c# 读取xml文件内容

Question

  <?xml version="1.0" encoding="utf-8" ?> 
- <response>
- <TAG>
  <tagid>105463687068</tagid> 
  <mac>00:18:8E:20:27:9C</mac> 
  <type>t301b</type> 
  <createdon>2009-05-19 17:27:16+0800</createdon> 
  <serialnumber>301B-0850-10141</serialnumber> 
  <swversion>1.1.0.50</swversion> 
  <name /> 
  <profile>slow</profile> 
  <custom>279c</custom> 
  <taggroupid>196</taggroupid> 
  <taggroupid>325</taggroupid> 
  <taggroupid>123</taggroupid> 
  <taggroupid>132</taggroupid> 
  <taggroupid>321</taggroupid>   <attributekey>defaultGroup</attributekey> 
  <attributevalue>325</attributevalue> 
  <pending>false</pending> 
  </TAG>
  </response>
怎么样得到这里面所有的taggroupid的值？

Answer 1

using System.Xml;private List<string> GetTagGroupIdList(string xmlfilePath)
{
    XmlTextReader textReader = null;
    List<string> listStr = new List<string>();    try
    {
        textReader = new XmlTextReader(xmlfilePath);        while (textReader.Read())
        {
            if (textReader.NodeType == XmlNodeType.Element && textReader.IsStartElement())
            {
                if (textReader.Name == "taggroupid")
                {
                    listStr.Add(textReader.ReadInnerXml());
                }
             }
         }         return listStr;
    }
    catch
    {
        textReader.Close();
        return null;
    }
}

Answer 2

//读节点1----------------------------------------------------------------
// XmlTextReader tr=new XmlTextReader("books.xml");
// while(tr.Read())
// {
// if(tr.NodeType==XmlNodeType.Text)
// listBox1.Items.Add(tr.Value);
// }
//----------------------------------------------------------------------
//读属性
FileStream fs=new FileStream("books.xml",FileMode.Open);
XmlTextReader tr=new XmlTextReader(fs);
while(tr.Read())
{
if(tr.MoveToContent()==XmlNodeType.Element)
{

for(int i=0;i<tr.AttributeCount;i++)
{
listBox1.Items.Add(tr.GetAttribute(i));
}
}

}
//读节点2----------------------------------------------------------------
private void button1_Click(object sender, System.EventArgs e)
{
FileStream fs=new FileStream("books.xml",FileMode.Open);
XmlTextReader tr=new XmlTextReader(fs);
while(!tr.EOF)
{
if(tr.MoveToContent()==XmlNodeType.Element&&tr.Name=="title")

{
//从元素中读取文本数据
//listBox1.Items.Add(tr.ReadElementString());//自动定位到下一个节点
LoadList(tr);//忽略异常

}
else
{
tr.Read();
}
}
}
private void LoadList(XmlReader reader)
{
try
{
listBox1.Items.Add(reader.ReadElementString());//自动定位到下一个节点
}
catch(XmlException er){}//ignore }
//----------------------------------------------------------------------

Answer 3

楼上得可以
  不过一般用dom读的比较好。

Answer 4

对XML的操作 接触得有点少    我顶了   学习啊

Answer 5

XmlDocument xml=new XmlDocument();
xml.LoadX("文件");
XmlNodeList nodeList=xml.SelectNodeList("response/tag/taggroupid");
foreach(Node n in nodeList){
 String t=node.FirstChild.Value;
}

Answer 6

太简单了DataSet ds_xml.ReadXml(path + @"\SystemKey.xml");
string ss="";
foreach (DataTable d in ds_xml.Tables)
{
    if (d.TableName == "taggroupid" && d.Rows.Count > 0)
    {
foreach (DataRows row in d.Rows)
{
 ss+=row["taggroupid_Text"].ToString()
}
    }
}messgebox.show(ss);

Answer 7

DataSet ds_xml=new DataSet(); ds_xml.ReadXml(path + @"\SystemKey.xml"); 写漏了

Answer 8

 XmlDocument xmlDocument = new XmlDocument();
            xmlDocument.Load(@"F:\SDK\XML\XML\XMLFile2.xml");//改成你文件的路径            XmlNodeList xmlNodeList=xmlDocument.SelectNodes("response/TAG/taggroupid[text()]");            for (int i = 0; i < xmlNodeList.Count; i++)
            {
               string str= xmlNodeList.Item(i).InnerText;
            }

Answer 9

        public List<string> ParseFile(string fileName)
        {
            List<string> strList = new List<string>();
            XmlDocument doc = new XmlDocument();
            try
            {
                doc.Load(fileName);
                XmlNodeList TAGNode = doc.SelectNodes("response/TAG/taggroupid");
                foreach (XmlNode node in TAGNode)
                {
                    strList.Add(node.InnerText);
                }
            }
            catch (Exception ex)
            {
            }
            return strList;
        }