100分 求问!C#读取XML的问题

Question

这是xml文件的一部分  总的很长 只贴一部分
<?xml version="1.0" encoding="utf-8" ?>
<Message>
<FormName Value="LoginForm">
<ID Value="0001">
<Type>Error</Type>
<Title>test</Title>
<Value>test1</Value>
</ID>
<ID Value="0002">
<Type>Error</Type>
<Title>test</Title>
<Value>test2</Value>
</ID>
</FormName>
<FormName Value="Common">
<ID Value="0001">
<Type>Error</Type>
<Title>test</Title>
<Value>test3</Value>
</ID>
</FormName>
</Message>现在我想在.cs中 ,取得 xml里面 formname的值 和 对应节点下面 的 value的值
请问怎么写!

Answer 1

using System.Xml;void ReadXml(string fileName)
{
    XmlDocument xd = new XmlDocument();
    xd.Load(fileName);
    foreach (XmlNode xn in xd.SelectNodes("Message/FormName"))
    {
        Console.WriteLine("FormName:" + xn.Attributes["Value"].Value);
        foreach (XmlNode childXN in xn.SelectNodes("ID"))
            Console.WriteLine("ID:" + childXN.Attributes["Value"].Value);
    }
}

Answer 2

xml相当于本地数据库，不知道下面这个对你是否有帮助：//xml文件信息
<abc>
  <Field>100</Field>
  <item>
    <id>1</id>
    <name>zhangsan</name>
    <sex>男</sex>
  </item>
  <item>
    <id>2</id>
    <name>lisi</name>
    <sex>男</sex>
  </item>
</abc>
 
//实体类。
 public class Information
    {
        private string id;
        public string Id
        {
            get { return id; }
            set { id = value; }
        }
        private string name;
        public string Name
        {
            get { return name; }
            set { name = value; }
        }
        private string sex;
        public string Sex
        {
            get { return sex; }
            set { sex = value; }
        }
        public Information()
        { 
            
        }
        public Information(string id,string name,string sex)
        {
            this.Id = id;
            this.Name = name;
            this.Sex = sex;
        }
    }
//读取xml里面的文件信息
 List<Information> list = new List<Information>();
            //实例化xml
            XmlDocument xml = new XmlDocument();
            //读取xml文件
            xml.Load(@"E:\C#\S2C#\DLCL\打印电脑\MyComputer\XulieHua\XML.xml");  //你的xml地址
            string id = "";
            string name = "";
            string sex = "";
            Information info = null;
            //////////*******下面开始循环读取xml文件信息********/
            ///////////////
            foreach (XmlNode node in xml.ChildNodes)
            {
                if (node.Name == "abc")
                {
                    foreach (XmlNode node1 in node.ChildNodes)
                    {
                        if (node1.Name == "item")
                        {
                            foreach (XmlNode node2 in node1.ChildNodes)
                            {
                                switch (node2.Name)
                                {
                                    case "id":
                                        id = node2.InnerText;
                                        break;
                                    case "name":
                                        name = node2.InnerText;
                                        break;
                                    default:
                                        sex = node2.InnerText;
                                        break;
                                }
                            }
                            info = new Information(id, name, sex);
                            //将信息保存至集合
                            list.Add(info);
                        }
                    }
                }
            }

Answer 3

这是经常测试的 部分代码:        static string[] GetXmlDocument( string path ) {
            string[] ss = default(string[]);
            List<string> ls = new List<string>();
            XDocument xd = XDocument.Load(path);
            foreach (var item in xd.Root.Elements()) {
                if (item.Name == "FormName") {
                    ls.Add(item.Attribute("Value").Value);
                }
            }
            ss = new string[ls.Count];
            ls.CopyTo(ss);
            return ss;                 
        }
//
输出 
LoginForm
Common

Answer 4

www.bbs180.com上有一些关于xml的解答。

Answer 5

楼主是不是要取<Value>节点的值而不是id属性的value值，我把这两个值都取出来了，如下
using System;
using System.Xml;
class read
{
static void Main()
{
XmlDocument xd = new XmlDocument();
string filename=@"f:\test\form.xml";
    xd.Load(filename);
    foreach (XmlNode xn in xd.SelectNodes("Message/FormName"))
    {
        Console.WriteLine("FormName:" + xn.Attributes["Value"].Value);  
        foreach (XmlNode childXN in xn.ChildNodes)
{
   Console.WriteLine("ID:" + childXN.Attributes["Value"].Value);
   Console.WriteLine("Value:" + childXN["Value"].InnerXml);
        
        }    
    }}
}

Answer 6

        public static XmlNodeList ReadRSS(string urlRSS)
        {
            XmlDocument doc = new XmlDocument();            doc.Load(urlRSS);            return doc.GetElementsByTagName("item");
        }

Answer 7

 public static void Outputxml(string filename)
        {
               XmlDocument xmldoc= new XmlDocument();
               xmldoc.Load(fileName);            try
            {
                XmlNodeList NodeList = xmldoc.SelectNodes("Message/FormName");
                foreach (XmlNode Node in NodeList)
                {                    richTextBox1.AppendText(   Node["FormName "].Attributes["Value"].Value+"\r\n");                }
                
            }
            catch (XmlException ex)
            {
               
            }
        }