我已经在这个问题上工作了大约一个星期，现在试图为我们的组织解决这个问题。
我们的财务部门使用 IRIS Exchequer，我们需要计算成本。使用上面的 PHP 代码，我设法让它在 Excel VBA 中使用以下代码（包括依赖函数）。如果下面没有正确归因，我从 www.sulprobil.com 获得了所有长 dec 到 bin 功能。如果将以下代码块复制并粘贴到模块中，则可以从单元格中引用我的 ExchequerDouble 函数。
在继续之前，我必须指出上面 C/PHP 代码中的一个错误。如果您查看 Significand 循环：
C/PHP: Significand = Significand + 2 ^ (-i)
VBA: Significand = Significand + 2 ^ (1 - i)
我在测试期间注意到答案非常接近，但通常是不正确的。再往下钻，我把它缩小到了 Significand。将代码从一种语言/方法翻译成另一种可能是一个问题，或者可能只是一个错字，但添加 (1 - i) 会产生很大的不同。
Function ExchequerDouble(Val1 As Integer, Val2 As Long) As Double
Dim Int2 As String
Dim Int4 As String
Dim Real48 As String
Dim Exponent As String
Dim Sign As String
Dim Significand As String
'Convert each value to binary
Int2 = LongDec2Bin(Val1, 16, True)
Int4 = LongDec2Bin(Val2, 32, True)
'Concatenate the binary strings to produce a 48 bit "Real"
Real48 = Int4 & Int2
'Calculate the exponent
Exponent = LongBin2Dec(Right(Real48, 8)) - 129
'Calculate the sign
Sign = Left(Real48, 1)
'Begin calculation of Significand
Significand = "1.0"
For i = 2 To 40
If Mid(Real48, i, 1) = "1" Then
Significand = Significand + 2 ^ (1 - i)
End If
Next i
ExchequerDouble = CDbl(((-1) ^ Sign) * Significand * (2 ^ Exponent))
End Function
Function LongDec2Bin(ByVal sDecimal As String, Optional lBits As Long = 32, Optional blZeroize As Boolean = False) As String
'Transforms decimal number into binary number.
'Reverse("moc.LiborPlus.www") V0.3 P3 16-Jan-2011
Dim sDec As String
Dim sFrac As String
Dim sD As String 'Internal temp variable to represent decimal
Dim sB As String
Dim blNeg As Boolean
Dim i As Long
Dim lPosDec As Long
Dim lLenBinInt As Long
lPosDec = InStr(sDecimal, Application.DecimalSeparator)
If lPosDec > 0 Then
If Left(sDecimal, 1) = "-" Then 'negative fractions later..
LongDec2Bin = CVErr(xlErrValue)
Exit Function
End If
sDec = Left(sDecimal, lPosDec - 1)
sFrac = Right(sDecimal, Len(sDecimal) - lPosDec)
lPosDec = Len(sFrac)
Else
sDec = sDecimal
sFrac = ""
End If
sB = ""
If Left(sDec, 1) = "-" Then
blNeg = True
sD = Right(sDec, Len(sDec) - 1)
Else
blNeg = False
sD = sDec
End If
Do While Len(sD) > 0
Select Case Right(sD, 1)
Case "0", "2", "4", "6", "8"
sB = "0" & sB
Case "1", "3", "5", "7", "9"
sB = "1" & sB
Case Else
LongDec2Bin = CVErr(xlErrValue)
Exit Function
End Select
sD = sbDivBy2(sD, True)
If sD = "0" Then
Exit Do
End If
Loop
If blNeg And sB <> "1" & String(lBits - 1, "0") Then
sB = sbBinNeg(sB, lBits)
End If
'Test whether string representation is in range and correct
'If not, the user has to increase lbits
lLenBinInt = Len(sB)
If lLenBinInt > lBits Then
LongDec2Bin = CVErr(x1ErrNum)
Exit Function
Else
If (Len(sB) = lBits) And (Left(sB, 1) <> -blNeg & "") Then
LongDec2Bin = CVErr(xlErrNum)
Exit Function
End If
End If
If blZeroize Then sB = Right(String(lBits, "0") & sB, lBits)
If lPosDec > 0 And lLenBinInt + 1 < lBits Then
sB = sB & Application.DecimalSeparator
i = 1
Do While i + lLenBinInt < lBits
sFrac = sbDecAdd(sFrac, sFrac) 'Double fractional part
If Len(sFrac) > lPosDec Then
sB = sB & "1"
sFrac = Right(sFrac, lPosDec)
If sFrac = String(lPosDec, "0") Then
Exit Do
End If
Else
sB = sB & "0"
End If
i = i + 1
Loop
LongDec2Bin = sB
Else
LongDec2Bin = sB
End If
End Function
Function LongBin2Dec(sBinary As String, Optional lBits As Long = 32) As String
'Transforms binary number into decimal number.
'Reverse("moc.LiborPlus.www") V0.3 PB 16-Jan-2011
Dim sBin As String
Dim sB As String
Dim sFrac As String
Dim sD As String
Dim sR As String
Dim blNeg As Boolean
Dim i As Long
Dim lPosDec As Long
lPosDec = InStr(sBinary, Application.DecimalSeparator)
If lPosDec > 0 Then
If (Left(sBinary, 1) = "1") And Len(sBin) >= lBits Then 'negative fractions later..
LongBin2Dec = CVErr(xlErrVa1ue)
Exit Function
End If
sBin = Left(sBinary, lPosDec - 1)
sFrac = Right(sBinary, Len(sBinary) - lPosDec)
lPosDec = Len(sFrac)
Else
sBin = sBinary
sFrac = ""
End If
Select Case Sgn(Len(sBin) - lBits)
Case 1
LongBin2Dec = CVErr(x1ErrNum)
Exit Function
Case 0
If Left(sBin, 1) = "1" Then
sB = sbBinNeg(sBin, lBits)
blNeg = True
Else
sB = sBin
blNeg = False
End If
Case -1
sB = sBin
blNeg = False
End Select
sD = "1"
sR = "0"
For i = Len(sB) To 1 Step -1
Select Case Mid(sB, i, 1)
Case "1"
sR = sbDecAdd(sR, sD)
Case "0"
'Do Nothing
Case Else
LongBin2Dec = CVErr(xlErrNum)
Exit Function
End Select
sD = sbDecAdd(sD, sD) 'Double sd
Next i
If lPosDec > 0 Then 'now the fraction
sD = "0.5"
For i = 1 To lPosDec
If Mid(sFrac, i, 1) = "1" Then
sR = sbDecAdd(sR, sD)
End If
sD = sbDivBy2(sD, False)
Next i
End If
If blNeg Then
LongBin2Dec = "-" & sR
Else
LongBin2Dec = sR
End If
End Function
Function sbDivBy2(sDecimal As String, blInt As Boolean) As String
'Divide sDecimal by two, blInt = TRUE returns integer only
'Reverse("moc.LiborPlus.www") V0.3 PB 16-Jan-2011
Dim i As Long
Dim lPosDec As Long
Dim sDec As String
Dim sD As String
Dim lCarry As Long
If Not blInt Then
lPosDec = InStr(sDecimal, Application.DecimalSeparator)
If lPosDec > 0 Then
'Without decimal point lPosDec already defines location of decimal point
sDec = Left(sDecimal, lPosDec - 1) & Right(sDecimal, Len(sDecimal) - lPosDec)
Else
sDec = sDecimal
lPosDec = Len(sDec) + 1 'Location of decimal point
End If
If ((1 * Right(sDec, 1)) Mod 2) = 1 Then
sDec = sDec & "0" 'Append zero so that integer algorithm calculates division exactly
End If
Else
sDec = sDecimal
End If
lCarry = 0
For i = 1 To Len(sDec)
sD = sD & Int((lCarry * 10 + Mid(sDec, i, 1)) / 2)
lCarry = (lCarry * 10 + Mid(sDec, i, 1)) Mod 2
Next i
If Not blInt Then
If Right(sD, Len(sD) - lPosDec + 1) <> String(Len(sD) - lPosDec + 1, "0") Then
'frac part Is non - zero
i = Len(sD)
Do While Mid(sD, i, 1) = "0"
i = i - 1 'Skip trailing zeros
Loop
'Insert decimal point again
sD = Left(sD, lPosDec - 1) _
& Application.DecimalSeparator & Mid(sD, lPosDec, i - lPosDec + 1)
End If
End If
i = 1
Do While i < Len(sD)
If Mid(sD, i, 1) = "0" Then
i = i + 1
Else
Exit Do
End If
Loop
If Mid(sD, i, 1) = Application.DecimalSeparator Then
i = i - 1
End If
sbDivBy2 = Right(sD, Len(sD) - i + 1)
End Function
Function sbBinNeg(sBin As String, Optional lBits As Long = 32) As String
'Negate sBin: take the 2's-complement, then add one
'Reverse("moc.LiborPlus.www") V0.3 PB 16-Jan-2011
Dim i As Long
Dim sB As String
If Len(sBin) > lBits Or sBin = "1" & String(lBits - 1, "0") Then
sbBinNeg = CVErr(xlErrValue)
Exit Function
End If
'Calculate 2 's-complement
For i = Len(sBin) To 1 Step -1
Select Case Mid(sBin, i, 1)
Case "1"
sB = "0" & sB
Case "0"
sB = "1" & sB
Case Else
sbBinNeg = CVErr(xlErrValue)
Exit Function
End Select
Next i
sB = String(lBits - Len(sBin), "1") & sB
'Now add 1
i = lBits
Do While i > 0
If Mid(sB, i, 1) = "1" Then
Mid(sB, i, 1) = "0"
i = i - 1
Else
Mid(sB, i, 1) = "1"
i = 0
End If
Loop
'Finally strip leading zeros
i = InStr(sB, "1")
If i = 0 Then
sbBinNeg = "0"
Else
sbBinNeg = Right(sB, Len(sB) - i + 1)
End If
End Function
Function sbDecAdd(sOne As String, sTwo As String) As String
'Sum up two string decimals.
'Reverse("moc.LiborPlus.www") V0.3 PB 16-Jan-2011
Dim lStrLen As Long
Dim s1 As String
Dim s2 As String
Dim sA As String
Dim sB As String
Dim sR As String
Dim d As Long
Dim lCarry As Long
Dim lPosDec1 As Long
Dim lPosDec2 As Long
Dim sF1 As String
Dim sF2 As String
lPosDec1 = InStr(sOne, Application.DecimalSeparator)
If lPosDec1 > 0 Then
s1 = Left(sOne, lPosDec1 - 1)
sF1 = Right(sOne, Len(sOne) - lPosDec1)
lPosDec1 = Len(sF1)
Else
s1 = sOne
sF1 = ""
End If
lPosDec2 = InStr(sTwo, Application.DecimalSeparator)
If lPosDec2 > 0 Then
s2 = Left(sTwo, lPosDec2 - 1)
sF2 = Right(sTwo, Len(sTwo) - lPosDec2)
lPosDec2 = Len(sF2)
Else
s2 = sTwo
sF2 = ""
End If
If lPosDec1 + lPosDec2 > 0 Then
If lPosDecl > lPosDec2 Then
sF2 = sF2 & String(lPosDec1 - lPosDec2, "0")
Else
sF1 = sFl & String(lPosDec2 - lPosDec1, "0")
lPosDec1 = lPosDec2
End If
sF1 = sbDecAdd(sF1, sF2) 'Add fractions as integer numbers
If Len(sF1) > lPosDecl Then
lCarry = 1
sF1 = Right(sF1, lPosDec1)
Else
lCarry = 0
End If
Do While lPosDec1 > 0
If Mid(sF1, lPosDec1, 1) <> "0" Then
Exit Do
End If
lPosDec1 = lPosDec1 - 1
Loop
sF1 = Left(sF1, lPosDec1)
Else
lCarry = 0
End If
lStrLen = Len(sl)
If lStrLen < Len(s2) Then
lStrLen = Len(s2)
sA = String(lStrLen - Len(s1), "0") & s1
sB = s2
Else
sA = s1
sB = String(lStrLen - Len(s2), "0") & s2
End If
Do While lStrLen > 0
d = 0 + Mid(sA, lStrLen, 1) + Mid(sB, lStrLen, 1) + lCarry
If d > 9 Then
sR = (d - 10) & sR
lCarry = 1
Else
sR = d & sR
lCarry = 0
End If
lStrLen = lStrLen - 1
Loop
If lCarry > 0 Then
sR = lCarry & sR
End If
If lPosDec1 > 0 Then
sbDecAdd = sR & Application.DecimalSeparator & sF1
Else
sbDecAdd = sR
End If
End Function
此代码有效，但有时（大约是我的测试数据的 1%）与 Excel 插件中的 Iris 的 EntDouble 函数相比，您最终会花费几美分。除非有人能弄清楚，否则我会将其归因于精确度。
最终让这个在 VBA 中工作是我检查一切工作的概念证明。此功能的预期平台是 SQL Server。如果您将 Excheque DB 链接到 SQL Server，您应该能够直接针对 Pervasive DB 中的数据运行此函数。在我的例子中，我们将把过去 2.5 年的交易数据转储到 SQL Server 上的静态表中，但我们每年只处理一次这些数据，所以这不是问题。以下两个功能应该可以解决您的问题。就精度而言，它们与上面的 VBA 代码相当，有时会差几美分，但似乎 99% 的时间完全相同。
CREATE FUNCTION dbo.FUNCTION_Exchequer_Double
(
@Val1 AS SmallInt,
@Val2 AS BigInt
)
RETURNS Decimal(38, 10)
AS
BEGIN
-- Declare and set decoy variables
DECLARE @Val1_Decoy AS SmallInt
DECLARE @Val2_Decoy AS BigInt
SELECT @Val1_Decoy = @Val1,
@Val2_Decoy = @Val2
-- Declare other variables
DECLARE @Val1_Binary AS Varchar(16)
DECLARE @Val2_Binary AS Varchar(32)
DECLARE @Real48_Binary AS Varchar(48)
DECLARE @Real48_Decimal AS BigInt
DECLARE @Exponent AS Int
DECLARE @Sign AS Bit
DECLARE @Significand AS Decimal(19, 10)
DECLARE @BitCounter AS Int
DECLARE @Two As Decimal(38, 10) -- Saves us casting inline in the code
DECLARE @Output AS Decimal(38, 10)
-- Convert values into two binary strings of the correct length (Val1 = 16 bits, Val2 = 32 bits)
SELECT @Val1_Binary = Replicate(0, 16 - Len(dbo.FUNCTION_Convert_To_Base(Cast(@Val1_Decoy AS Binary(2)), 2)))
+ dbo.FUNCTION_Convert_To_Base(Cast(@Val1_Decoy AS Binary(2)), 2),
@Val2_Binary = Replicate(0, 32 - Len(dbo.FUNCTION_Convert_To_Base(Cast(@Val2_Decoy AS Binary(4)), 2)))
+ dbo.FUNCTION_Convert_To_Base(Cast(@Val2_Decoy AS Binary(4)), 2)
-- Find the decimal value of the new 48 bit number and its binary value
SELECT @Real48_Decimal = @Val2_Decoy * Power(2, 16) + @Val1_Decoy
SELECT @Real48_Binary = @Val2_Binary + @Val1_Binary
-- Determine the Exponent (takes the first 8 bits and subtracts 129)
SELECT @Exponent = Cast(@Real48_Decimal AS Binary(1)) - 129
-- Determine the Sign
SELECT @Sign = Left(@Real48_Binary, 1)
-- A bit of setup for determining the Significand
SELECT @Significand = 1,
@Two = 2,
@BitCounter = 2
-- Determine the Significand
WHILE @BitCounter <= 40
BEGIN
IF Substring(@Real48_Binary, @BitCounter, 1) Like '1'
BEGIN
SELECT @Significand = @Significand + Power(@Two, 1 - @BitCounter)
END
SELECT @BitCounter = @BitCounter + 1
END
SELECT @Output = Power(-1, @Sign) * @Significand * Power(@Two, @Exponent)
-- Return the output
RETURN @Output
END
CREATE FUNCTION dbo.FUNCTION_Convert_To_Base
(
@value AS BigInt,
@base AS Int
)
RETURNS Varchar(8000)
AS
BEGIN
-- Code from http://dpatrickcaldwell.blogspot.co.uk/2009/05/converting-decimal-to-hexadecimal-with.html
-- some variables
DECLARE @characters Char(36)
DECLARE @result Varchar(8000)
-- the encoding string and the default result
SELECT @characters = '0123456789abcdefghijklmnopqrstuvwxyz',
@result = ''
-- make sure it's something we can encode. you can't have
-- base 1, but if we extended the length of our @character
-- string, we could have greater than base 36
IF @value < 0 Or @base < 2 Or @base > 36
RETURN Null
-- until the value is completely converted, get the modulus
-- of the value and prepend it to the result string. then
-- devide the value by the base and truncate the remainder
WHILE @value > 0
SELECT @result = Substring(@characters, @value % @base + 1, 1) + @result,
@value = @value / @base
-- return our results
RETURN @result
END
随意使用我的 VBA 或 SQL 代码。真正艰苦的工作是由上面将其转换为 PHP 的人完成的。如果有人发现任何改进方法，请告诉我，以便我们使这段代码尽可能完美。
谢谢！