看到有人写
XML数据与数据集之间的相互转化
我好象用不着XML这个,不如直接搞个RSS吧,实用些。
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using System;
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using System.Collections.Generic;
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using System.Text;
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using System.Data;
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using Rss;
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using System.Xml;
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namespace buildRss
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{
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class buildRss
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{
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/// <summary>
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/// 此类是将传入的DataTable对象进行转化为RSS对象RSSfeed
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/// 用writeRSS()方法输出到设备流,这里主要是输出到服务器返回流
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/// getRss(string fileName)方法用来将DataTable对象以RSS保存到文件中
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/// DataTable对象的列要求,1:ID,2:名称,3:作者,4:描述,5:连接地址
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/// </summary>
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private DataTable dt;
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private Rss.RssFeed rfeed;
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public buildRss()
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{ }
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public buildRss(System.Data.DataTable DT)
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{
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dt = DT;
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}
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/// <summary>
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/// 将DataTable对象写入RSSfeed对象
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/// </summary>
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/// <returns></returns>
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private Rss.RssFeed RssFactory()
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{
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if (dt == null)
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throw new Exception("传入的DataTable对象为空!");
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if (dt.Rows.Count < 1)
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throw new Exception("DataTable对象中无数据!");
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if (dt.Columns.Count < 5)
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throw new Exception("传入的DataTable对象中列数不符合规定!");
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Rss.RssFeed rf = new RssFeed();
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rf.Encoding = System.Text.Encoding.GetEncoding("gb2312");
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rf.Version = RssVersion.RSS20;
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RssItem ri = new RssItem();
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RssChannel rc = new RssChannel();
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rc.Title = "Test Channel Title 1";
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rc.Description = "Test Channel Description 1";
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rc.Link = new Uri("http://www.lutoo.com");
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rc.PubDate = DateTime.Now;
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for (int i = 0; i < dt.Rows.Count;i++ )
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{
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ri.Author = dt.Rows[i][2].ToString();
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ri.Title = dt.Rows[i][1].ToString();
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ri.Description = dt.Rows[i][3].ToString();
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ri.Link = new Uri(dt.Rows[i][4].ToString());
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ri.PubDate = DateTime.Now;
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rc.Items.Add(ri);
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}
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rf.Channels.Add(rc);
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return rf;
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}
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/// <summary>
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/// 传入文件路径带文件名,将RSS写入该文件
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/// </summary>
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/// <param name="fileName">文件名</param>
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public void getRss(string fileName)
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{
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try
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{
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rfeed = this.RssFactory();
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rfeed.Write(fileName);
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}
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catch(Exception e)
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{
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System.Console.Out.WriteLine(e.Message.ToString());
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}
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}
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/// <summary>
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/// 返回RSS字符串
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/// </summary>
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/// <returns>返回RSS字符串</returns>
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public string getRss()
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{
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try
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{
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rfeed = this.RssFactory();
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}
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catch (Exception e)
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{
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System.Console.Out.WriteLine(e.Message.ToString());
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}
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return rfeed.ToString();
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}
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/// <summary>
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///把RSS内容写到流中
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/// </summary>
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/// <param name="stream">要写入的流,指Response.OutputStream</param>
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public void writeRSS(System.IO.Stream stream)
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{
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try
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{
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rfeed = this.RssFactory();
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rfeed.Write(stream);
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}
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catch (Exception e)
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{
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System.Console.Out.WriteLine(e.Message.ToString());
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}
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}
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/*RssFeed feed = new RssFeed();
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feed.Channels.Add(channel);
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Response.ContentType = "text/xml";
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feed.Write(Response.OutputStream);
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Response.End();
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*/
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}
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}
需要RSS.NET.dll

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