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This is the code for button on .aspx page when i click on this button
<asp:Button ID="Button_Login" runat="server" CssClass="button_Login" Text="Login" OnClientClick="show()" />
the java script functions
document.getElementById('light').style.display = 'block';
document.getElementById('fade').style.display = 'block'
are called BUT IT DOES NOT STAY !!!
<style type="text/css">
.black_overlay{
display:none;
position: absolute;
top: 0%;
left: 0%;
width: 100%;
height: 100%;
background-color:black;
z-index:1001;
-moz-opacity: 0.8;
opacity:.80;
filter: alpha(opacity=80);
}
.white_content {
display:none;
position: absolute;
top: 25%;
left: 35%;
width: 35%;
padding: 0px;
border: 0px solid #a6c25c;
background-color: white;
z-index:1002;
overflow: auto;
}
.headertext{
font-family:Arial, Helvetica, sans-serif;
font-size:14px;
color:#f19a19;
font-weight:bold;
}
.textfield
{
border:1px solid #a6c25c;
width:135px;
}
.button2
{
 background-color:#a6c25c;
 color:White;
 font-size:11px;
 font-weight:bold;
 border:1px solid #7f9db9;
 width:68px;
}
</style>
<script type="text/javascript" >
    function show() {
    document.getElementById('light').style.display = 'block';
    document.getElementById('fade').style.display = 'block';
 
     }
</script> 
Posted 5-Feb-13 21:27pm
sr_24484
Edited 5-Feb-13 22:22pm
v4
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Solution 1

if You're Doing PostBack On Button Click Event then Jquery will lose its state after postback means Login Pop-up will Disappear.
 
use this Thread-
 
http://stackoverflow.com/questions/1477219/jquery-fancybox-not-working-after-postback?rq=1[^]
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Solution 2

May be because of Postback your code is not working.
 
Look at code below....
 
<script type="text/javascript">
    function show() {
    document.getElementById('light').style.display = 'block';
    document.getElementById('fade').style.display = 'block';
 
    return false;
     }
    </script>
 
Here I included "return false". Which avoids postback for a button click and your code will work fine...
 
Hope it works...
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