C#读取XML

Question

读取这样的XML。  现在要读xCode节点的值，要怎么读呢？<?xml version="1.0" encoding="gb2312"?>
<ActionResult>
<xMsgID>1265268631</xMsgID>
<xCode>1002</xCode>
<xMessage>不存在或已过期</xMessage>
<xSign>0671f0c1b2b377f2dce9f3f1e9cea72e</xSign>
<xValue>12</xValue>
</ActionResult>关键部分代码高手可以提供一下。谢谢。。

Answer 1

try->
  XmlDocument doc = new XmlDocument();
            doc.Load("test1.xml");
            XmlNode node = doc.SelectSingleNode("//xCode");
            if (node != null)
            {
                string str= node.InnerText;
            }

Answer 2

XmlDocument xml = new XmlDocument();
            xml.Load(@"F:\3.xml");
            XmlNode xl = xml.SelectSingleNode("ActionResult/xCode");
            if (xl != null)
            {
                Console.WriteLine(xl.InnerText);
                
            }

Answer 3

如果可以用.NET3.5,
用LINQ TO XML 最容易了.

Answer 4

用XmlDocument来加载 然后SelectSingleNode

Answer 5

把我写的代码给你吧 public void ListRead()
        {
            string fn = XmlPath;//这里就是xml文件地址
            if (File.Exists(fn))
            {
                XmlTextReader reader = null;
                try
                {
                    reader = new XmlTextReader(fn);
                    while (reader.Read())
                    {
                        if (reader.NodeType == XmlNodeType.Element)
                        {
                            switch (reader.LocalName)
                            {
                                case "xCode":
                                    string str = reader.ReadString();  //得到你要的东西了
                                    break;
                                  //这里加上case 语句，可以得到别的东西了
                            }
                        }
                    }
                }
                catch (Exception ex)
                {
                    MessageBox.Show("未知错误！\n" + ex.Message, "error", MessageBoxButtons.OK);
                }
                finally
                {
                    reader.Close();
                }
            }
        }           

Answer 6

XmlDocument doc = new XmlDocument();
doc.Load("a.xml");
string str;
XmlNode node = doc.SelectSingleNode("ActionResult/xCode");
if (node != null)
{
   str=node.InnerText;
}

Answer 7

哎 我都放弃xml了3.5之后用json还可以直接序列化成对象

Answer 8

  XmlDocument doc = new XmlDocument();
            doc.Load("test1.xml");
            XmlNode node = doc.SelectSingleNode("./xCode");
            if (node != null)
            {
                string str= node.InnerText;
            }