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Re: border-radius

From: Brad Kemper <brkemper.comcast@gmail.com>
Date: Thu, 4 Sep 2008 10:09:15 -0700
Cc: fantasai <fantasai.lists@inkedblade.net>, www-style@w3.org
Message-Id: <85D403D5-323D-4A54-84A1-0711BD5D5895@gmail.com>
To: Nick_Hofstede@inventivegroup.com
On Sep 4, 2008, at 12:34 AM, Nick_Hofstede@inventivegroup.com wrote:

>
> >> The proposed "innermost radius wins" rule doesn't have this problem
> >> in addition to being consistent with the color rule.
>
> > So far, the innermost radius is zero. Are you saying that if the TD
> > has a zero radius and the table has a non-zero radius, that the zero
> > should always win because it is smaller?
> >
> > I think the largest should win.
>
> No, with "innermost radius wins", I mean that the radius of the cell  
> should win.

Understood. But that means if the cell has zero radius and the table  
(or tbody or tr) has a large radius, that the cell wins, and there is  
no corner radii at all. Because no radius = border-radius:0.


> Let the largest (or smallest) radius win is another rule we should  
> consider.
>
> Nick
>
>
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Received on Thursday, 4 September 2008 17:09:54 GMT

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