                {
                    "id": 10,
                    "name": "D",
                    "children": null
                }
            ]
        }
    ] }

你可以使用

var jsonDeserialized = JsonConvert.DeserializeObject<Data>(json);

public partial class Data
{
    [JsonProperty("children")]
    public List<Child> Children { get; set; }
}