Avatar of andybrooke
andybrooke
 asked on

Only showing numbers in cell

Hi,

a previous question I wanted only number to be shown from a field this worked great code sample below. I want to expand on this and only show numbers the are more than 3 digits together.

For example I have have a cell like "58     YBS1    QWS/ENQ/1501510 BGC". From this I would only want 1501510 to show....

Thanks
Option Explicit

Function RemoveAlpha(Rng As String) As String
    Dim Tmp As String
    Dim i As Integer
    Dim Alpha As String
    
    Tmp = Rng
    'Numbers are 48-57
    For i = 1 To 255
        If i >= 48 And i <= 57 Then GoTo skip
        Alpha = Chr(i)
        Tmp = Application.Substitute(Tmp, Alpha, "")
skip:
    Next i
    RemoveAlpha = Tmp
End Function

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Microsoft Excel

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Last Comment
dsacker

8/22/2022 - Mon
dsacker

Function RemoveAlphaKeep3Digit(ByVal Rng As String) As String
    Dim Alpha As String
    Dim Char As String
    Dim Cnt As Integer
    Dim i As Integer
    Dim Hold As String
    
    Cnt = 0
    Hold = ""
    For i = 1 To Len(Rng)
        Char = Mid(Rng, i, 1)
        Select Case Char
            Case "0" To "9"
                Cnt = Cnt + 1
                Hold = Hold + Char
            Case Len(Hold) < 3
                Hold = ""
                Cnt = 0
            Case Else
                Alpha = Alpha + Hold
                Hold = ""
                Cnt = 0
        End Select
    Next i
    If Len(Hold) >= 3 Then
        Alpha = Alpha + Hold
    End If
    RemoveAlphaKeep3Digit = Alpha
End Function

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ASKER CERTIFIED SOLUTION
dsacker

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Patrick Matthews

What should the result be for, say, "58     YBS1    QWS/ENQ/1501510 BGC 1234XYZ5678"?
andybrooke

ASKER
yes I though of this, unfrotunaley I have to go through those manuall but there isnt masses of them. Thanks dsacker seems to be working great!
Your help has saved me hundreds of hours of internet surfing.
fblack61
SafetyFish

Option Explicit

Function RemoveAlpha(Rng As String) As String
    Dim Tmp As String
    Dim i As Integer
    Dim Alpha As String
   
    Dim Left As String
    Dim LeftLength As Integer
    Dim RightLength As Integer
    Dim ExitLoop As Boolean
   
    ExitLoop = False
   
    Tmp = Rng
    'Numbers are 48-57
    For i = 1 To 255
        If i >= 48 And i <= 57 Then GoTo skip
        Alpha = Strings.Chr(i)
        Tmp = Application.Substitute(Tmp, Alpha, " ")
skip:
    Next i
   
    LeftLength = InStr(1, Tmp, " ") - 1
    RightLength = Strings.Len(Tmp) - LeftLength
   
    Do While ExitLoop = False
        If LeftLength > 3 Then
            RemoveAlpha = Strings.Left(Tmp, LeftLength)
            ExitLoop = True
        Else
            Tmp = Strings.Right(Tmp, Strings.Len(Tmp) - LeftLength)
            If InStr(1, Tmp, " ") = 1 Then
                LeftLength = 1
            Else
                LeftLength = InStr(1, Tmp, " ") - 1
            End If
            RightLength = Strings.Len(Tmp) - LeftLength
            ExitLoop = False
        End If
    Loop
   
    RemoveAlpha = Tmp
End Function
Patrick Matthews

andybrooke,

You might also want to consider an approach using Regular Expressions, as outlined here:

https://www.experts-exchange.com/Programming/Languages/Visual_Basic/A_1336-Using-Regular-Expressions-in-Visual-Basic-for-Applications-and-Visual-Basic-6.html

Depending on the exact behavior you want, you can use either the RegExpFind or RegExpReplace functions from that article:

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
        Else
            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
                Else
                    
                    ' 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
                    Else
                        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
            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
                    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
    
    Else
        RegExpFind = ""
    End If
    
Cleanup:
    ' Release object variables
    
    Set TheMatches = Nothing
    
End Function

Function RegExpReplace(LookIn As String, PatternStr As String, Optional ReplaceWith As String = "", _
    Optional ReplaceAll As Boolean = True, Optional MatchCase As Boolean = True, _
    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, and replace parts of the string
    ' matching the specified pattern with another string.  The optional argument ReplaceAll
    ' controls whether all instances of the matched string are replaced (True) or just the first
    ' instance (False)
    
    ' If you need to replace the Nth match, or a range of matches, then use RegExpReplaceRange
    ' instead
    
    ' By default, RegExp is case-sensitive in pattern-matching.  To keep this, omit MatchCase or
    ' set it to True
    
    ' If you use this function from Excel, you may substitute range references for all the arguments
    
    ' 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
    
    If RegX Is Nothing Then Set RegX = CreateObject("VBScript.RegExp")
    With RegX
        .Pattern = PatternStr
        .Global = ReplaceAll
        .IgnoreCase = Not MatchCase
        .MultiLine = MultiLine
    End With
    
    RegExpReplace = RegX.Replace(LookIn, ReplaceWith)
    
End Function

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To grab just the first instance of 4+ consecutive digits:

=RegExpFind(A2,"\d{4,}")

To remove all characters from a string except for blocks of 4+ consecutive digits:

=RegExpReplace(A2,"((^|\D)\d{1,3}($|\D))|\D")

Note that that second expression seems to be returning the same results as dsacker's code.

The advantage to RegExp is that is is extremely flexible, but to each his/her own :)

Patrick
dsacker

Thanks, Andy. Glad it worked.
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