color conversion
in [Visual Basic]
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From: Dmitriy Antonov on 30 Sep 2006 01:49 "Larry Serflaten" <serflaten(a)usinternet.com> wrote in message news:eLhxxKF5GHA.668(a)TK2MSFTNGP02.phx.gbl... > > "Jim Y" <jj819stuffNOSPAM(a)comcast.net> wrote >> For example, for the light red, it requires the RGB values of 255, 224, >> 224 which correspond >> (close, not exact) to the hexadecimal values of C0C0FF. How do I obtain >> 224 from C0? 255 from FF? >> If I remember, C is 13 and 0 is 10, F is 15. > > Form1.BackColor = &HC0C0FF > Form1.BackColor = RGB(&HFF, &HC0, &HC0) > Form1.BackColor = RGB(255, 224, 224) > > These three statements produce the exact same color. You already > have the hex values, so use them in your calls. Unless you need > to send text, hex and decimal values are equivalent / interchangeable. > > In the Hex notation, the order is BBGGRR > In the RGB function, the order is RR, GG, BB > > Experiment with it a little, it is not difficult.... > > LFS > > ?RGB(&HFF, &HC0, &HC0) 12632319 ?RGB(255, 224, 224) 14737663 &HC0 is equal to 192, so correct equivalent should be ?RGB(255, 192, 192) 12632319 Dmitriy.
From: Larry Serflaten on 30 Sep 2006 01:59 "Dmitriy Antonov" <antonovdima(a)netzero.net_NOT_FOR_SPAM> wrote > > Form1.BackColor = &HC0C0FF > > Form1.BackColor = RGB(&HFF, &HC0, &HC0) > > Form1.BackColor = RGB(255, 224, 224) > > > > These three statements produce the exact same color. Thanks Dmitriy, I should have tested that. I took the OP at his word that those were the correct values. > &HC0 is equal to 192, so correct equivalent should be > > ?RGB(255, 192, 192) > 12632319 I stand corrected... LFS
From: Dmitriy Antonov on 30 Sep 2006 02:05 "Larry Serflaten" <serflaten(a)usinternet.com> wrote in message news:uQsxhVF5GHA.4996(a)TK2MSFTNGP04.phx.gbl... > > "Dmitriy Antonov" <antonovdima(a)netzero.net_NOT_FOR_SPAM> wrote > >> > Form1.BackColor = &HC0C0FF >> > Form1.BackColor = RGB(&HFF, &HC0, &HC0) >> > Form1.BackColor = RGB(255, 224, 224) >> > >> > These three statements produce the exact same color. > > Thanks Dmitriy, I should have tested that. I took the OP at > his word that those were the correct values. Jim was so insisting on that equivalency that you decided: "Let it be" <g> Dmitriy.
From: Michael C on 30 Sep 2006 02:32 "Jim Y" <jj819stuffNOSPAM(a)comcast.net> wrote in message news:uYCdndgesrWERYDYnZ2dnUVZ_v6dnZ2d(a)comcast.com... > VB6 uses hexadecimals to define color of forms and text. Is there a > program or some way that I can obtain the following for these 4 colors? > > Light Blue = &H00FFFFC0& > Light Yellow = &H00C0FFFF& > Light Red = &H00C0C0FF& > Light Green = &H00C0FFC0& > > I need the Hue, Sat, Lum, Red, Green and Blue numerical values of each > color. > > I am creating some small graphics and want to color match the background > when the graphic is placed on an Image Control. Consider the graphic to > be a silver oval in a rectangle. Unfortunately, I cannot make the > background transparent with the software that I am using. I would prefer > a method that will permit me to use any color in the future. Here's how to get a color from hue, saturation and luminance in c#. Unfortunately I don't have code to convert an RGB color to HSL because c# has built in functionality for this. This should be pretty straight forward to translate to VB. You can consider HSL to be a type with member called Hue Saturation and Luminance each defined as single (float in c#). public static explicit operator Color(HSL Value) { float r, g, b; if (Value.Saturation == 0) { // gray values r = g = b = Value.Luminance; } else { float v1, v2; v2 = (Value.Luminance < 0.5f) ? (Value.Luminance * (1 + Value.Saturation)) : ((Value.Luminance + Value.Saturation) - (Value.Luminance * Value.Saturation)); v1 = 2f * Value.Luminance - v2; r = HueToRGB(v1, v2, Value.Hue + (1.0f / 3)); g = HueToRGB(v1, v2, Value.Hue); b = HueToRGB(v1, v2, Value.Hue - (1.0f / 3)); } return Color.FromArgb(Value.Alpha, RoundIt(r), RoundIt(g), RoundIt(b)); } private static float HueToRGB(float v1, float v2, float vH) { if(vH < 0f) vH += 1f; if(vH > 1f) vH -= 1f; if(6f * vH < 1f) return (v1 + (v2 - v1) * 6f * vH); if((2f * vH) < 1f) return v2; if((3f * vH) < 2f) return (v1 + (v2 - v1) * ((2.0f / 3f) - vH) * 6f); return v1; } private static int RoundIt(float Value) { return (int)Math.Round(Value * 255.0, 0); } > > Thank you, > Jim Y > >
From: Sneha Menon on 30 Sep 2006 02:32
Jim Y wrote: > How do I get the color value? I have the hex value. > > Jim Y > > -------------------------------------------------------------------------------------- Hi Jim This is something dug out of my learning-vb time files. Havent modified or optimized thereafter. Anyway this will suit your need. Put 4 Text boxes and command button on a form Enter Hex value in Text1 and press the button You will get the RGB values in the other --------------Code------------------------------- Private Sub Command1_Click() Dim k, rH, rR, rG, rB, HeStr, thisCh HeStr = "0123456789ABCDEF" rH = UCase(Text1.Text) For k = 1 To 6 thisCh = Mid(rH, k, 1) If InStr(HeStr, thisCh) < 1 Then MsgBox ("Invalid Characters in the Hex field!") Else rR = (((InStr(1, HeStr, Mid(rH, 1, 1)) - 1) * 16)) + (InStr(1, HeStr, Mid(rH, 2, 1)) - 1) rG = (((InStr(1, HeStr, Mid(rH, 3, 1)) - 1) * 16)) + (InStr(1, HeStr, Mid(rH, 4, 1)) - 1) rB = (((InStr(1, HeStr, Mid(rH, 5, 1)) - 1) * 16)) + (InStr(1, HeStr, Mid(rH, 5, 1)) - 1) Text2.Text = rR 'Red value Text3.Text = rG 'Green value Text4.Text = rB 'Blue End If Next k End Sub ----------------------------Code End------------------ Let me know whether it solved your problem Regards Sneha --------------------------------------------------------------------------------------- |