How can I draw a “soft” line in WPF (presumably using a LinearGradientBrush)?

问题

I'm trying to draw a line with soft edges, regardless of the slope.

Here's the code I have so far:

<Line   HorizontalAlignment="Stretch" VerticalAlignment="Center"
        Stretch="Uniform" StrokeThickness="5" X1="0" Y1="0" X2="1" Y2="0">
    <Shape.Stroke>
        <LinearGradientBrush StartPoint="0,0" EndPoint="0,1">
            <GradientStop Color="Transparent" Offset="0" />
            <GradientStop Color="Green" Offset="0.5" />
            <GradientStop Color="Transparent" Offset="1" />
        </LinearGradientBrush>
    </Shape.Stroke>
</Line>

This makes sense to me, since the line is horizontal, and the linear gradient is vertical, with the edges being transparent and the middle of the line being solid green.

The result is pleasing:

Zoomed in so you can see the gradient:
http://img225.imageshack.us/img225/5027/horizontalsoftlinezoomeb.png

However, when the line is no longer horizontal, the gradient is calculated based on the line's bounding rectangle, rather than on the geometry of the line itself. The result is a slanted line that is shaded vertically, instead of the gradient being perpendicular to the line:

Does anyone know how WPF handles soft edges? I can't find anything on Google or MSDN, and I know there is a way to do this somewhow...

回答1:

Well, I do not know if that is applicable to your scenario but you could simply rotate the horizontal line using LayoutTransform and the gradient will be okay.

<Line   HorizontalAlignment="Stretch" VerticalAlignment="Center"
    Stretch="Uniform" StrokeThickness="5" X1="0" Y1="0" X2="1" Y2="0">
<Shape.Stroke>
    <LinearGradientBrush StartPoint="0,0" EndPoint="0,1">
        <GradientStop Color="Transparent" Offset="0" />
        <GradientStop Color="Green" Offset="0.5" />
        <GradientStop Color="Transparent" Offset="1" />
    </LinearGradientBrush>
</Shape.Stroke>
    <Line.LayoutTransform>
        <RotateTransform Angle="40"/>
    </Line.LayoutTransform>

回答2:

Try to use a shape instead of a line

<Path Data="M0,0 L25,25 z" Fill="#FFF4F4F5" StrokeThickness="5" Canvas.Left="122" Canvas.Top="58">
<Path.Stroke>
    <LinearGradientBrush EndPoint="1.135,0.994" StartPoint="-0.177,-0.077">
        <GradientStop Color="Black"/>
        <GradientStop Color="#FF68A8FF" Offset="1"/>
    </LinearGradientBrush>
</Path.Stroke>

Tomer

回答3:

You can stack a lot of paths with increasing thickness and decreasing color tints, drawing one over the other.

For all the paths to have the same Geometry, you should use Element Binding to the Data property of one of them.

Most probably some code-behind would be useful to generate the paths and color gradients dynamically, if needed.

回答4:

You could set MappingMode="Absolute" on your 'LinearGradientBrush'. Then your brush start/end coordinates aren't relative to the bounding box. Of course, you'd have to crunch some trigonometry to get the right points...

https://docs.microsoft.com/en-us/dotnet/api/system.windows.media.gradientbrush.mappingmode?view=netframework-4.7.2#System_Windows_Media_GradientBrush_MappingMode

来源：https://stackoverflow.com/questions/1412833/how-can-i-draw-a-soft-line-in-wpf-presumably-using-a-lineargradientbrush

标签

.net

wpf

xaml

.net-3.5

gradient