c# - 将嵌套的 JSON 转换为简单的 JSON

Question

我正在尝试通过递归遍历将嵌套的 json 转换为简单的 json。(输入json结构未知)

比如我想要这样的json

{
    "FirstName": "Rahul",
    "LastName": "B",
    "EmpType": {
        "RID": 2,
        "Title": "Full Time"
    },
    "CTC": "3.5",
    "Exp": "1",
    "ComplexObj": {
        "RID": 3,
        "Title": {
            "Test": "RID",
            "TWO": {
                "Test": 12
            }
        }
    }
}

转换成这样

{
    "FirstName": "Rahul",
    "LastName": "B",
    "EmpType__RID": 2,
    "EmpType__Title": "Full Time",
    "CTC": "3.5",
    "Exp": "1",
    "ComplexObj__RID": 3,
    "ComplexObj__Title__Test": "RID",
    "ComplexObj__Title__TWO__Test": 12
}

嵌套对象中的每个字段将更改为表示其实际路径的键。

这就是我到目前为止所做的。

    public static void ConvertNestedJsonToSimpleJson(JObject jobject, ref JObject jobjectRef, string currentNodeName = "", string rootPath = "")
    {
        string propName = "";
        if (currentNodeName.Equals(rootPath))
        {
            propName = currentNodeName;
        }
        else
        {
            propName = (rootPath == "" && currentNodeName == "") ? rootPath + "" + currentNodeName : rootPath + "__" + currentNodeName;
        }

        foreach (JProperty jprop in jobject.Properties())
        {
            if (jprop.Children<JObject>().Count() == 0)
            {
                jobjectRef.Add(propName == "" ? jprop.Name : propName + "__" + jprop.Name, jprop.Value);
            }
            else
            {
                currentNodeName = jprop.Name;
                rootPath = rootPath == "" ? jprop.Name : rootPath;
                ConvertNestedJsonToSimpleJson(JObject.Parse(jprop.Value.ToString()), ref jobjectRef, currentNodeName, rootPath);
            }
        }
    }

得到错误的结果

{
    "FirstName": "Rahul",
    "LastName": "B",
    "EmpType__RID": 2,
    "EmpType__Title": "Full Time",
    "CTC": "3.5",
    "Exp": "1",
    "EmpType__ComplexObj__RID": 3,
    "EmpType__Title__Test": "RID",
    "EmpType__two__Test": 12
}

将感谢任何有关更正我的代码的帮助，或任何其他存档方法。

Answer 1

• 您不需要每次都将属性的值转换为字符串然后重新解析它——只需将其转换为JObject
• 您不需要复杂的条件逻辑来生成属性名称 - 只需使用:prefix + jprop.Name + "__"

代码:

public static void FlattenJson(JObject node, JObject result, string prefix = "")
{
    foreach (var jprop in node.Properties())
    {
        if (jprop.Children<JObject>().Count() == 0)
        {
            result.Add(prefix + jprop.Name, jprop.Value);
        }
        else
        {
            FlattenJson((JObject)jprop.Value, $"{prefix}{jprop.Name}__", result);
        }
    }
}

你可以这样调用它:

var node = JObject.Parse(/* the input string */);
var result = new JObject();
FlattenJson(node, result);