在ASP.NET 2.0中,对于跨页提交已经有了非常合理的解决方案
下面是一个示例:
BeginPage.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 Name
{
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
="请输入姓名"
Width
="183px"
></
asp:Label
>
<
asp:TextBox
ID
="TextBox1"
runat
="server"
></
asp:TextBox
>
<
asp:Button
ID
="Button1"
runat
="server"
Text
="提交"
PostBackUrl
="~/EndPage.aspx"
/></
div
>
</
form
>
</
body
>
</
html
>
EndPage.aspx:请注意PreviousPageType的属性设置
<%
@ 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.Name;
}
</
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
>
OK,就通过这么简单的两个属性设置,就可以非常方便的得到跨页提交的特性
本文转自高海东博客园博客,原文链接:http://www.cnblogs.com/ghd258/archive/2005/10/26/262511.html,如需转载请自行联系原作者