Theme confusion in SpreadsheetML.
- by dmaruca
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