I've been fighting this all day. Inside my styles.
xml file I have color information given like so:
<fgColor theme="0" tint="-0.249977111117893" /
ECMA 376 defines a theme color reference as:
Index into the <clrScheme collection,
referencing a particular <sysClr or
<srgbClr value expressed in the Theme
part.
Ok, that sounds easy. Here is an excerpt from my clrScheme xml:
<a:clrScheme name="Office"
<a:dk1
<a:sysClr val="windowText" lastClr="000000" /
</a:dk1
<a:lt1
<a:sysClr val="window" lastClr="FFFFFF" /
</a:lt1
Index zero is black, and they are wanting to darken it? I can tell you that after the tint is applied, the color should be #F2F2F2.
My confusion is what does theme="0" really mean? It can't possible mean to darken #000000. Checking MSDN only confuses me even more. From http://msdn.microsoft.com/en-us/library/dd560821.aspx
note that the theme color integer
begins counting from left to right in
the palette starting with zero. Theme
color 3 is the dark 2 text/background
color.
Actually, if you start counting at zero the third entry is Light 2. Dark 2 is the second one. Can anyone here shed some light on this subject for me? What does theme="0" really mean?
Here is the VB6 code I have been working with to apply the tint. You can paste it into your vba editor and run the test sub.
Public Type tRGB
R As Byte
G As Byte
B As Byte
End Type
Public Type tHSL
H As Double
S As Double
L As Double
End Type
Sub TestRgbTint()
Dim c As tRGB
RGB_Hex2Type "ffffff", c
RGB_ApplyTint c, -0.249977111117893
Debug.Print Hex(c.R) & Hex(c.G) & Hex(c.B)
End Sub
Public Sub RGB_Hex2Type(ByVal HexString As String, RGB As tRGB)
'Remove the alpha channel if it exists
If Len(HexString) = 8 Then
HexString = mID(HexString, 3)
End If
RGB.R = CByte("&H" & Left(HexString, 2))
RGB.G = CByte("&H" & mID(HexString, 3, 2))
RGB.B = CByte("&H" & Right(HexString, 2))
End Sub
Public Sub RGB_ApplyTint(RGB As tRGB, tint As Double)
Const HLSMAX = 1#
Dim HSL As tHSL
If tint = 0 Then Exit Sub
RGB2HSL RGB, HSL
If tint < 0 Then
HSL.L = HSL.L * (1# + tint)
Else
HSL.L = HSL.L * (1# - tint) + (HLSMAX - HLSMAX * (1# - tint))
End If
HSL2RGB HSL, RGB
End Sub
Public Sub HSL2RGB(HSL As tHSL, RGB As tRGB)
HSL2RGB_ByVal HSL.H, HSL.S, HSL.L, RGB
End Sub
Private Sub HSL2RGB_ByVal(ByVal H As Double, ByVal S As Double, ByVal L As Double, RGB As tRGB)
Dim v As Double
Dim R As Double, G As Double, B As Double
'Default color to gray
R = L
G = L
B = L
If L < 0.5 Then
v = L * (1# + S)
Else
v = L + S - L * S
End If
If v > 0 Then
Dim m As Double, sv As Double
Dim sextant As Integer
Dim fract As Double, vsf As Double, mid1 As Double, mid2 As Double
m = L + L - v
sv = (v - m) / v
H = H * 6#
sextant = Int(H)
fract = H - sextant
vsf = v * sv * fract
mid1 = m + vsf
mid2 = v - vsf
Select Case sextant
Case 0
R = v
G = mid1
B = m
Case 1
R = mid2
G = v
B = m
Case 2
R = m
G = v
B = mid1
Case 3
R = m
G = mid2
B = v
Case 4
R = mid1
G = m
B = v
Case 5
R = v
G = m
B = mid2
End Select
End If
RGB.R = R * 255#
RGB.G = G * 255#
RGB.B = B * 255#
End Sub
Public Sub RGB2HSL(RGB As tRGB, HSL As tHSL)
Dim R As Double, G As Double, B As Double
Dim v As Double, m As Double, vm As Double
Dim r2 As Double, g2 As Double, b2 As Double
R = RGB.R / 255#
G = RGB.G / 255#
B = RGB.B / 255#
'Default to black
HSL.H = 0
HSL.S = 0
HSL.L = 0
v = IIf(R > G, R, G)
v = IIf(v > B, v, B)
m = IIf(R < G, R, G)
m = IIf(m < B, m, B)
HSL.L = (m + v) / 2#
If HSL.L < 0 Then
Exit Sub
End If
vm = v - m
HSL.S = vm
If HSL.S > 0 Then
If HSL.L <= 0.5 Then
HSL.S = HSL.S / (v + m)
Else
HSL.S = HSL.S / (2# - v - m)
End If
Else
Exit Sub
End If
r2 = (v - R) / vm
g2 = (v - G) / vm
b2 = (v - B) / vm
If R = v Then
If G = m Then
HSL.H = 5# + b2
Else
HSL.H = 1# - g2
End If
ElseIf G = v Then
If B = m Then
HSL.H = 1# + r2
Else
HSL.H = 3# - b2
End If
Else
If R = m Then
HSL.H = 3# + g2
Else
HSL.H = 5# - r2
End If
End If
HSL.H = HSL.H / 6#
End Sub
