先看ASPX：
復制代碼 代碼如下:
<!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>Petter Liu demo</title>
<script src="http://img.jb51.net/jslib/jquery/jquery14.js" type="text/javascript"></script>
<script type="text/javascript">
$(function() {
$("input:submit").click(function() {
$("#HiddenField1").val($(this).attr("id")
+ " 引起一個 postback");
});
});
</script>
</head>
<body>
<form id="form1" runat="server">
<div>
<asp:Button ID="Button1" runat="server" Text="Button1" />
<asp:Button ID="Button2" runat="server" Text="Button2" />
<asp:Button ID="Button3" runat="server" Text="Button3" />
<asp:HiddenField ID="HiddenField1" runat="server" />
</div>
</form>
</body>
</html>
然後在Sever端這麼寫：
復制代碼 代碼如下:
/// <summary>
/// Handles the Load event of the Page control.
/// </summary>
/// <param name="sender">The source of the event.</param>
/// <param name="e">The <see cref="System.EventArgs"/> instance containing the event data.</param>
/// <remarks>Author Petter Liu http://wintersun.cnblogs.com </remarks>
protected void Page_Load(object sender, EventArgs e)
{
Response.Write(HiddenField1.Value);
}
很簡單的CODE.
希望這篇POST對您有幫助。
