C#中读取XML方法（一）—— DataSet方式获取

转自：http://hi.baidu.com/pao8041/blog/item/73dca638df6680c3d462254e.html

 

 

（一）首先创建XML文件“XMLFile.xml”：

<?xml version="1.0" encoding="utf-8" ?>
<people>
     <name>Edison Liang</name>
     <age>23</age>
     <sex>male</sex>
</people>

（二）编写页面脚本文件：

<%@ Page Language="C#" AutoEventWireup="true"     CodeFile="Default.aspx.cs" Inherits="_Default" %>

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">

<html xmlns="http://www.w3.org/1999/xhtml" >
<head runat="server">
       <title>无标题页</title>
</head>
<body>
       <form id="form1" runat="server">
       <div>
           <asp:GridView ID="gv_xml" runat="server">
           </asp:GridView>
    
       </div>
           <asp:Button ID="Button1" runat="server" OnClick="Button1_Click" Text="Button" />
       </form>
</body>
</html>

（三）读取XML文件内容：

using System;
using System.Data;
using System.Configuration;
using System.Web;
using System.Web.Security;
using System.Web.UI;
using System.Web.UI.WebControls;
using System.Web.UI.WebControls.WebParts;
using System.Web.UI.HtmlControls;

public partial class _Default : System.Web.UI.Page
{
       protected void Page_Load(object sender, EventArgs e)
       {

       }

       protected void Button1_Click(object sender, EventArgs e)
       {
           DataSet ds = new DataSet();
           ds.ReadXml(Server.MapPath("XMLFile.xml"));
           gv_xml.DataSource = ds.Tables["people"].DefaultView;
           gv_xml.DataBind();
       }
}

posted on 2008-08-09 19:50  王丹小筑  阅读(795)  评论(0)    收藏  举报

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