跨页传送-PreviousPage

 

SourcePage.aspx: 请注意Button1的PostBackUrl属性设置

<%@ 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">
    public string YourName
    {
        get
       {
            return this.TextBox1.Text;
        }
    }
</script>
<html xmlns="http://www.w3.org/1999/xhtml" >
<head runat="server">
    <title>Untitled Page</title>
</head>
<body>
    <form id="form1" runat="server">
    <div>
        <asp:Label ID="Label1" runat="server" Text="请输入您的姓名" Width="183px"></asp:Label>
        <asp:TextBox ID="TextBox1" runat="server"></asp:TextBox>
        <asp:Button ID="Button1" runat="server"  Text="提交" PostBackUrl="~/TargetPage.aspx" /></div>
    </form>
</body>
</html>
  TargetPage.aspx:请注意PreviousPageType的属性设置

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<%@ Page Language="C#" %>

<%@ PreviousPageType VirtualPath="~/SourcePage.aspx" %>
<!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)
   {
        this.Label1.Text = PreviousPage.YourName;
    }
</script>
<html xmlns="http://www.w3.org/1999/xhtml" >
<head runat="server">
    <title>Untitled Page</title>
</head>
<body>
    <form id="form1" runat="server">
    <div>
        <asp:Label ID="Label1" runat="server" ></asp:Label>
   
    </div>
    </form>
</body>
</html>

 

PreviousPage!=Page.PreviousPage,

当需要FindControl("")两个都可以用,

当需要得到前一个页面的公共属性时,只能用前一个.

posted @ 2009-12-23 10:06  t_l  阅读(210)  评论(0编辑  收藏  举报