我开始使用Json。NET将JSON格式的字符串转换为对象,反之亦然。在Json中我不确定。NET框架,它是可能的转换字符串在JSON到XML格式,反之亦然?
当前回答
下面是将xml转换为json的完整c#代码
public static class JSon
{
public static string XmlToJSON(string xml)
{
XmlDocument doc = new XmlDocument();
doc.LoadXml(xml);
return XmlToJSON(doc);
}
public static string XmlToJSON(XmlDocument xmlDoc)
{
StringBuilder sbJSON = new StringBuilder();
sbJSON.Append("{ ");
XmlToJSONnode(sbJSON, xmlDoc.DocumentElement, true);
sbJSON.Append("}");
return sbJSON.ToString();
}
// XmlToJSONnode: Output an XmlElement, possibly as part of a higher array
private static void XmlToJSONnode(StringBuilder sbJSON, XmlElement node, bool showNodeName)
{
if (showNodeName)
sbJSON.Append("\"" + SafeJSON(node.Name) + "\": ");
sbJSON.Append("{");
// Build a sorted list of key-value pairs
// where key is case-sensitive nodeName
// value is an ArrayList of string or XmlElement
// so that we know whether the nodeName is an array or not.
SortedList<string, object> childNodeNames = new SortedList<string, object>();
// Add in all node attributes
if (node.Attributes != null)
foreach (XmlAttribute attr in node.Attributes)
StoreChildNode(childNodeNames, attr.Name, attr.InnerText);
// Add in all nodes
foreach (XmlNode cnode in node.ChildNodes)
{
if (cnode is XmlText)
StoreChildNode(childNodeNames, "value", cnode.InnerText);
else if (cnode is XmlElement)
StoreChildNode(childNodeNames, cnode.Name, cnode);
}
// Now output all stored info
foreach (string childname in childNodeNames.Keys)
{
List<object> alChild = (List<object>)childNodeNames[childname];
if (alChild.Count == 1)
OutputNode(childname, alChild[0], sbJSON, true);
else
{
sbJSON.Append(" \"" + SafeJSON(childname) + "\": [ ");
foreach (object Child in alChild)
OutputNode(childname, Child, sbJSON, false);
sbJSON.Remove(sbJSON.Length - 2, 2);
sbJSON.Append(" ], ");
}
}
sbJSON.Remove(sbJSON.Length - 2, 2);
sbJSON.Append(" }");
}
// StoreChildNode: Store data associated with each nodeName
// so that we know whether the nodeName is an array or not.
private static void StoreChildNode(SortedList<string, object> childNodeNames, string nodeName, object nodeValue)
{
// Pre-process contraction of XmlElement-s
if (nodeValue is XmlElement)
{
// Convert <aa></aa> into "aa":null
// <aa>xx</aa> into "aa":"xx"
XmlNode cnode = (XmlNode)nodeValue;
if (cnode.Attributes.Count == 0)
{
XmlNodeList children = cnode.ChildNodes;
if (children.Count == 0)
nodeValue = null;
else if (children.Count == 1 && (children[0] is XmlText))
nodeValue = ((XmlText)(children[0])).InnerText;
}
}
// Add nodeValue to ArrayList associated with each nodeName
// If nodeName doesn't exist then add it
List<object> ValuesAL;
if (childNodeNames.ContainsKey(nodeName))
{
ValuesAL = (List<object>)childNodeNames[nodeName];
}
else
{
ValuesAL = new List<object>();
childNodeNames[nodeName] = ValuesAL;
}
ValuesAL.Add(nodeValue);
}
private static void OutputNode(string childname, object alChild, StringBuilder sbJSON, bool showNodeName)
{
if (alChild == null)
{
if (showNodeName)
sbJSON.Append("\"" + SafeJSON(childname) + "\": ");
sbJSON.Append("null");
}
else if (alChild is string)
{
if (showNodeName)
sbJSON.Append("\"" + SafeJSON(childname) + "\": ");
string sChild = (string)alChild;
sChild = sChild.Trim();
sbJSON.Append("\"" + SafeJSON(sChild) + "\"");
}
else
XmlToJSONnode(sbJSON, (XmlElement)alChild, showNodeName);
sbJSON.Append(", ");
}
// Make a string safe for JSON
private static string SafeJSON(string sIn)
{
StringBuilder sbOut = new StringBuilder(sIn.Length);
foreach (char ch in sIn)
{
if (Char.IsControl(ch) || ch == '\'')
{
int ich = (int)ch;
sbOut.Append(@"\u" + ich.ToString("x4"));
continue;
}
else if (ch == '\"' || ch == '\\' || ch == '/')
{
sbOut.Append('\\');
}
sbOut.Append(ch);
}
return sbOut.ToString();
}
}
要将给定的XML字符串转换为JSON,只需如下所示调用XmlToJSON()函数。
string xml = "<menu id=\"file\" value=\"File\"> " +
"<popup>" +
"<menuitem value=\"New\" onclick=\"CreateNewDoc()\" />" +
"<menuitem value=\"Open\" onclick=\"OpenDoc()\" />" +
"<menuitem value=\"Close\" onclick=\"CloseDoc()\" />" +
"</popup>" +
"</menu>";
string json = JSON.XmlToJSON(xml);
// json = { "menu": {"id": "file", "popup": { "menuitem": [ {"onclick": "CreateNewDoc()", "value": "New" }, {"onclick": "OpenDoc()", "value": "Open" }, {"onclick": "CloseDoc()", "value": "Close" } ] }, "value": "File" }}
其他回答
是的。使用包含辅助方法的JsonConvert类来实现这个精确的目的:
// To convert an XML node contained in string xml into a JSON string
XmlDocument doc = new XmlDocument();
doc.LoadXml(xml);
string jsonText = JsonConvert.SerializeXmlNode(doc);
// To convert JSON text contained in string json into an XML node
XmlDocument doc = JsonConvert.DeserializeXmlNode(json);
这里的文档:使用JSON在JSON和XML之间转换。网
Cinchoo ETL -一个开放源码库,只需几行代码就可以轻松地将Xml转换为JSON
Xml -> JSON:
using (var p = new ChoXmlReader("sample.xml"))
{
using (var w = new ChoJSONWriter("sample.json"))
{
w.Write(p);
}
}
JSON -> Xml
using (var p = new ChoJsonReader("sample.json"))
{
using (var w = new ChoXmlWriter("sample.xml"))
{
w.Write(p);
}
}
样本提琴:https://dotnetfiddle.net/enUJKu
请查看CodeProject文章以获得更多帮助。
声明:我是这个库的作者。
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……
我花了很长时间寻找公认解决方案的替代代码,希望不使用外部程序集/项目。感谢DynamicJson项目的源代码,我想到了以下内容:
public XmlDocument JsonToXML(string json)
{
XmlDocument doc = new XmlDocument();
using (var reader = JsonReaderWriterFactory.CreateJsonReader(Encoding.UTF8.GetBytes(json), XmlDictionaryReaderQuotas.Max))
{
XElement xml = XElement.Load(reader);
doc.LoadXml(xml.ToString());
}
return doc;
}
注意:出于xPath目的,我希望使用XmlDocument而不是XElement。 此外,这段代码显然只能从JSON转换为XML,有各种相反的方法。
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 } };
}