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How do I extract targeted characters from a field value in a Microsoft Access query?

I need to run a query and extract a numeric value from a field in a transactions table using Access 2007. The value that I need is part of a larger string of alpha-numeric characters. The new query results, therefore, will contain the extracted value that I will use to look up other values in other tables.
 Here are the details:
1)  The value in the source field is variable length and alphanumeric in the form of ABC123, ABC12, ABC1, AB123, AB12, or AB1.
2) I need to extract ALL the numeric characters from right to left.  In the examples above, for example, the query returns 123, 12, 1, 123, 12, 1, respectively.
In query design mode, what should I enter in the calculated FIELD: and CRITERIA: to return the value that I need?
I would sincerely appreciate any help offered.   I'm a rookie trying to learn on the job!

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2 Solutions
Patrick MatthewsCommented:
This may be a good slot for Regular Expressions.  Please see my article:


From that article, add the RegExpFind function to a regular VBA module (source code below).

You can then use it in a query like this:

SELECT OriginalColumn, RegExpFind(OriginalColumn, "\d+", 1) AS DigitsOnly
FROM SomeTable

Please see the attached file for an example.
Function RegExpFind(LookIn As String, PatternStr As String, Optional Pos, _
    Optional MatchCase As Boolean = True, Optional ReturnType As Long = 0, _
    Optional MultiLine As Boolean = False)
    ' Function written by Patrick G. Matthews.  You may use and distribute this code freely,
    ' as long as you properly credit and attribute authorship and the URL of where you
    ' found the code
    ' For more info, please see:
    ' http://www.experts-exchange.com/articles/Programming/Languages/Visual_Basic/Using-Regular-Expressions-in-Visual-Basic-for-Applications-and-Visual-Basic-6.html
    ' This function relies on the VBScript version of Regular Expressions, and thus some of
    ' the functionality available in Perl and/or .Net may not be available.  The full extent
    ' of what functionality will be available on any given computer is based on which version
    ' of the VBScript runtime is installed on that computer
    ' 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 = 1                   : the first match
    ' Pos = 2                   : the second match
    ' Pos = <positive integer>  : the Nth match
    ' Pos = 0                   : the last match
    ' Pos = -1                  : the last match
    ' Pos = -2                  : the 2nd to last match
    ' Pos = <negative integer>  : the Nth to last match
    ' If Pos is non-numeric, or if the absolute value of Pos is greater than the number of
    ' matches, the function returns an empty string.  If no match is found, the function returns
    ' an empty string.  (Earlier versions of this code used zero for the last match; this is
    ' retained for backward compatibility)
    ' 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]).
    ' ReturnType indicates what information you want to return:
    ' ReturnType = 0            : the matched values
    ' ReturnType = 1            : the starting character positions for the matched values
    ' ReturnType = 2            : the lengths of the matched values
    ' 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()
    ' Note: RegExp counts the character positions for the Match.FirstIndex property as starting
    ' at zero.  Since VB6 and VBA has strings starting at position 1, I have added one to make
    ' the character positions conform to VBA/VB6 expectations
    ' Normally as an object variable I would set the RegX variable to Nothing; however, in cases
    ' where a large number of calls to this function are made, making RegX a static variable that
    ' preserves its state in between calls significantly improves performance
    Static RegX As Object
    Dim TheMatches As Object
    Dim Answer()
    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
            Pos = CLng(Pos)
        End If
    End If
    ' Evaluate ReturnType
    If ReturnType < 0 Or ReturnType > 2 Then
        RegExpFind = ""
        Exit Function
    End If
    ' Create instance of RegExp object if needed, and set properties
    If RegX Is Nothing Then Set RegX = CreateObject("VBScript.RegExp")
    With RegX
        .Pattern = PatternStr
        .Global = True
        .IgnoreCase = Not MatchCase
        .MultiLine = MultiLine
    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)
        ' Test to see if Pos is negative, which indicates the user wants the Nth to last
        ' match.  If it is, then based on the number of matches convert Pos to a positive
        ' number, or zero for the last match
        If Not IsMissing(Pos) Then
            If Pos < 0 Then
                If Pos = -1 Then
                    Pos = 0
                    ' If Abs(Pos) > number of matches, then the Nth to last match does not
                    ' exist.  Return a zero-length string
                    If Abs(Pos) <= TheMatches.Count Then
                        Pos = TheMatches.Count + Pos + 1
                        RegExpFind = ""
                        GoTo Cleanup
                    End If
                End If
            End If
        End If
        ' 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)
            For Counter = 0 To UBound(Answer)
                Select Case ReturnType
                    Case 0: Answer(Counter) = TheMatches(Counter)
                    Case 1: Answer(Counter) = TheMatches(Counter).FirstIndex + 1
                    Case 2: Answer(Counter) = TheMatches(Counter).Length
                End Select
            RegExpFind = Answer
        ' User wanted the Nth match (or last match, if Pos = 0).  Get the Nth value, if possible
            Select Case Pos
                Case 0                          ' Last match
                    Select Case ReturnType
                        Case 0: RegExpFind = TheMatches(TheMatches.Count - 1)
                        Case 1: RegExpFind = TheMatches(TheMatches.Count - 1).FirstIndex + 1
                        Case 2: RegExpFind = TheMatches(TheMatches.Count - 1).Length
                    End Select
                Case 1 To TheMatches.Count      ' Nth match
                    Select Case ReturnType
                        Case 0: RegExpFind = TheMatches(Pos - 1)
                        Case 1: RegExpFind = TheMatches(Pos - 1).FirstIndex + 1
                        Case 2: RegExpFind = TheMatches(Pos - 1).Length
                    End Select
                Case Else                       ' Invalid item number
                    RegExpFind = ""
            End Select
        End If
    ' If there are no matches, return empty string
        RegExpFind = ""
    End If
    ' Release object variables
    Set TheMatches = Nothing
End Function

Open in new window

Try this:

SELECT txtFld, Val(StrReverse(CStr(Val(StrReverse(txtFld)))))
Patrick MatthewsCommented:

That is positively *brilliant*.

Won't work if the string is like ABC123H, as mine will, but then again the Asker did not indicate that the string may go <letters><digits><letters>; all of the example were <letters><digits>.

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Patrick MatthewsCommented:
BTW, I think hnasr's formula can be reduced:

SELECT txtFld, Val(StrReverse(Val(StrReverse(txtFld))))
thutchinsonAuthor Commented:
Thanks guys.  I'm getting a syntax error.  Will you please take a look.  See attached pdf.
Patrick MatthewsCommented:
You need to switch to SQL view, and then paste in hnasr's code (changing for actual column/table names, of course).
thutchinsonAuthor Commented:
It worked! That is so cool.  Thank you gentlemen.  I really appreciate the help.

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