请问怎么读写XML中某个节点的值呢？

[www_datasky_cn]

环境：vs2005 、c#、asp.net，
我想取得“系统001”，“对象002”里面的“orderType”的值，请问如何读取呢？
如果我现在想增加一个“系统003”，然后再在“系统003”里面增加“对象008”应该怎么增加节点呢？
总体就是下面这样的XML怎么进行节点的读、改、增、删呢？
请提供较为详细的代码，万分感谢！
<myStyle>
<system sysName="系统001">
<object objName="对象001">
<orderField>id</orderField>
<orderType>0</orderType>
</object>
<object objName="对象002">
<orderField>id</orderField>
<orderType>0</orderType>
</object>
</system>
<system sysName="系统002">
<object objName="对象003">
<orderField>id</orderField>
<orderType>0</orderType>
</object>
<object objName="对象004">
<orderField>id</orderField>
<orderType>0</orderType>
</object>
</system>
</myStyle>

[jinpeng_dotnet]

希望能有些帮助:
/// <summary>
/// 保存Web.config
/// </summary>
/// <param name="strKeyName">节点名</param>
/// <param name="strKeyValue">要保存的值</param>
/// <param name="strWhich">哪个配置文件</param>
public static void SaveSetting(string strKeyName, string strKeyValue,string strWhich)
{
System.Xml.XmlDocument XMLWebSetting = new System.Xml.XmlDocument();
XMLWebSetting.Load(System.Web.HttpContext.Current.Server.MapPath("~/"+ strWhich +""));
System.Xml.XmlNodeList XmlNodeList = XMLWebSetting.SelectSingleNode("//appSettings").ChildNodes;
try
{
foreach (System.Xml.XmlNode xn in XmlNodeList)
{
System.Xml.XmlElement xe = ((System.Xml.XmlElement)(xn));
if (xe.Attributes["key"].InnerText == strKeyName)
{
xe.Attributes["value"].InnerText = strKeyValue;
XMLWebSetting.Save(System.Web.HttpContext.Current.Server.MapPath("~/"+ strWhich +""));
}
}
}
catch (System.Exception Ex)
{
throw Ex;
}
}



/// <summary>
/// 读取 XML中指定节点值
/// </summary>
/// <param name="strKeyName">节点键名</param>
/// <returns>指定节点键值</returns>
public static string GetConfig(string strKeyName)
{
//创建XML文档实例
System.Xml.XmlDocument XMLWebSetting=new System.Xml.XmlDocument();

//打开XML文档
XMLWebSetting.Load(System.Web.HttpContext.Current.Server.MapPath("~/System.config"));

//查找节点所在位置
System.Xml.XmlNodeList XmlNodeList=XMLWebSetting.SelectSingleNode("//appSettings").ChildNodes;

//查找指定节点
string strKeyValue="";
foreach(System.Xml.XmlNode xn in XmlNodeList)
{
System.Xml.XmlElement xe=(System.Xml.XmlElement)xn;
if(xe.Attributes["key"].InnerText==strKeyName)
{
strKeyValue=xe.Attributes["value"].InnerText;
break;
}
}

return strKeyValue;
}

[wangsaokui]

http://community.csdn.net/Expert/topic/5617/5617526.xml?temp=.1240045

[jimu8130]

http://www.cnblogs.com/ocean2000/archive/2006/08/08/471029.html
http://www.cnblogs.com/ocean2000/archive/2006/07/25/459226.html
http://www.cnblogs.com/ocean2000/archive/2006/07/20/455514.html
http://www.cnblogs.com/ocean2000/archive/2006/07/20/455504.html

看看这几篇blog，一定能解决你的问题

[zhangweisjz]

查找一下XPath表达式，非常容易。:)
刚刚做过。

[balenoww]

mark

[hanguoji84]

简单的方法是利用DS读取XML然后像操作表格一样操作XML然后再写出XML文件。