SvgColourServer with translucent color?

[svg-net]

SvgColourServer with translucent color?

[tebjan]

Hello,

svg has an extra attribute for the opacity value, or to be precise a few:
http://www.w3.org/TR/SVG/masking.html#ObjectAndGroupOpacityProperties

Its a float in the range 0..1 where 0 is fully transparent and 1 fully opaque.

So instead of o.Fill, you should write a value to o.FillOpacity. There is also o.Opacity (overall opacity) and o.StrokeOpacity (opacity for strokes).

I've set the FIllOpacity and Opacity on all SvgVisualElements to 1 already (did not include that part in my sample code, sorry).

I looked through the source and it appears that the SvgColourServer attempts to extract and re-calculate the alpha channel to use for the brush (https://github.com/vvvv/SVG/blob/master/Source/Painting/SvgColourServer.cs line 27).

I'm going to run some tests and see what's happening here. But it seems like a Color instance with an alpha channel is meant to be supported.

[tebjan]

Hm... Yes, there seems to be some work in that direction. But its as far as i know not in the official svg spec. To control the transparency you have to set the opacity attributes. In your case the FillOpacity to 0.25f.

But anyways the code looks somehow correct. But there could be a problem with the division. Since both values are Integer, i think its always 0 because Colour.A is smaller than 255. it should be:

 int alpha = (int)((opacity * (this.Colour.A/255.0f) ) * 255);

I fixed that by 457a82f. If you can compile the code you can check that right away. Otherwise you have to wait for the next Nuget package.

[tebjan]

Nuget 1.0.1 is now also available.

I think what I was trying to do would result in deeper-nested elements becoming more and more transparent, so it's not right anyway. I appreciate the attempted fix, though.

What I'm doing is programatically recoloring SVGs. So what I'll do when I need to make one translucent is move all of the SVG's child elements into a new group element with the opacity/fill set appropriately.