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how to extract a number from a string containng text and special character including white space

how to extract a number from a string containng text and special character (like , . / ? etc) including white space
2 Solutions
Do you know at what position the number is? Is there a particular character that always follows or prefixes the number, give me an example
You don't say whether the text string has any structure at all, what type of number you are looking for (interger or decimal point), neither where you need to use th extracted number (form, report, query).

However, try the following: Place the code below into a VBA module in your database. Then you can use this function in a query like this:

SELECT other_fields, ExtractFirstNumber(source_field_name) FROM table_name

This somewhat crude function will extract the first number contained within the string. It can be either integer or decimal and can be preceeded by the + or - signs.
Private Function Min(ByVal valueA As Long, ByVal valueB As Long) As Long
    If valueA < valueB Then
        Min = valueA
        Min = valueB
    End If
End Function
Public Function ExtractFirstNumber(ByVal strSourceString As String) As Variant
Dim strNumber As String
Dim l As Long
Dim c As String * 1
Dim i As Long
Dim j As Long
Dim last As Long
    l = Len(strSourceString)
    For i = 1 To l
        c = Mid(strSourceString, i, 1)
        If IsNumeric(c) Or ((c = "." Or c = "+" Or c = "-") And IsNumeric(Mid(strSourceString, Min(i + 1, l), 1))) Then
            strNumber = c
            For j = i + 1 To Len(strSourceString)
                c = Mid(strSourceString, j, 1)
                If Not IsNumeric(strNumber & c) Then Exit For
                strNumber = strNumber & c
            Next j
            ExtractFirstNumber = CDec(strNumber)
            Exit Function
        End If
    Next i
End Function

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Patrick MatthewsCommented:

The UDF below uses Regular Expressions to provide very flexible and powerful text parsing capabilities.

For example, if you want to extract a number from a string like the below:

Th/is col#umn has a 12345 number em?bedded in it plus $pecial char.acters

You would do this:

SELECT SomeColumn, RegExpFind(SomeColumn, "\d+", 1) AS ReturnVal
FROM SomeTable

It would return 12345 as a string.  To return it as a number:

SELECT SomeColumn, Val(RegExpFind(SomeColumn, "\d+", 1)) AS ReturnVal
FROM SomeTable

If your needs are different, then please clarify.

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 = <positive integer>  : 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()
    Static 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
            Pos = CLng(Pos)
        End If
    End If
    ' Create instance of RegExp object
    If RegX Is Nothing Then 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)
            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
                    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
        RegExpFind = ""
    End If
    ' Release object variables
    Set TheMatches = Nothing
End Function

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