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json转换为简单json。（输入json的结构未知）

例如，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
            }
        }
    }
}

被转换成这样

{
    "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都将更改为代表其实际路径的键。

这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);
            }
        }
    }

并得到错误的结果

{
    "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
}

将感谢您提供有关纠正我的代码的帮助，或任何其他存档方法。

Answer 1

• 您无需将属性的值转换为字符串，然后每次都再次解析-只需将其转换为 JObject
• 您不需要复杂的条件逻辑来生成属性的名称-只需使用以下命令： 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);
        }
    }
}

您可以这样称呼它：

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