看到一篇老外的介绍，介绍如何在弹出的窗口中引发父窗口的postback事件，现转载如下：
首先，建立main.aspx文件，写入如下代码，其中的一个按钮设置为visiable,只是为了效果而设置。
<script language="javascript">
//To cause postback "as" the Button
function PostBackOnMainPage(){
<%=Page.GetPostBackEventReference(btnPostBack)%>
}
//Helper just to open popup
function openPopUp(){
window.open('popup.aspx','popup','width=400,height=100');
//Maybe handling something else also, like giving extra arguments etc etc
}
</script>
<script runat="server" language="VB">
'This is to react to the postback
Protected Sub btnPostBack_Click(ByVal sender As Object, ByVal e As EventArgs)
lblShowPostInfo.Text = "Postback happened: " & DateTime.Now.ToString()
End Sub
</script>
<form id="Form1" method="post" runat="server">
<input type="button" onclick="openPopUp()" value="Open the popup">
<asp:Button ID="btnPostback" runat="server" Visible="false" OnClick="btnPostBack_Click" />
<asp:Label ID="lblShowPostInfo" runat="server" />
</form>
可以看到，button按钮引发openpopup的事件，弹出窗口。为了明显，设置一个文本lblshowpostinfo，显示引发postback的时间.而在popup.aspx中，代码如下：
<input type=button onclick="opener.PostBackOnMainPage()" value="Cause a postback on opening page" />
作用是点这个按钮，引发父窗口的postbackonmainpage()事件
