SIMPLEST IsNumber() function

Function to test an Unsign Whole Number: (cannot be simpler)

Code:

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Function IsDigits(ByVal Num As Variant) As Boolean
  IsDigits = (Len(Num) > 0) And (Num Like String(Len(Num), "#"))
End Function

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'Simplest' IsNumer() function for all possible combination cases: Please test and comment.
You need to change to match with your local setting such as . , $

Code:

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Function IsNumber(var As Variant, Optional AllowDot As Boolean, _
                                  Optional AllowComma As Boolean, _
                                  Optional AllowSgnL As Boolean, _
                                  Optional AllowSgnR As Boolean, _
                                  Optional AllowPars As Boolean, _
                                  Optional AllowCurr As Boolean, _
                                  Optional AllowSpcL As Boolean, _
                                  Optional AllowSpcR As Boolean, _
                                  Optional AllowSci As Boolean, _
                                  Optional AllowHex As Boolean, _
                                  Optional AllowOct As Boolean) As Boolean
  If Not IsNumeric(var) Then
  ElseIf AllowDot > -UBound(Split(var, ".")) Then '-- see Note below
  ElseIf Not AllowComma And var Like "*,*" Then  '-- thousand seperator
  ElseIf Not AllowSgnL And var Like "[-+]*" Then  '-- leading sign
  ElseIf Not AllowSgnR And var Like "*[-+]" Then  '-- trailing sign
  ElseIf Not AllowPars And var Like "*(*" Then    '-- parentheses as negative
  ElseIf Not AllowCurr And var Like "*$*" Then    '-- currency
  ElseIf Not AllowSpcL And var Like " *" Then    '-- leading spaces
  ElseIf Not AllowSpcR And var Like "* " Then    '-- trailing spaces
  ElseIf Not AllowSci And var Like "*[DE]*" Then  '-- Scientific
  ElseIf Not AllowHex And var Like "*H*" Then    '-- Hex
  ElseIf Not AllowOct And var Like "*O*" Then    '-- Oct
  Else
      IsNumber = True
  End If
End Function

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Code:

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'-- Note: ElseIf AllowDot > -UBound(Split(var, ".")) Then
'-- equivalent with
'  Dim d As Integer
'  d = UBound(Split(var, ".")) '-- count number of dots
'  ... ...
'  ElseIf Not AllowDot And d > 0 Then
'  ElseIf AllowDot And d > 1 Then

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Code:

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Function IsDigits(ByVal Num As Variant) As Boolean
  IsDigits = (Len(Num) > 0) And (Num Like String(Len(Num), "#"))
End Function

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Faster:

Code:

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Function IsDigits(ByVal Num As Variant) As Boolean
  IsDigits = (LenB(Num) > 0) And (Num Like String(Len(Num), "#"))
End Function

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Quote:

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Originally Posted by Zach_VB6

Code:

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Function IsDigits(ByVal Num As Variant) As Boolean
  IsDigits = (Len(Num) > 0) And (Num Like String(Len(Num), "#"))
End Function

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Faster:

Code:

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Function IsDigits(ByVal Num As Variant) As Boolean
  IsDigits = (LenB(Num) > 0) And (Num Like String(Len(Num), "#"))
End Function

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How faster with a single call? ±1*10^-10 second different?.
Have you actually test it?

This is a bit faster if you want:

Code:

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Function IsDigits(ByVal Num As Variant) As Boolean
  Dim n As Long
  n = Len(Num)
  IsDigits = (n > 0) And (Num Like String$(n, "#"))
End Function

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The biggest problem is use of Variant, I don't see a need for it so:

Code:

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' shorter by one character, two times faster
Function IsDigits(ByVal Num As String) As Boolean
    If LenB(Num) Then IsDigits = Num Like String$(Len(Num), "#")
End Function

' shorter by three characters but two times slower
Function IsDigits(ByVal Num As String) As Boolean
    IsDigits = Num = Format$(Val(Num), String$(Len(Num), "0"))
End Function

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Good use of default Boolean value (False) and ByVal String, Merri.