Could this be supported with a background color brush type?

[lukewire129]

I can't use the LinearGradientBrush type because I only specify it as BackgroundColor(Color).

LinearGradientBrush type because I can't set the LinearGradientBrush type.
support is possible?

[taublast]

Hi.
Actually, the properties SkiaShape is using are not WPF-inspired brushes but rather those inspired by Figma, to allow its designs to be easily recreated:
So you have properties:

• BackgroundColor
• FillGradient
• StrokeGradient (+StrokeWidth, StrokeColor)

So looking at your code it can rewritten:

    BackgroundColor="Black" whatever
   <draw:SkiaShape.FillGradient>
        <draw:SkiaGradient
            EndXRatio="1"
            EndYRatio="0"
            StartXRatio="0"
            StartYRatio="1"
            Type="Linear">
            <draw:SkiaGradient.Colors>
                <Color>#0A5125</Color>
                <Color>#00FFFF</Color>
                <Color>#00FF00</Color>
            </draw:SkiaGradient.Colors>
            <draw:SkiaGradient.ColorPositions>
                <x:Double>0.1</x:Double>
                <x:Double>0.5</x:Double>
                <x:Double>1.0</x:Double>
            </draw:SkiaGradient.ColorPositions>
        </draw:SkiaGradient>
    </draw:SkiaShape.FillGradient>

It is not excluded at all that we start using WPF-like brushes instead of this approach at some point, just need time to evaluate all the pros and cons.

[lukewire129]

Thank you!