String Extraction Memo Field

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Is there a way to get the first 2 lines of a memo field?  The lines would be split by a hard return.
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Try to use the InStr() function together with Chr(13), something like:

InStr(1,[YourMemoField], Chr(13))



This just gives me a number and not the string on the first and second lines.
Hamed NasrRetired IT Professional

Try this: memo field: rt

Private Sub Command8_Click()
    Dim s As String   ' string to hold memo text
    s = Me.rt 'memo field
    Dim x As Variant ' array to hold lines of text
    x = Split(s, Chr(13) & Chr(10))
    Dim l1 As String  'line 1
    Dim l2 As String
    l1 = Replace(x(0), Chr(13) & Chr(10), "")
    l2 = Replace(x(1), Chr(13) & Chr(10), "")
End Sub
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Top Expert 2010
Try adding the RegExpFind function from my article

Then, use it in a query like this:

SELECT [ID], RegExpFind([MemoColumn],"^.*(\n.*)?",1,False,0,True) AS FirstTwoLines
FROM [SomeTable]

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That returns the first two lines, or just the first line if the memo has only one line.

The source code for RegExpFind:

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:
    ' 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

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Actually need to use it in a query...
Hamed NasrRetired IT Professional

play with chr(10) and see the effect on the second line.
Top Expert 2010


Your apporach will throw an error if there is only one line in the memo column.




matthewspatrick - That works!
Top Expert 2010


Glad to help!  If you have not already done so, I would really appreciate it if you could please return to my article
and click 'Yes' for the 'Was this helpful?' voting.


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