CSS RGBA border / background alpha double
- by stockli
I'm working on a website that has a lot of transparency involved, and I thought I would try to build it entirely in RGBA and then do fallbacks for IE. I need a "facebox" style border effect, where the outer border is rounded and is less opaque than the background of the box it surrounds.
The last example from http://24ways.org/2009/working-with-rgba-colour seems to suggest that it's possible, but I can't seem to get it to work. When I try the following:
<!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" xml:lang="en" lang="en">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8"/>
<title>RGBA Test</title>
<style type='text/css'>
body {
background: #000;
color: #fff;
}
#container {
width: 700px;
margin: 0 auto;
background: rgba(255, 255, 255, 0.2);
border: 10px solid rgba(255, 255, 255, 0.1);
padding: 20px;
}
</style>
</head>
<body>
<div id='container'>
This should look like a facebox.
</div>
</body></html>
It seems like the background "extends" underneath the border of the element, which causes the pixel values to get added together. Thus, when both the background and the border are semi-transparent, the border will ALWAYS be more opaque than the background of the element. This is exactly the opposite of what I am trying to achieve, but it seems like it should be possible based on the examples I've seen.
I should also add that I can't use another element inside the container, because I'm also going to use a border-radius on the container to get rounded corners, and webkit squares the corners of the child elements if they have a background assigned, which would essentially mean a rounded outer border with square contents.
Sorry I can't post an image of this... Apparently I don't have enough rep to post an image.
