Dav*_*own 416
是.为了这个精确的目的,使用包含帮助器方法的JsonConvert类:
// 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);
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这里的文档:使用Json.NET在JSON和XML之间进行转换
DaF*_*Fi4 43
是的,你可以做到(我这样做)但是在转换时要注意一些悖论,并妥善处理.您不能自动符合所有接口的可能性,并且控制转换的内置支持有限 - 许多JSON结构和值不能自动双向转换.请记住,我使用Newtonsoft JSON库和MS XML库的默认设置,因此您的里程可能会有所不同:
{}或嵌套数组[ {} {} ...].您将在JavaScript等中使用这两种不同的方式.符合相同模式的XML的不同示例可以通过这种方式生成实际上不同的JSON结构.您可以将属性json:Array ='true'添加到元素中,以便在某些(但不一定是所有)情况下解决此问题.一个新的更新改变了这一点(感谢Jon Story指出):https://www.newtonsoft.com/json/help/html/T_Newtonsoft_Json_NullValueHandling.htm
请随意提及您注意到的任何其他问题,我已经开发了自己的自定义例程来准备和清理字符串,因为我来回转换.您的情况可能会或可能不会要求准备/清理.正如StaxMan所提到的,你的情况实际上可能需要你在对象之间进行转换......这可能需要适当的接口和一堆case语句/ etc来处理我上面提到的警告.
Ter*_*nja 32
您也可以使用.NET Framework进行这些转换:
JSON to XML:使用System.Runtime.Serialization.Json
var xml = XDocument.Load(JsonReaderWriterFactory.CreateJsonReader(
Encoding.ASCII.GetBytes(jsonString), new XmlDictionaryReaderQuotas()));
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XML到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 } };
}
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Sta*_*Man 29
我不确定这种转换是否有意义(是的,许多人这样做,但主要是通过圆孔强制方形钉) - 结构阻抗不匹配,转换是有损的.所以我建议反对这种格式到格式的转换.
但是如果你这样做,首先从json转换为object,然后从object转换为xml(反之亦然,反向).进行直接转换会导致输出难看,信息丢失或两者兼而有之.
Dav*_*.ca 27
感谢David Brown的回答.在我的JSON.Net 3.5的情况下,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);
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为了不使用外部组件/项目,我搜索了很长时间以找到已接受解决方案的替代代码.由于DynamicJson项目的源代码,我想出了以下内容:
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;
}
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注意:我想要一个XmlDocument而不是XElement用于xPath目的.此外,这段代码显然只是从JSON到XML,有各种方法可以做相反的事情.
以下是将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();
}
}
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要将给定的XML字符串转换为JSON,只需调用XmlToJSON()函数,如下所示.
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" }}
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