日期:2014-05-18 浏览次数:20619 次
<%@ Page Language="C#" %>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<script runat="server">
protected void Page_Load(object sender, EventArgs e)
{
if (!Page.IsPostBack)
{
System.Data.DataTable dt = new System.Data.DataTable();
System.Data.DataRow dr;
dt.Columns.Add(new System.Data.DataColumn("postId", typeof(System.String)));
dt.Columns.Add(new System.Data.DataColumn("postTime", typeof(System.String)));
dt.Columns.Add(new System.Data.DataColumn("postTitle", typeof(System.String)));
for (int i = 0; i < 8; i++)
{
dr = dt.NewRow();
dr[0] = i.ToString();
dr[1] = (i * i).ToString();
dr[2] = (i * i * i).ToString();
dt.Rows.Add(dr);
}
GridView1.DataSource = dt;
GridView1.DataBind();
}
}
</script>
<html xmlns="http://www.w3.org/1999/xhtml">
<head id="Head1" runat="server">
</head>
<body>
<form id="form1" runat="server">
<asp:GridView ID="GridView1" runat="server" AutoGenerateColumns="false">
<Columns>
<asp:HyperLinkField DataNavigateUrlFields="postId,postTime" DataNavigateUrlFormatString="http://dotnet.aspx.cc/Default.aspx?postId={0}&postTime={1}"
DataTextField="postTitle" HeaderText="标题">
</asp:HyperLinkField>
</Columns>
</asp:GridView>
</form>
</body>
</html>
------解决方案--------------------