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Number format regional settings in VB6

I have an app that is developed in vb6. When I do calculations I format numbers based on a '.' for the decimals. I now have customers in the Netherlands who format using ',' and when I do any calculations or submit the data I get errors.

How can I change this code for example:

myTotal = format(mysubtotal + mytax,"0.00")

Another One

rs!total = format(myTotal, "0.00')
I don't want to use the comma but would rather like code to do the math using the local regional settings.
Not all values are currency other some are qty's .

Please can anyone help


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1 Solution
If you want to use format function is to change Regional Settings of the. You might want to save current settings before calling the format function and restore them back afterward

you can find a code for changing  reg.settings here: http://www.vbcode.com/Asp/showsn.asp?theID=8337

Another solution (better) would be to create your own function that format the numbers the way you need.
It is not recommended changing Regional Settings on another users machine.
You can format a number using U.S. English - LCID-1033, or any other supported LCID regardless of the users configurations using the following code:

Option Explicit
Private Declare Function GetLocaleInfoA Lib "kernel32" (ByVal Locale As Long, ByVal LCType As Long, ByVal lpData As Any, ByVal cchData As Long) As Long
Private Declare Function GetNumberFormatA Lib "kernel32" (ByVal Locale As Long, ByVal dwFlags As Long, ByVal lpValue As String, ByRef lpFormat As NUMBERFMT, ByVal lpNumberStr As String, ByVal cchNumber As Long) As Long
Private Type NUMBERFMT
        NumDigits       As Long     ' number of decimal digits
        LeadingZero     As Long     ' if leading zero in decimal fields
        Grouping        As Long     ' group size left of decimal
        lpDecimalSep    As String   ' ptr to decimal separator string
        lpThousandSep   As String   ' ptr to thousand separator string
        NegativeOrder   As Long     ' negative number ordering
End Type
Private Sub Form_Load()
   Debug.Print FormatNumberByLCID("123456789", 1033) 'U.S. English
   Debug.Print FormatNumberByLCID("123456789", 1043) 'Dutch
End Sub
Public Function FormatNumberByLCID(ByVal sNumber As String, _
   Optional ByVal LCID As Long) As String
   Const LOCALE_SDECIMAL As Long = &HE
   Const LOCALE_STHOUSAND As Long = &HF&
   Const LOCALE_SGROUPING As Long = &H10
   Const LOCALE_IDIGITS As Long = &H11
   Const LOCALE_ILZERO As Long = &H12
   Const LOCALE_INEGNUMBER As Long = &H1010
   Dim Buffer          As String * 255
   With NF
      .NumDigits = Val(pfGLI(LCID, LOCALE_IDIGITS))
      .Grouping = Val(pfGLI(LCID, LOCALE_SGROUPING))
      .lpDecimalSep = Left$(pfGLI(LCID, LOCALE_SDECIMAL), 1)
      .lpThousandSep = Left$(pfGLI(LCID, LOCALE_STHOUSAND), 1)
      .NegativeOrder = Val(pfGLI(LCID, LOCALE_INEGNUMBER))
      .LeadingZero = Val(pfGLI(LCID, LOCALE_ILZERO))
   End With
   GetNumberFormatA LCID, 0, sNumber, NF, Buffer, 255
   FormatNumberByLCID = StripNull(Buffer)
End Function
Public Function pfGLI(ByVal LCID As Long, ByVal reqInfo As Long) As String
   Dim nRet             As Long
   Dim Buffer           As String
   nRet = GetLocaleInfoA(LCID, reqInfo, Buffer, 0&)
   If nRet Then
      Buffer = Space$(nRet)
      nRet = GetLocaleInfoA(LCID, reqInfo, Buffer, Len(Buffer))
      pfGLI = Left$(Buffer, nRet - 1)
   End If
End Function
Public Function StripNull(ByVal StrIn As String) As String
   Dim nul              As Long
   nul = InStr(StrIn, vbNullChar)
   Select Case nul
      Case Is > 1
         StripNull = Left$(StrIn, nul - 1)
      Case 1
         StripNull = ""
      Case 0
         StripNull = Trim$(StrIn)
   End Select
End Function

Open in new window

first check that is there comma ,
InStr(1, IP, ",", vbTextCompare)

and then replace , with .
using replace function
>>first check that is there comma ,
>>InStr(1, IP, ",", vbTextCompare)

>>and then replace , with .
>>using replace function

What if regional setting are using comma as thousand separator? Then 1000 will be represented as 1,000.00
After applying your formula you will get 1.000.00 which is not legal decimal number at all.

danaseaman's solution is the way to do it right
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