Asp.net Jquery Dialog 2

来源:互联网 发布:linux 退出gdb调试 编辑:程序博客网 时间:2024/06/05 05:18

<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 openModalDiv(divname) {
        $('#' + divname).dialog({
                autoOpen: false,

                buttons: {

                    Ok: function () {

                        $("[id*=Button1]").click();

                    },

                    Close: function () {

                        $(this).dialog('close');

                    }

                }

           });
        $('#' + divname).dialog('open');
        $('#' + divname).parent().appendTo($("form:first"));
    }

    function closeModalDiv(divname) {
        $('#' + divname).dialog('close');
    }
</script>


</head>

<body>

    <form id="form1" runat="server">

 <input id="Button3" type="button" value="Open 1" onclick="javascript:openModalDiv('Div1');" />

    <div id="dialog" style="display:none">

    Click OK to do PostBack!

    </div>

 <div id="Div1" title="Basic dialog" style="display:none">
    <asp:UpdatePanel ID="UpdatePanel1" runat="server">
       <ContentTemplate>
          postback test<br />
   <input id="Ok" type="text" value="OK"/>
        </ContentTemplate>
    <asp:UpdatePanel>
</div>

    <asp:Button ID="Button1" runat="server" Text="Button" style = "display:none" OnClick = "Button1_Click" />

    </form>

</body>

</html>

原创粉丝点击