Question

我有一个像这样的xml字符串

string stxml="<Status>Success</Status>";

我还创建了一个xml文档

  XmlDocument doc = new XmlDocument();
  XmlNode docNode = doc.CreateXmlDeclaration("1.0", "UTF-8", null);
  doc.AppendChild(docNode);
  XmlNode rootNode = doc.CreateElement("StatusList");
  doc.AppendChild(rootNode);

我需要这样的输出。

  <StatusList>
  <Status>Success</Status>
  </StatusList>

如何实现这一点。如果我们使用innerhtml，它会插入。但我想将xml字符串作为xmlnode本身插入

Answer 1

实现目标的一种非常简单的方法是使用经常被忽视的XmlDocumentFragment类：

  XmlDocument doc = new XmlDocument();
  XmlNode docNode = doc.CreateXmlDeclaration("1.0", "UTF-8", null);
  doc.AppendChild(docNode);
  XmlNode rootNode = doc.CreateElement("StatusList");
  doc.AppendChild(rootNode);

  //Create a document fragment and load the xml into it
  XmlDocumentFragment fragment = doc.CreateDocumentFragment();
  fragment.InnerXml = stxml;
  rootNode.AppendChild(fragment);

Answer 2

使用Linq to XML：

string stxml = "<Status>Success</Status>";
XDocument doc = new XDocument(new XDeclaration("1.0", "utf-8", "yes"),
             new XElement("StatusList", XElement.Parse(stxml)));

Answer 3

您可以改为使用XElement类：

string stxml = "<Status>Success</Status>";
var status = XElement.Parse(stxml);
var statusList = new XElement("StatusList", status);

var output = statusList.ToString(); // looks as you'd like

如果要将新的statusList内容写入文件：

statusList.Save(@"C:\yourFile.xml", SaveOptions.None);

Answer 4

您可以使用xmlwriter

进行尝试

using (XmlWriter writer = XmlWriter.Create("new.xml"))
{
        writer.WriteStartDocument();
        writer.WriteStartElement("StatusList");
        writer.WriteElementString("Status", "Success");   // <-- These are new
        writer.WriteEndDocument();
}

Answer 5

using System;
using System.Collections.Generic;
using System.Xml;
using System.Xml.Serialization;
using System.IO;
using System.Reflection;
using System.ComponentModel;


public class MyClass
{
    public static void RunSnippet()
    {
        XmlNode node = default(XmlNode);
        if(node == null)
            Console.WriteLine(bool.TrueString);
        if(node != null)
            Console.WriteLine(bool.FalseString);

        XmlDocument doc = new  XmlDocument();

        node = doc.CreateNode (XmlNodeType.Element,"Query", string.Empty);

        node.InnerXml=@"<Where><Eq><FieldRef Name=""{0}"" /><Value Type=""{1}"">{2}</Value></Eq></Where>";

        string xmlData = ToXml<XmlNode>(node);

        Console.WriteLine(xmlData);

        XmlNode node1 = ConvertFromString(typeof(XmlNode), @"<Query><Where><Eq><FieldRef Name=""{0}"" /><Value Type=""{1}"">{2}</Value></Eq></Where></Query>") as XmlNode;
        if(node1 == null)
            Console.WriteLine(bool.FalseString);
        if(node1 != null)
            Console.WriteLine(bool.TrueString);

        string xmlData1 = ToXml<XmlNode>(node1);
        Console.WriteLine(xmlData1);
    }
    public static string ToXml<T>(T t)
    {
        string Ret = string.Empty;
        XmlSerializer s = new XmlSerializer(typeof(T));
        using (StringWriter Output = new StringWriter(new System.Text.StringBuilder()))
        {
            s.Serialize(Output, t);
            Ret = Output.ToString();
        }
        return Ret;
    }
        public static object ConvertFromString(Type t, string sourceValue)
        {
            object convertedVal = null;

            Type parameterType = t;
            if (parameterType == null) parameterType = typeof(string);
            try
            {

                // Type t = Type.GetType(sourceType, true);
                TypeConverter converter = TypeDescriptor.GetConverter(parameterType);
                if (converter != null && converter.CanConvertFrom(typeof(string)))
                {
                    convertedVal = converter.ConvertFromString(sourceValue);
                }
                else
                {
                    convertedVal = FromXml(sourceValue, parameterType);
                }
            }
            catch { }
            return convertedVal;
        }
              public static object FromXml(string Xml, Type t)
        {
            object obj;
            XmlSerializer ser = new XmlSerializer(t);
            using (StringReader stringReader = new StringReader(Xml))
            {
                using (System.Xml.XmlTextReader xmlReader = new System.Xml.XmlTextReader(stringReader))
                {
                    obj = ser.Deserialize(xmlReader);
                }
            }
            return obj;
        }

    #region Helper methods

    public static void Main()
    {
        try
        {
            RunSnippet();
        }
        catch (Exception e)
        {
            string error = string.Format("---\nThe following error occurred while executing the snippet:\n{0}\n---", e.ToString());
            Console.WriteLine(error);
        }
        finally
        {
            Console.Write("Press any key to continue...");
            Console.ReadKey();
        }
    }

    private static void WL(object text, params object[] args)
    {
        Console.WriteLine(text.ToString(), args);   
    }

    private static void RL()
    {
        Console.ReadLine(); 
    }

    private static void Break() 
    {
        System.Diagnostics.Debugger.Break();
    }

    #endregion
}

Answer 6

根据我的经验，使用唯一身份证明总是更好我建议您首先查看该情况然后返回此页面，如果您还没有解决方案，请研究/定位我的代码。我刚刚为我自己的项目完成了它，我已经修改了abit以便为你的项目看起来更加集成。祝好运。很抱歉迟到的回复; - ）

                XmlDocument xDoc = new XmlDocument();
                string Bingo = "Identification code";
                xDoc.Load(pathFile);
                XmlNodeList idList = xDoc.GetElementsByTagName("id");
                XmlNodeList statusList = xDoc.GetElementsByTagName("Status");         

                for (int i = 0; i < idList.Count; i++)
                {
                    StatusNode = "<Status>fail</Status>";
                    XmlDocumentFragment fragment = xDoc.CreateDocumentFragment();
                    fragment.InnerXml = StatusNode;
                    statusList[i].InnerXml = "";
                    statusList[i].AppendChild(fragment);
                    if (statusList[i].InnerText == Bingo)
                    {
                    StatusNode = "<Status>Succes!</Status>";
                    fragment.InnerXml = Status;
                    statusList[i].InnerXml = "";
                    statusList[i].AppendChild(fragment);


                    }


                }
                xDoc.Save(pathFile);

将xmlstring转换为XmlNode

6 个答案: