VBA. How to find position of first digit in string - vba

Question

I have string "ololo123".
I need get position of first digit - 1.
How to set mask of search ?

Answer 1

Here is a lightweight and fast method that avoids regex/reference additions, thus helping with overhead and transportability should that be an advantage.
Public Function GetNumLoc(xValue As String) As Integer
For GetNumLoc = 1 To Len(xValue)
If Mid(xValue, GetNumLoc, 1) Like "#" Then Exit Function
Next
GetNumLoc = 0
End Function

Answer 2

Something like this should do the trick for you:
Public Function GetPositionOfFirstNumericCharacter(ByVal s As String) As Integer
For i = 1 To Len(s)
Dim currentCharacter As String
currentCharacter = Mid(s, i, 1)
If IsNumeric(currentCharacter) = True Then
GetPositionOfFirstNumericCharacter = i
Exit Function
End If
Next i
End Function
You can then call it like this:
Dim iPosition as Integer
iPosition = GetPositionOfFirstNumericCharacter("ololo123")

Answer 3

Not sure on your environment, but this worked in Excel 2010
'Added reference for Microsoft VBScript Regular Expressions 5.5
Const myString As String = "ololo123"
Dim regex As New RegExp
Dim regmatch As MatchCollection
regex.Pattern = "\d"
Set regmatch = regex.Execute(myString)
MsgBox (regmatch.Item(0).FirstIndex) ' Outputs 5

Answer 4

I actually have that function:
Public Function GetNumericPosition(ByVal s As String) As Integer
Dim result As Integer
Dim i As Integer
Dim ii As Integer
result = -1
ii = Len(s)
For i = 1 To ii
If IsNumeric(Mid$(s, i, 1)) Then
result = i
Exit For
End If
Next
GetNumericPosition = result
End Function

Answer 5

You could try regex, and then you'd have two problems. My VBAfu is not up to snuff, but I'll give it a go:
Function FirstDigit(strData As String) As Integer
Dim RE As Object REMatches As Object
Set RE = CreateObject("vbscript.regexp")
With RE
.Pattern = "[0-9]"
End With
Set REMatches = RE.Execute(strData)
FirstDigit = REMatches(0).FirstIndex
End Function
Then you just call it with FirstDigit("ololo123").

Answer 6

If speed is an issue, this will run a bit faster than Robs (noi Rob):
Public Sub Example()
Const myString As String = "ololo123"
Dim position As Long
position = GetFirstNumeric(myString)
If position > 0 Then
MsgBox "Found numeric at postion " & position & "."
Else
MsgBox "Numeric not found."
End If
End Sub
Public Function GetFirstNumeric(ByVal value As String) As Long
Dim i As Long
Dim bytValue() As Byte
Dim lngRtnVal As Long
bytValue = value
For i = 0 To UBound(bytValue) Step 2
Select Case bytValue(i)
Case vbKey0 To vbKey9
If bytValue(i + 1) = 0 Then
lngRtnVal = (i \ 2) + 1
Exit For
End If
End Select
Next
GetFirstNumeric = lngRtnVal
End Function

Answer 7

An improved version of spere's answer (can't edit his answer), which works for any pattern
Private Function GetNumLoc(textValue As String, pattern As String) As Integer
For GetNumLoc = 1 To (Len(textValue) - Len(pattern) + 1)
If Mid(textValue, GetNumLoc, Len(pattern)) Like pattern Then Exit Function
Next
GetNumLoc = 0
End Function
To get the pattern value you can use this:
Private Function GetTextByPattern(textValue As String, pattern As String) As String
Dim NumLoc As Integer
For NumLoc = 1 To (Len(textValue) - Len(pattern) + 1)
If Mid(textValue, NumLoc, Len(pattern)) Like pattern Then
GetTextByPattern = Mid(textValue, NumLoc, Len(pattern))
Exit Function
End If
Next
GetTextByPattern = ""
End Function
Example use:
dim bill as String
bill = "BILLNUMBER 2202/1132/1 PT2200136"
Debug.Print GetNumLoc(bill , "PT#######")
'Printed result:
'24
Debug.Print GetTextByPattern(bill , "PT#######")
'Printed result:
'PT2200136