本文介绍如何利用jQuery和jQuery UI库创建弹出对话框，并通过设置按钮事件来触发服务器端回发操作，实现用户交互与服务器响应之间的无缝衔接。
<html
xmlns="http://www.w3.org/1999/xhtml">
<head
id="Head1"
runat="server">
<title>Untitled Page</title>
<script type="text/javascript"
src="http://ajax.googleapis.com/ajax/libs/jquery/1.7.2/jquery.min.js"></script>
<script src="http://ajax.aspnetcdn.com/ajax/jquery.ui/1.8.9/jquery-ui.js"
type="text/javascript"></script>
<link href="http://ajax.aspnetcdn.com/ajax/jquery.ui/1.8.9/themes/start/jquery-ui.css"
rel="stylesheet"
type="text/css" />
<script type="text/javascript">
$(function () {
$("#dialog").dialog({
buttons: {
Ok:
function () {
$("[id*=Button1]").click();
},
Close:
function () {
$(this).dialog('close');
}
}
});
});
</script>
</head>
<body>
<form id="form1"
runat="server">
<div id="dialog"
style="display:none">
Click OK to do PostBack!
</div>
<asp:Button
ID="Button1"
runat="server" Text="Button"
style =
"display:none"
OnClick =
"Button1_Click" />
</form>
</body>
</html>
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