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gmayor · 05-28-2015, 12:44 AM

First of all the first set of brackets is superfluous
Find: ^13([a-z])
Replace: \1
would achieve the same result - http://www.gmayor.com/replace_using_wildcards.htm

As for a macro, select the text that you wish to process and run

Code:

Sub RemoveParaOrLineBreaks()
Dim oRng As Range
Dim oFind As Range
    Set oRng = Selection.Range
    Set oFind = Selection.Range
    With oFind.Find
        Do While .Execute(FindText:="[^13^l]{1,}", MatchWildcards:=True)
            If oFind.InRange(oRng) Then
                oFind.Text = ""
            End If
        Loop
    End With
    Set oFind = oRng
    'Optionally remove additional spaces
    With oFind.Find
        Do While .Execute(FindText:="[ ]{2,}", MatchWildcards:=True)
            If oFind.InRange(oRng) Then
                oFind.Text = Chr(32)
                oFind.Collapse 0
            End If
        Loop
    End With
    'end of option
lbl_Exit:
    Set oRng = Nothing
    Set oFind = Nothing
    Exit Sub
End Sub