How to generate random color names in C#

Question

I need to generate random color names e.g. \"Red\", \"White\" etc. How can I do it? I am able to generate random color like this:

Random randonGen = new Ran         


        

Answer 1

Or you could try out this: For .NET 4.5

public Windows.UI.Color GetRandomColor()
{
            Random randonGen = new Random();
            Windows.UI.Color randomColor = 
                Windows.UI.Color.FromArgb(
                (byte)randonGen.Next(255), 
                (byte)randonGen.Next(255),
                (byte)randonGen.Next(255), 
                (byte)randonGen.Next(255));
            return randomColor;
}

Answer 2

private string getRandColor()
        {
            Random rnd = new Random();
            string hexOutput = String.Format("{0:X}", rnd.Next(0, 0xFFFFFF));
            while (hexOutput.Length < 6)
                hexOutput = "0" + hexOutput;
            return "#" + hexOutput;
        }

Answer 3

generate a random number and cast it to KnownColor Type

((KnownColor)Random.Next());

Answer 4

Sounds like you just need a random color from the KnownColor enumeration.

Answer 5

If you wanted to get random color from specific colors list,
then try below one

Color[] colors = new[] { Color.Red, Color.Blue, Color.FromArgb(128, 128, 255) };    //Add your favorite colors list
Random randonGen = new Random();
Color randomColor = colors[Generator.Next(0, colors.Length)];    //It will choose random color from "colors" array

Answer 6

To clear up the syntax errors in the original question, it's

Color.FromRgb, and

(Byte)random2.Next(255)

converts the integer to a byte value needed by Color:

    Random random2 = new Random();
    public int nn = 0x128;
    public int ff = 0x512;
    Color randomColor = Color.FromRgb((Byte)random2.Next(nn, ff), (Byte)random2.Next(nn, ff), (Byte)random2.Next(nn, ff));