如何将JSON中的键值对更改为XML属性而不是节点

使用Newtonsoft时不需要开箱即用。相反,您必须创建包含所有节点的XML,然后将其重写,将所有节点移动到属性,并删除子节点。

您可以使用这两种方法来实现这一点 XmlDocument 和 XNode

string json = @"{
    ""ItemDetails"": [
    {
        ""ItemNo"": ""0001"",
        ""Desc"": ""Office Supplies"",
        ""Note"": """",
        ""Units"": ""20""
    }
    ]}";

XNode node = JsonConvert.DeserializeXNode(json, "root");

// select all ItemDetails
var itemDetails = node.XPathSelectElements("//ItemDetails");

foreach (XElement item in itemDetails)
{
    foreach (XNode childNode in item.Nodes().ToList())
    {
        // add attribute to node
        var element = childNode as XElement;
        item.SetAttributeValue(element.Name, element.Value);

        // remove the childnode
        element.Remove();
    }
}

Console.WriteLine(node.Document.ToString());

试运行: https://dotnetfiddle.net/EVDwHN

XmlDocument示例

string json = @"{
    ""ItemDetails"": [
    {
        ""ItemNo"": ""0001"",
        ""Desc"": ""Office Supplies"",
        ""Note"": """",
        ""Units"": ""20""
    }
    ]}";

// using xmldocument
XmlDocument doc = JsonConvert.DeserializeXmlNode(json, "root", true);

// select all ItemDetails
var itemDetails = doc.SelectNodes("//ItemDetails");

foreach (XmlNode item in itemDetails)
{
    foreach (XmlNode childNode in item.ChildNodes.Cast<XmlNode>().ToList())
    {
        var attribute = doc.CreateAttribute(childNode.Name);
        attribute.Value = childNode.InnerText;

        // add attribute to node
        item.Attributes.Append(attribute);

        // remove the childnode
        item.RemoveChild(childNode);
    }
}

Console.WriteLine(doc.InnerXml);

https://dotnetfiddle.net/9HtdsU