how can i retrieve the rgb code from a selected color?
I,m sending you this project hope this will bring you an idea how to retrieve RGB color from this project
Best of luck
Nice little program but the OP (tripes) asked how to break down a selected color into the rgb triplet values. What tripes did not tell us is how they are getting those value and off the top of my head I can think of a couple of ways. GetPixel API, Point, and the common dialog show color method. Each of these methods returns a long value and the following shows how to break down that long into each of the rgb triplet values.
Dim MyColor As Long, R As Integer, G As Integer, B As Integer, Temp As Integer MyColor = RGB(124, 45, 36) Temp = MyColor Mod 65536 MyColor = MyColor - Temp B = MyColor / 65536 MyColor = Temp Temp = MyColor Mod 256 MyColor = MyColor - Temp G = MyColor / 256 R = Temp Debug.Print R, G, B
Option Explicit Private Const RGB_R = 1 Private Const RGB_G = 2 Private Const RGB_B = 3 Private Sub cmdFindColor_Click() Dim lngColor as Long, intRGB as Integer, lngRGBValue as Long ' Code to find RGB value ' Blah, Blah, Blah ' intRGB would be RGB_R, RGB_B, etc. ' lngRGBValue would be the RGB Color value ' lngColor would be the Red, Green, or Blue Value you wish to find lngColor = Return_RGB_VALUE(lngRGBValue, intRGB) End Sub Private Function Return_RGB_VALUE(RGB_Long As Long, RGB_Color As Integer) As Long If (RGB_Long > 0) And (RGB_Long < 16777216) Then ' Return the Red, Green, or Blue value based on the RGB Value and ' the Color requested. Return_RGB_VALUE = (RGB_Long / 256) ^ ((RGB_Color - 1) And 255) ' Convert the RGB value into an 8 bit value and take that value ' the exponential value of the bitwise comparison of ((Color - 1) And 255) ' The And operator in this case returns the bitwise comparison of 2 numbers. Else ' Not a valid RGB value Return_RGB_VALUE = -1 End If End Function
Visual Basic really doesn't have any operators that deal with bit shifting directly like C (<< or >>).
But in the C language the
COLORREF data type contains the
RGB value with a high order byte of zero and it's in a 32 bit value with the following format: 0x00bbggrr. And C has three special color macros:
GetBValue which do not exist in VB6: C uses these three macros to extract the red, green, and the blue values. C uses bit shift operators in these macros.
COLORREF data type is a hexadecimal number which is formed from the
RGB value and the high order byte
There are 4 bytes in that number: the low order red byte (rr or red), the second byte (gg or green), the third byte (bb or green), and the high order byte must always be zero.
Visual Basic only gives us the first 3 bytes. But that's all we need anyway.
The highest RGB Value would be from the values of 255, 255, 255
lngValue = RGB(255, 255, 255)
The lngValue would be 16777215
Hex$(16777215) would yield FFFFFF
Red would be
However, if you had a small value like 25 for the RGB VALUE, that would pose problems for finding the hex values:
hex$(25) would yield the string "19." You have no values for the green or the blue.
So, one way to get around this problem is to just add (maximum value + 1) to the RGB value.
The max hex value for the RGB value is
That value plus one is
Thus, if you just add the hex value of the RGB
(hex(RGB(Red, Green, Blue)), to the max value you would always be able to have a hex string that you could work with to extract the red green and Blue values.
&H1000000 + &H25 ----------------- &H1000025
Just convert that Hex number to a
HEX$ and use some VB string functions to extract the 'rr', the 'gg', and the 'bb' from that hexadecimal string.
dim strColorRef as String Dim strColorRed as String, strColorBlue as String, strColorGreen as String Dim lngRGB as long Dim lngColorReg as long Dim lngRed as long, lngGreen as long, lngBlue as long lngRGB = 67 ' Since &H1000000 = 16777216 ' just add 16777216 to lngRGB lngColorRef = 16777216 + lngRGB ' convert that value to a hexadecimal string strColorRef = hex$(lngColorRef) ' strColorRef here would be "1000043" ' Red strColorRed = Right(strColorRef,2) lngRed = clng("&H" & strColorRed) ' Green strColorGreen = left(strColorRef, 5) strColorGreen = Right(strColorGreen, 2) lngGreen = clng("&H" & strColorGreen) ' Blue strColorBlue = left(strColorRef, 3) strColorBlue = Right(strColorBlue, 2) lngBlue = clng("&H" & strColorBlue)
Edited 3 Years Ago by mike_2000_17: Fixed formatting
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