8

我整天都在和这个作斗争。在我的 styles.xml 文件中，我给出了如下颜色信息：

<fgColor theme="0" tint="-0.249977111117893" />

ECMA 376 将主题颜色参考定义为：

<clrScheme> 集合的索引，引用主题部分中表示的特定 <sysClr> 或 <srgbClr> 值。

好的，这听起来很容易。这是我的 clrScheme xml 的摘录：

<a:clrScheme name="Office">
<a:dk1>
<a:sysClr val="windowText" lastClr="000000" />
</a:dk1>
<a:lt1>
<a:sysClr val="window "lastClr="FFFFFF" />
</a:lt1>

索引零是黑色的，他们想让它变暗吗？我可以告诉你，在应用了色调之后，颜色应该是#F2F2F2。

我的困惑是 theme="0" 的真正含义是什么？它不可能意味着变暗#000000。检查 MSDN 只会让我更加困惑。来自http://msdn.microsoft.com/en-us/library/dd560821.aspx

请注意，主题颜色整数从零开始在调色板中从左到右开始计数。主题颜色 3 是深色 2 文本/背景颜色。

实际上，如果您从零开始计数，则第三个条目是 Light 2。 Dark 2 是第二个条目。这里的任何人都可以为我阐明这个主题吗？theme="0" 的真正含义是什么？

这是我一直在使用的用于应用色调的 VB6 代码。您可以将其粘贴到您的 vba 编辑器中并运行测试子程序。

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
4

2 回答 2

8

我从 odf-converter.sourceforge.net 查看了 xsl，看起来 0 和 1 已切换，2 和 3 已切换。这是 xsl 部分：

    <xsl:variable name="theme">
      <xsl:choose>
        <xsl:when test="@theme = 0">
          <xsl:text>1</xsl:text>
        </xsl:when>
        <xsl:when test="@theme = 1">
          <xsl:text>0</xsl:text>
        </xsl:when>
        <xsl:when test="@theme = 2">
          <xsl:text>3</xsl:text>
        </xsl:when>
        <xsl:when test="@theme = 3">
          <xsl:text>2</xsl:text>
        </xsl:when>
        <xsl:otherwise>
          <xsl:value-of select="@theme"/>
        </xsl:otherwise>
      </xsl:choose>
    </xsl:variable>

现在一切都在匹配。注意 0=1, 1=0, 2=3, 3=2 开关。

于 2010-05-03T21:32:35.973 回答
3

是的，它是主题颜色的索引。0 是“第一个”，这意味着 1。看起来你正在做 Excel。对于 PowerPoint，请注意色调/阴影算法是不同的 - 它不是基于 HSL，而是基于线性 RGB。

于 2010-05-04T16:39:54.487 回答