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RGB value of selected shaded cell

Posted on 2009-05-03
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Last Modified: 2012-05-06
Dear Experts:

I would like to retrieve the RGB value of the selected shaded cell of a word table and display it in a msgbox.
Help is much appreciated. Thank you very much in advance. Regards, Andreas
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Question by:AndreasHermle
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11 Comments
 
LVL 59

Expert Comment

by:Chris Bottomley
Comment Utility
Hello AndreasHermle,

SOmething like the following perhaps:

Regards,
Chris
Sub cellRGB()
Dim str As String 
str = "Color is :" & vbCrLf & vbCrLf
With Application.ActiveCell.Interior
    str = str & "Red:   " & CStr(&HFF& And .Color) & vbCrLf
    str = str & "Green: " & CStr((&HFF00& And .Color) \ 256) & vbCrLf
    str = str & "Blue:  " & CStr(.Color \ 65536) & vbCrLf
End With
MsgBox str, vbOKOnly, "Cell Interior RGB" 
End Sub

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LVL 76

Assisted Solution

by:GrahamSkan
GrahamSkan earned 150 total points
Comment Utility
Hello Andreas,

Chris has done a good job on splitting the colour components, but I think that he has confused Word and Excel.

This code reports the numeric values for the foreground and background values for the selected cells. You need to provide a range that includes a cell. I wopuld suggest something like:


MsgBox RGBValue(selection.range)
Function RGBValue(rng As Range) As String

    Dim cl As Word.Cell

    Dim RGBValueF As Long

    Dim RGBValueB As Long

    Set cl = rng.Cells(1)

    RGBValueF = cl.Shading.ForegroundPatternColor

    RGBValueB = cl.Shading.BackgroundPatternColor

    RGBValue = "Foreground; " & RGBValueF & vbCrLf & "Background; " & RGBValueB

End Function

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LVL 59

Expert Comment

by:Chris Bottomley
Comment Utility
Keeping the structure from Grahams solution and merging it with the RGB display produces for example the following:

Chris
Function RGBValue(rng As Range) As String

    Dim cl As Word.Cell

    Dim RGBValueF As Long

    Dim RGBValueB As Long

    Set cl = rng.Cells(1)

    RGBValueF = cl.Shading.ForegroundPatternColor

    RGBValueB = cl.Shading.BackgroundPatternColor

    RGBValue = "Foreground; " & RGB(RGBValueF) & vbCrLf & "Background; " & RGB(RGBValueB)

End Function
 

Function RGB(lng As Long) As String

    lng = Abs(lng)

    RGB = "Color is :" & vbCrLf & vbCrLf

    RGB = RGB & "Red:   " & CStr(&HFF& And lng) & vbCrLf

    RGB = RGB & "Green: " & CStr((&HFF00& And lng) \ 256) & vbCrLf

    RGB = RGB & "Blue:  " & CStr(lng \ 65536) & vbCrLf

End Function

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Author Comment

by:AndreasHermle
Comment Utility
Dear Graham and Chris,

thank you very much for your swift help. I do not know how to call up this function. I guess I need to call it from an existing macro. If the latter is true, this macro should be just a stand-alone macro with no more code.
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LVL 59

Expert Comment

by:Chris Bottomley
Comment Utility
How would you like to call it? ... for example run via the tools macro run menu using the activecell?

Chris
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LVL 76

Expert Comment

by:GrahamSkan
Comment Utility
Other options are to attach the macro to a new button on a toolbar and/or to create a keystroke shortcut. These are both done via the Customise dialogue.
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Author Comment

by:AndreasHermle
Comment Utility
Dear Chris and Graham,
I guess you did not understand me correctly. Am I right, but this function must be 'embedded' in a macro starting with Sub someName() and End Sub? If running this macro (the cursor resides in some selected cell of some table) a msgbox should come up, telling me which RGB value the selected cell has.

I hope I could make myself clear.

Regards, Andreas
0
 
LVL 59

Accepted Solution

by:
Chris Bottomley earned 350 total points
Comment Utility
Okay then try the following change.

Put the cursor in a cell and press alt + F8 then select RGBValue

Chris
Sub RGBValue()

    Dim cl As Word.Cell

    Dim RGBValueF As Long

    Dim RGBValueB As Long

    If Not Selection.Information(wdWithInTable) Then Exit Sub

    Set cl = Application.Selection.Range.Cells(1)

    RGBValueF = cl.Shading.ForegroundPatternColor

    RGBValueB = cl.Shading.BackgroundPatternColor

    MsgBox "Foreground; " & RGB(RGBValueF) & vbCrLf & "Background; " & RGB(RGBValueB), vbOKOnly, "Activecell Color VAlues"

End Sub
 

Function RGB(lng As Long) As String

    lng = Abs(lng)

    RGB = "Color is :" & vbCrLf & vbCrLf

    RGB = RGB & "Red:   " & CStr(&HFF& And lng) & vbCrLf

    RGB = RGB & "Green: " & CStr((&HFF00& And lng) \ 256) & vbCrLf

    RGB = RGB & "Blue:  " & CStr(lng \ 65536) & vbCrLf

End Function

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Author Comment

by:AndreasHermle
Comment Utility
Dear Chris,
that's it. Thank you very much for your professional help. I am not quite sure about how to distribute the points. Graham also contributed to the solution, at least at the beginning. Would it be ok to do a 350 to 150 split or 400 to 100 with you getting the bigger share?

Regards, Andreas  
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LVL 59

Expert Comment

by:Chris Bottomley
Comment Utility
Absolutely right that Graham should get a share so whichever split you like ...

But hes a genius and i'm a mere master so perhaps 499 to me and 1 to him ;o) nah joshing aside I would say 350 : 150 is the fairer of the options you proposed.

Chris
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Author Closing Comment

by:AndreasHermle
Comment Utility
Again, thank you for your professional help. Regards, Andreas
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