[请教]asp.net(c#)传送xml

Question

请问，在asp.net如何与一个服务器传送和接受xml文档，谢谢

Answer 1

bcastr.xml的内容格式是:
<?xml version="1.0" encoding="utf-8"?>
<bcaster autoPlayTime="2">
  <item item_url="newspic/834e7edd-3580-4ec0-9d2f-0e037bf2bbca.jpg" link="news.aspx?id=20090301190410">
  </item>
  <item item_url="newspic/3cdcdb2a-3a68-449e-bc1e-9c7cdc6c1a9c.jpg" link="news.aspx?id=20090301190340">
  </item>
  <item item_url="newspic/34eff785-f8e6-4f53-bf6e-a8b09813231b.jpg" link="news.aspx?id=20090301190300">
  </item>
  <item item_url="newspic/0da0aadd-ffb8-4f7a-8ce9-aa3ce88159e2.jpg" link="news.aspx?id=20090301190056">
  </item>
  <item item_url="newspic/d272622d-9e6d-4532-bb1c-53c2b96b3e49.jpg" link="news.aspx?id=20090301190011">
  </item>
</bcaster>/// <summary>
    /// 载入数据到xml
    /// </summary>
    private void LoadXml()
    {        XmlDocument document = new XmlDocument();
        document.Load(Server.MapPath("bcastr.xml"));
        XmlNode root = document.DocumentElement;
        XmlNodeList child = root.ChildNodes;
        News[] data = this.ReadData();
        try
        {
            for (int i = 0; i < child.Count; i++)
            {
                child[i].Attributes[0].InnerText = data[i].Pic;
                child[i].Attributes[1].InnerText = data[i].Url;
            }
        }
        catch
        {
        }
        document.Save(Server.MapPath("xml/bcastr.xml"));
    }    private News[] ReadData()
    {
        News[] Result = new News[5];
        int i = 0;
        using (SqlDataReader read = SqlHelper.ExecuteReader(SqlHelper.ConnectionStringLocalTransaction, CommandType.Text, "select top 5 * from news_table where pic<>''", null))//把数据库里面的数据写入xml文件中.
        {
            while (read.Read())
            {
                Result[i] = new News(read["pic"].ToString().Trim(), "news.aspx?id=" + read["id"].ToString().Trim());
                i++;
            }
        }
        return Result;
    }
}public class News
{
    private string picPath;
    private string urlAddress;
    public News()
    {
    }    public News(string pic, string url)
    {
        picPath = pic;
        urlAddress = url;
    }    public string Pic
    {
        get
        {
            return picPath;
        }
        set
        {
            picPath = value;
        }
    }    public string Url
    {
        get
        {
            return urlAddress;
        }
        set
        {
            urlAddress = value;
        }
    }

Answer 2

我是把 xml 读成dataset 进行传输 接收端再 读取dataset 或者 把dataset 当作xml
因为 dataset 原理 就是XML。方便安全