我已经使用下面的方法将JSON转换为XML

List <Item> items;
public void LoadJsonAndReadToXML() {
  using(StreamReader r = new StreamReader(@ "E:\Json\overiddenhotelranks.json")) {
    string json = r.ReadToEnd();
    items = JsonConvert.DeserializeObject <List<Item>> (json);
    ReadToXML();
  }
}

And

public void ReadToXML() {
  try {
    var xEle = new XElement("Items",
      from item in items select new XElement("Item",
        new XElement("mhid", item.mhid),
        new XElement("hotelName", item.hotelName),
        new XElement("destination", item.destination),
        new XElement("destinationID", item.destinationID),
        new XElement("rank", item.rank),
        new XElement("toDisplayOnFod", item.toDisplayOnFod),
        new XElement("comment", item.comment),
        new XElement("Destinationcode", item.Destinationcode),
        new XElement("LoadDate", item.LoadDate)
      ));

    xEle.Save("E:\\employees.xml");
    Console.WriteLine("Converted to XML");
  } catch (Exception ex) {
    Console.WriteLine(ex.Message);
  }
  Console.ReadLine();
}

我使用名为Item的类来表示元素

public class Item {
  public int mhid { get; set; }
  public string hotelName { get; set; }
  public string destination { get; set; }
  public int destinationID { get; set; }
  public int rank { get; set; }
  public int toDisplayOnFod { get; set; }
  public string comment { get; set; }
  public string Destinationcode { get; set; }
  public string LoadDate { get; set; }
}

它的工作原理……

感谢David Brown的回答。在JSON的例子中。Net 3.5中，转换方法在JsonConvert静态类下:

XmlNode myXmlNode = JsonConvert.DeserializeXmlNode(myJsonString); // is node not note
// or .DeserilizeXmlNode(myJsonString, "root"); // if myJsonString does not have a root
string jsonString = JsonConvert.SerializeXmlNode(myXmlNode);

我已经使用下面的方法将JSON转换为XML

List <Item> items;
public void LoadJsonAndReadToXML() {
  using(StreamReader r = new StreamReader(@ "E:\Json\overiddenhotelranks.json")) {
    string json = r.ReadToEnd();
    items = JsonConvert.DeserializeObject <List<Item>> (json);
    ReadToXML();
  }
}

And

public void ReadToXML() {
  try {
    var xEle = new XElement("Items",
      from item in items select new XElement("Item",
        new XElement("mhid", item.mhid),
        new XElement("hotelName", item.hotelName),
        new XElement("destination", item.destination),
        new XElement("destinationID", item.destinationID),
        new XElement("rank", item.rank),
        new XElement("toDisplayOnFod", item.toDisplayOnFod),
        new XElement("comment", item.comment),
        new XElement("Destinationcode", item.Destinationcode),
        new XElement("LoadDate", item.LoadDate)
      ));

    xEle.Save("E:\\employees.xml");
    Console.WriteLine("Converted to XML");
  } catch (Exception ex) {
    Console.WriteLine(ex.Message);
  }
  Console.ReadLine();
}

我使用名为Item的类来表示元素

public class Item {
  public int mhid { get; set; }
  public string hotelName { get; set; }
  public string destination { get; set; }
  public int destinationID { get; set; }
  public int rank { get; set; }
  public int toDisplayOnFod { get; set; }
  public string comment { get; set; }
  public string Destinationcode { get; set; }
  public string LoadDate { get; set; }
}

它的工作原理……

试试这个函数。我刚刚写了它，还没有太多机会测试它，但我的初步测试是有希望的。

public static XmlDocument JsonToXml(string json)
{
    XmlNode newNode = null;
    XmlNode appendToNode = null;
    XmlDocument returnXmlDoc = new XmlDocument();
    returnXmlDoc.LoadXml("<Document />");
    XmlNode rootNode = returnXmlDoc.SelectSingleNode("Document");
    appendToNode = rootNode;

    string[] arrElementData;
    string[] arrElements = json.Split('\r');
    foreach (string element in arrElements)
    {
        string processElement = element.Replace("\r", "").Replace("\n", "").Replace("\t", "").Trim();
        if ((processElement.IndexOf("}") > -1 || processElement.IndexOf("]") > -1) && appendToNode != rootNode)
        {
            appendToNode = appendToNode.ParentNode;
        }
        else if (processElement.IndexOf("[") > -1)
        {
            processElement = processElement.Replace(":", "").Replace("[", "").Replace("\"", "").Trim();
            newNode = returnXmlDoc.CreateElement(processElement);
            appendToNode.AppendChild(newNode);
            appendToNode = newNode;
        }
        else if (processElement.IndexOf("{") > -1 && processElement.IndexOf(":") > -1)
        {
            processElement = processElement.Replace(":", "").Replace("{", "").Replace("\"", "").Trim();
            newNode = returnXmlDoc.CreateElement(processElement);
            appendToNode.AppendChild(newNode);
            appendToNode = newNode;
        }
        else
        {
            if (processElement.IndexOf(":") > -1)
            {
                arrElementData = processElement.Replace(": \"", ":").Replace("\",", "").Replace("\"", "").Split(':');
                newNode = returnXmlDoc.CreateElement(arrElementData[0]);
                for (int i = 1; i < arrElementData.Length; i++)
                {
                    newNode.InnerText += arrElementData[i];
                }

                appendToNode.AppendChild(newNode);
            }
        }
    }

    return returnXmlDoc;
}

你也可以用.NET Framework做这些转换:

JSON到XML:使用System.Runtime.Serialization.Json

var xml = XDocument.Load(JsonReaderWriterFactory.CreateJsonReader(
    Encoding.ASCII.GetBytes(jsonString), new XmlDictionaryReaderQuotas()));

XML转JSON:使用System.Web.Script.Serialization

var json = new JavaScriptSerializer().Serialize(GetXmlData(XElement.Parse(xmlString)));

private static Dictionary<string, object> GetXmlData(XElement xml)
{
    var attr = xml.Attributes().ToDictionary(d => d.Name.LocalName, d => (object)d.Value);
    if (xml.HasElements) attr.Add("_value", xml.Elements().Select(e => GetXmlData(e)));
    else if (!xml.IsEmpty) attr.Add("_value", xml.Value);

    return new Dictionary<string, object> { { xml.Name.LocalName, attr } };
}

对于转换JSON字符串到XML尝试:

    public string JsonToXML(string json)
    {
        XDocument xmlDoc = new XDocument(new XDeclaration("1.0", "utf-8", ""));
        XElement root = new XElement("Root");
        root.Name = "Result";

        var dataTable = JsonConvert.DeserializeObject<DataTable>(json);
        root.Add(
                 from row in dataTable.AsEnumerable()
                 select new XElement("Record",
                                     from column in dataTable.Columns.Cast<DataColumn>()
                                     select new XElement(column.ColumnName, row[column])
                                    )
               );


        xmlDoc.Add(root);
        return xmlDoc.ToString();
    }

要将XML转换为JSON，请尝试以下方法:

    public string XmlToJson(string xml)
    {
       XmlDocument doc = new XmlDocument();
       doc.LoadXml(xml);

       string jsonText = JsonConvert.SerializeXmlNode(doc);
       return jsonText;
     }