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Re: [css3-images] exactly 2 adjacent colors stops

From: Alan Gresley <alan@css-class.com>
Date: Tue, 17 May 2011 16:51:36 +1000
Message-ID: <4DD21AF8.9090906@css-class.com>
To: "Tab Atkins Jr." <jackalmage@gmail.com>
CC: Brian Manthos <brianman@microsoft.com>, www-style list <www-style@w3.org>
On 17/05/2011 11:20 AM, Tab Atkins Jr. wrote:
> On Mon, May 16, 2011 at 6:11 PM, Alan Gresley<alan@css-class.com>  wrote:
>> FYI, The code *-gradient(red 10px, blue 10px) when used with
>> repeating-linear-gradient can be abused to create solid uniform color
>> (opaque ~ transparent) shapes. Currently IE10 preview wrongly shows red.
> [snip]
>> I would prefer to see an non hacky way to do this. Perhaps a property called
>> 'solid-color()'
> Less abusively, you could just do "linear-gradient(blue, blue)".

Your correct. I made this assumption when using something like this.

   linear-gradient(rgba(0,0,255,1), rgba(0,0,255,1))

> Somewhat less hackily (mechanically, it's hacky, but it *looks* good
> when you do it), just use "image(blue)" once image() is supported.
> image()'s main functionality is letting you specify multiple urls and
> fallback when they don't work, and you can supply a color as the final
> fallback for the case when all the images fail.  "image(blue)" just
> forgoes the images and skips straight to the fallback.  I think it
> looks appropriate to create a blue image with "image(blue)".
> ~TJ

OK, please see image().


I'm am looking and I not sure where I should look. What I want to know 
is if image() allows for comma separated values.

Here is a test using linear-gradient() as comma separated values for 


A question. Is my syntax correct since this is only working as intended 
in Chrome11?

FF4, Aurora and IE10 preview have two white blocks (above and to the 
left of the gold block) missing as well as the final background layer of 

Alan Gresley
Received on Tuesday, 17 May 2011 06:52:07 UTC

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