我一整天都在和这个战斗。在styles.xml文件中,提供了如下颜色信息:
<fgcolor theme=“0”tint=”-0.249977111117893“/>
ECMA 376将主题颜色引用定义为:
索引到<clrscheme>集合中,
引用特定的<sysclr>或
<srgbclr>主题中表示的值
部分。
好吧,听起来很简单。以下是我的clrscheme xml的摘录:
<a:clrscheme name=“office”>
<a:dk1>
<a:sysclr val=“windowtext”lastcrl=“000000”/>
</A:dk1>
<a:lt1>
<a:sysclr val=“window”lastcrl=“ffffff”/>
</A:lt1>
索引0是黑色的,他们想把它变暗吗?我可以告诉你,色调应用后,颜色应该是f2f2f2。
我的困惑是,theme=“0”到底是什么意思?它不可能意味着变暗000000。查看msdn只会让我更加困惑。从
http://msdn.microsoft.com/en-us/library/dd560821.aspx
注意主题颜色整数
开始从左到右计数
调色板从零开始。主题
颜色3是深色2文本/背景
颜色。
实际上,如果你从零开始计数,第三个条目是light 2。黑暗2是第二个。这里有人能为我解释一下这个问题吗?theme=“0”到底是什么意思?
这是我一直在使用的VB6代码来应用色调。您可以将其粘贴到VBA编辑器中并运行测试子组件。
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