Hello,
I created 2 lightboxes on my page. When I just had one, everything worked fine.
The 2 lightboxes are not displayed at the same time. There are 2 links that open each respectly.
<a href="#" class="lightbox1">Open box 1</a>
<a href="#" class="lightbox2">Open box 2</a>
The following is the HTML/CSS:
<div class="backdrop"></div>
<div id="wapper" class="centered">
<div class="box1">
<div class="close">x</div>
<div class="clear"></div>
<form name="form1" id="form1">
...
</form>
</div>
<div class="box2">
<div class="close">x</div>
<div class="clear"></div> <!-- end of div clear -->
<div id="responce">
</div> <!-- end of div response -->
</div> <!-- end of box2 -->
</div><!--end of wapper -->
<style>
.backdrop
{
position:absolute;
top:0px;
left:0px;
width:100%;
height:100%;
background:#000;
opacity: .0;
filter:alpha(opacity=0);
z-index:50;
display:none;
}
.box1, .box2
{
position:absolute;
top:20%;
left:30%;
width:500px;
height:300px;
background:#ffffff;
z-index:51;
padding:10px;
-webkit-border-radius: 5px;
-moz-border-radius: 5px;
border-radius: 5px;
-moz-box-shadow:0px 0px 5px #444444;
-webkit-box-shadow:0px 0px 5px #444444;
box-shadow:0px 0px 5px #444444;
display:none;
}
.close
{
float:right;
margin-right:6px;
cursor:pointer;
}
</style>
I need help with jquery/javascript.
What I want to do is something like:
<script type="text/javascript">
$(document).ready(function(<lightbox>, <box>){
$('<lightbox1>').click(function(){
$('.backdrop, <box>').animate({'opacity':'.50'}, 300, 'linear');
$('<box>').animate({'opacity':'1.00'}, 300, 'linear');
$('.backdrop, <box>').css('display', 'block');
});
$('.close').click(function(){
close_box(<box>);
});
$('.backdrop').click(function(){
close_box(<box>);
});
});
function close_box(<box>)
{
$('.backdrop, <box>').animate({'opacity':'0'}, 300, 'linear', function(){
$('.backdrop, <box>').css('display', 'none');
});
}
</script>
I know this is not possible. I tried using if-else but I can't seem to get the syntax right with document.ready.
Can someone please HELP!!!
Apprecite any help as always!
drjay