Question : Formula to find variable text string within another string

I have a column of data where I am looking for a formula to isolate a text string of variable characters in this format:

00-000-000

This could either be at the beginning, somewhere in the middle or end of the entire string. Here are some examples:

Existing string          Desired result

31-260-033U          31-260-033
04VS0636 48-464-345W          48-464-354
9032901914 None          (does not exist so blank cell)

Thank you for your advice.

Glenn

   

Answer : Formula to find variable text string within another string

Hello Glen,

1) Add this UDF to your VBA project:

Function RegExpFind(LookIn As String, PatternStr As String, Optional Pos, _
    Optional MatchCase As Boolean = True)

    ' For more info see: http://vbaexpress.com/kb/getarticle.php?kb_id=841

    ' This function uses Regular Expressions to parse a string (LookIn), and return matches to a
    ' pattern (PatternStr).  Use Pos to indicate which match you want:
    ' Pos omitted               : function returns a zero-based array of all matches
    ' Pos = 0                   : the last match
    ' Pos = 1                   : the first match
    ' Pos = 2                   : the second match
    ' Pos =  : the Nth match
    ' If Pos is greater than the number of matches, is negative, or is non-numeric, the function
    ' returns an empty string.  If no match is found, the function returns an empty string
   
    ' If MatchCase is omitted or True (default for RegExp) then the Pattern must match case (and
    ' thus you may have to use [a-zA-Z] instead of just [a-z] or [A-Z]).
   
    ' If you use this function in Excel, you can use range references for any of the arguments.
    ' If you use this in Excel and return the full array, make sure to set up the formula as an
    ' array formula.  If you need the array formula to go down a column, use TRANSPOSE()
   
    Dim RegX As Object
    Dim TheMatches As Object
    Dim Answer() As String
    Dim Counter As Long
   
    ' Evaluate Pos.  If it is there, it must be numeric and converted to Long
    If Not IsMissing(Pos) Then
        If Not IsNumeric(Pos) Then
            RegExpFind = ""
            Exit Function
        Else
            Pos = CLng(Pos)
        End If
    End If
   
    ' Create instance of RegExp object
    Set RegX = CreateObject("VBScript.RegExp")
    With RegX
        .Pattern = PatternStr
        .Global = True
        .IgnoreCase = Not MatchCase
    End With
       
    ' Test to see if there are any matches
    If RegX.test(LookIn) Then
       
        ' Run RegExp to get the matches, which are returned as a zero-based collection
        Set TheMatches = RegX.Execute(LookIn)
       
        ' If Pos is missing, user wants array of all matches.  Build it and assign it as the
        ' function's return value
        If IsMissing(Pos) Then
            ReDim Answer(0 To TheMatches.Count - 1) As String
            For Counter = 0 To UBound(Answer)
                Answer(Counter) = TheMatches(Counter)
            Next
            RegExpFind = Answer
       
        ' User wanted the Nth match (or last match, if Pos = 0).  Get the Nth value, if possible
        Else
            Select Case Pos
                Case 0                          ' Last match
                    RegExpFind = TheMatches(TheMatches.Count - 1)
                Case 1 To TheMatches.Count      ' Nth match
                    RegExpFind = TheMatches(Pos - 1)
                Case Else                       ' Invalid item number
                    RegExpFind = ""
            End Select
        End If
   
    ' If there are no matches, return empty string
    Else
        RegExpFind = ""
    End If
   
    ' Release object variables
    Set RegX = Nothing
    Set TheMatches = Nothing
   
End Function



2) Use it in a formula like this:


=RegExpFind(A2,"\d{2}-\d{3}-\d{3}",1)

It will return the first matching string if at least one is found, or an empty string if there is no match.

Regards,

Patrick