- From: Andrew Fedoniouk <news@terrainformatica.com>
- Date: Fri, 14 Aug 2009 18:02:26 -0700
- To: "Tab Atkins Jr." <jackalmage@gmail.com>
- CC: Brad Kemper <brad.kemper@gmail.com>, www-style list <www-style@w3.org>
Tab Atkins Jr. wrote: > On Fri, Aug 14, 2009 at 6:06 PM, Andrew > Fedoniouk<news@terrainformatica.com> wrote: > >> What if that element has background-image (real one) too? >> > > Exactly the same as you would if you specified two backgrounds that > were both url()s. > > background: url(),url(); > is *exactly the same* as > background: url(),linear-gradient(); > in terms of how to treat it. > That is not the answer on my question. How would you remove gradient if element has gradient and background image with (e.g. no-repeat)? I suspect that you will propose something nice as this: background: url(??????),none; But what shall I put instead of ?????? if I don't know that URL(it is defined in other file out of my influence)? In short: gradients shall be uniquely addressable. solid-color() and linear-gradient() are mutually exclusive as they define value of the same thing. So gradient should be addressable by background-color, border-color, etc. I cannot imagine real life situation when you would want to use solid color and gradient color at the same time. That is why I am insisting that gradients are such colors (color fill/distribution functions to be precise). As I said, there are many cases when you will want to have gradients on boxes other than background box of the element. Think about these cases: background-color: linear-gradient(top to bottom / yellow, blue); border-top-color: yellow; border-left-color: linear-gradient(top to bottom / yellow, blue); border-right-color: linear-gradient(top to bottom / yellow, blue); border-bottom-color: blue; column-rule-color: linear-gradient(top to bottom / yellow, blue); -- Andrew Fedoniouk. http://terrainformatica.com
Received on Saturday, 15 August 2009 01:03:11 UTC