convert from Color to brush

C#Wpf

C# Problem Overview


How do I convert a Color to a Brush in C#?

C# Solutions


Solution 1 - C#

This is for Color to Brush....

you can't convert it, you have to make a new brush....

SolidColorBrush brush = new SolidColorBrush( myColor );

now, if you need it in XAML, you COULD make a custom value converter and use that in a binding

Solution 2 - C#

Brush brush = new SolidColorBrush(color);

The other way around:

if (brush is SolidColorBrush colorBrush)
    Color color = colorBrush.Color;

Or something like that.

Point being not all brushes are colors but you could turn all colors into a (SolidColor)Brush.

Solution 3 - C#

SolidColorBrush brush = new SolidColorBrush( Color.FromArgb(255,255,139,0) )

Solution 4 - C#

you can use this:

new SolidBrush(color)

where color is something like this:

Color.Red

or

Color.FromArgb(36,97,121))

or ...

Solution 5 - C#

If you happen to be working with a application which has a mix of Windows Forms and WPF you might have the additional complication of trying to convert a System.Drawing.Color to a System.Windows.Media.Color. I'm not sure if there is an easier way to do this, but I did it this way:

System.Drawing.Color MyColor = System.Drawing.Color.Red;
System.Windows.Media.Color = ConvertColorType(MyColor);

System.Windows.Media.Color ConvertColorType(System.Drawing.Color color)
{
  byte AVal = color.A;
  byte RVal = color.R;
  byte GVal = color.G;
  byte BVal = color.B;

  return System.Media.Color.FromArgb(AVal, RVal, GVal, BVal);
}

Then you can use one of the techniques mentioned previously to convert to a Brush.

Solution 6 - C#

I had same issue before, here is my class which solved color conversions Use it and enjoy :

Here U go, Use my Class to Multi Color Conversion

using System;
using System.Windows.Media;
using SDColor = System.Drawing.Color;
using SWMColor = System.Windows.Media.Color;
using SWMBrush = System.Windows.Media.Brush;

//Developed by امین امیری دربان
namespace APREndUser.CodeAssist
{
    public static class ColorHelper
    {
        public static SWMColor ToSWMColor(SDColor color) => SWMColor.FromArgb(color.A, color.R, color.G, color.B);
        public static SDColor ToSDColor(SWMColor color) => SDColor.FromArgb(color.A, color.R, color.G, color.B);
        public static SWMBrush ToSWMBrush(SDColor color) => (SolidColorBrush)(new BrushConverter().ConvertFrom(ToHexColor(color)));
        public static string ToHexColor(SDColor c) => "#" + c.R.ToString("X2") + c.G.ToString("X2") + c.B.ToString("X2");
        public static string ToRGBColor(SDColor c) => "RGB(" + c.R.ToString() + "," + c.G.ToString() + "," + c.B.ToString() + ")";
        public static Tuple<SDColor, SDColor> GetColorFromRYGGradient(double percentage)
        {
            var red = (percentage > 50 ? 1 - 2 * (percentage - 50) / 100.0 : 1.0) * 255;
            var green = (percentage > 50 ? 1.0 : 2 * percentage / 100.0) * 255;
            var blue = 0.0;
            SDColor result1 = SDColor.FromArgb((int)red, (int)green, (int)blue);
            SDColor result2 = SDColor.FromArgb((int)green, (int)red, (int)blue);
            return new Tuple<SDColor, SDColor>(result1, result2);
        }
    }

}

Solution 7 - C#

Here is the Easiest way, No Helpers, No Converters, No weird Media references. Just 1 line:

System.Drawing.Brush _Brush = new SolidBrush(System.Drawing.Color.FromArgb(this.ForeColor.R, this.ForeColor.G, this.ForeColor.B));

Solution 8 - C#

It's often sufficient to use sibling's or parent's brush for the purpose, and that's easily available in wpf via retrieving their Foreground or Background property.

ref: Control.Background

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionkartalView Question on Stackoverflow
Solution 1 - C#Muad'DibView Answer on Stackoverflow
Solution 2 - C#H.B.View Answer on Stackoverflow
Solution 3 - C#TruthOf42View Answer on Stackoverflow
Solution 4 - C#Omid-RHView Answer on Stackoverflow
Solution 5 - C#Jerry PileView Answer on Stackoverflow
Solution 6 - C#Amin AmiriDarbanView Answer on Stackoverflow
Solution 7 - C#JhollmanView Answer on Stackoverflow
Solution 8 - C#Alexey KhoroshikhView Answer on Stackoverflow