.net 操作xml --移除注释节点

 1 /// <summary> 2 ????????/// xml字符串转xml文档 ?忽略注释信息 3 ????????/// </summary> 4 ????????/// <param name="sXml">xml字符串</param> 5 ????????/// <returns></returns> 6 ????????public ?XmlDocument XmlFromStrIgnoreComment(string sXml) 7 ????????{ 8 ????????????try 9 ????????????{10 ????????????????XmlDocument _XmlDocument = new XmlDocument();11 ????????????????_XmlDocument.LoadXml(sXml);12 ????????????????return this.XmlRemoveComment(_XmlDocument);13 ????????????}14 ????????????catch (Exception)15 ????????????{16 ????????????????return null;17 ????????????}18 19 ????????}20 ????????/// <summary>21 ????????/// xml文档移除 注释节点22 ????????/// </summary>23 ????????/// <param name="sXml">xml文档</param>24 ????????/// <returns></returns>25 ????????public ?XmlDocument XmlRemoveComment(XmlDocument sXml)26 ????????{27 ????????????List<XmlNode> dXmlNodeList = new List<XmlNode>();28 ????????????this.GetXmlRemoveComment(sXml.ChildNodes, dXmlNodeList);29 ????????????foreach (XmlNode vXmlNode in dXmlNodeList)30 ????????????{31 ????????????????vXmlNode.ParentNode.RemoveChild(vXmlNode);32 ????????????}33 ????????????return sXml;34 ????????}35 36 ????????/// <summary>37 ????????/// 获取xml中的 注释节点38 ????????/// </summary>39 ????????/// <param name="xmlNodeList">待查找结点</param>40 ????????/// <param name="dXmlNodeList">注释节点</param>41 ????????public void GetXmlRemoveComment(XmlNodeList xmlNodeList, List<XmlNode> dXmlNodeList)42 ????????{43 ????????????foreach (XmlNode xmlNode in xmlNodeList)44 ????????????{45 ????????????????if (xmlNode.NodeType == XmlNodeType.Comment)46 ????????????????{47 ????????????????????dXmlNodeList.Add(xmlNode);48 ????????????????}49 ????????????????else50 ????????????????{51 ????????????????????if (xmlNode.HasChildNodes)52 ????????????????????{53 ????????????????????????GetXmlRemoveComment(xmlNode.ChildNodes, dXmlNodeList);54 ????????????????????}55 ????????????????}56 ????????????}57 ????????}