读取XML文件问题？

Question

<?xml version="1.0" encoding="utf-8" ?>
<UserConment>
  <News>
    <ID>1786</ID>
    <Conment>这个很好
    </Conment>
  </News>
  <News>
    <ID>1968</ID>
    <Conment>
      我认为这个还有待改进
    </Conment>
  </News>
  <News>
    <ID>1786</ID>
    <Conment>
      这个不行 效果不好
    </Conment>
  </News>
  ... ... ...
  ... ... ...
</UserConment>
我要根据ID号读取某个ID的所有结点信息，这个XML文件数据量很大，采用什么什么方式？XmlReader只能读，XmlDocument使用XPath很慢，那么采用什么解决方法好呢？

Answer 1

  XmlDocument document = new XmlDocument();
            document.Load(@"path");
            XmlNode basicnode = document.DocumentElement;
            if (basicnode.Name == "UserConment")
            {
                foreach (XmlNode node in basicnode.ChildNodes)
                {
                    if (node.Name == "News")
                    {
                        foreach (XmlNode cnode in node.ChildNodes)
                        {
                            if (cnode.Name == "Id" && cnode.InnerText == "1786")
                            {
                                ID = cnode.InnerText;
                            }
                            if (cnode.Name == "Conment")
                            {
                                Conment = cnode.InnerText;
                            }
                        }
                    }
                }            }
写同样.   if (cnode.Name == "Id" && cnode.InnerText == "1786")
  {
      cnode.InnerText = ID;
  }
在后面加个.
  document.Save(path)就可以了.

Answer 2

using System;
using System.Collections.Generic;
using System.ComponentModel;
using System.Data;
using System.Drawing;
using System.Text;
using System.Windows.Forms;
using System.Xml; namespace WindowsApplication211
{
    public partial class Form1 : Form
    {
        public Form1()
        {
            InitializeComponent();            XmlDocument Doc = new XmlDocument();
            Doc.LoadXml("<?xml version=\"1.0\" encoding=\"utf-8\" ?>"
                + "<UserConment>"
                + "<News>"
                + "<ID>1786</ID>"
                + "<Conment>这个很好"
                + "</Conment>"
                + "</News>"
                + "<News>"
                + "<ID>1968</ID>"
                + "<Conment>"
                + "我认为这个还有待改进"
                + "</Conment>"
                + "</News>"
                + "</UserConment>");            XmlNode Node=Doc.SelectSingleNode("//UserConment/News/ID[.=\"1786\"]");
            MessageBox.Show(Node.ChildNodes[0].InnerText);  
        }
    }
}

Answer 3

修改下:
 foreach (XmlNode cnode in node.ChildNodes)
{
       if (cnode.Name == "Id" && cnode.InnerText != "1786")
     {
      continue;
     }
    if(code.Name="Id")
    {
      ID = cnode.InnerText;    }
    if (cnode.Name == "Conment")
     {
       Conment = cnode.InnerText;
     }
 }

Answer 4

如果是.net3.5的，用Linq好了， var result1 = from p in doc.Descendants("News")
                          .Where(u => u.Element("ID").Value == "1723")
                          select p;
            XElement xel1 = result.FirstOrDefault();

Answer 5

用的是2.0  XmlDocument对于大数据量的查询 很慢啊

Answer 6

为啥不放到DB里啊 数据量大的话还是DB好 没有办法的话也就只能是XPath了