XML父节点下多个同名子节点读取

[dengyi422]

<?xml version="1.0" encoding="utf-8" ?>
<list>
  <tablename id="WorkOrder" name="工单">
    <dataFiled>filed1</dataFiled>
    <dataNote>字段1</dataNote>
    <dataFiled>filed2</dataFiled>
    <dataNote>字段2</dataNote>
    <dataFiled>filed3</dataFiled>
    <dataNote>字段3</dataNote>
    <dataFiled>filed4</dataFiled>
    <dataNote>字段4</dataNote>
    <dataFiled>filed5</dataFiled>
    <dataNote>字段5</dataNote>
  </tablename>
  <tablename id="Other" name="其他">
    <dataFiled>filed1</dataFiled>
    <dataNote>字段1</dataNote>
    <dataFiled>filed2</dataFiled>
    <dataNote>字段2</dataNote>
    <dataFiled>filed3</dataFiled>
    <dataNote>字段3</dataNote>
    <dataFiled>filed4</dataFiled>
    <dataNote>字段4</dataNote>
    <dataFiled>filed5</dataFiled>
    <dataNote>字段5</dataNote>
  </tablename>
</list>

XML文件中格式是这样的 如何读取tablename下所有dataFiled和dataNote的所有信息。用逗号分隔。
谢谢

[dengyi422]

XmlDocument xmlDoc = new XmlDocument(); string strPath = HttpContext.Current.Server.MapPath("xml_Data.xml"); xmlDoc.Load(strPath); XmlNodeList nodes = xmlDoc.SelectNodes("list/tablename"); foreach (XmlNode node in nodes) { if (node.Attributes["name"].Value == "工单") { XmlNodeList nodelist = node.SelectNodes("dataFiled"); foreach (XmlNode child in nodelist) strFiled += child.InnerText + ","; } } 我弄好了，欢迎接分 速度

[全栈极简]

string result = string.Empty;

            XmlDocument xmlDocument = new XmlDocument();
            xmlDocument.Load("test.xml");

            XmlNodeList nodeList = xmlDocument.SelectNodes("list/tablename");
            foreach (XmlNode node in nodeList)
            {
                if (node.HasChildNodes)
                {
                    XmlNodeList temp = node.ChildNodes;
                    foreach (XmlNode item in temp)
                    {
                        result += item.InnerText + ",";
                    }
                }
            }

            Console.WriteLine(result);

[全栈极简]

逻辑基本差不多