Smart quotations aren't recognized by InStr() - vba

Question

I have a code like so:
Sub MoveToBeginningSentence()
Application.ScreenUpdating = False
Dim selectedWords As Range
Dim selectedText As String
Const punctuation As String = " & Chr(145) & "
On Error GoTo ErrorReport
' Cancel macro when there's no text selected
Selection.Cut
Selection.MoveLeft Unit:=wdSentence, Count:=1, Extend:=wdMove
Selection.MoveRight Unit:=wdCharacter, Count:=1, Extend:=wdExtend
Set selectedWords = Selection.Range
selectedText = selectedWords
If InStr(selectedText, punctuation) = 0 Then
Selection.MoveLeft Unit:=wdSentence, Count:=1, Extend:=wdMove
Selection.Paste
Else
Selection.MoveLeft Unit:=wdSentence, Count:=1, Extend:=wdMove
Selection.Paste
Selection.Paste
Selection.Paste
Selection.Paste
End If
ErrorReport:
End Sub
Basically, it help me move whatever text I have selected to the beginning of the sentence in Word. If there's no quotation mark, then paste once. If there is a quote mark, paste 4 times.
The problem is regardless of whether there's any quotation there or not, it will only paste once. If I set the macro to detect any other character, it will work fine. But every single time I try to force it to detect smart quotations, it will fail.
Is there any way to fix it?

Answer 1

Working with the Selection object is always a bit chancy; on the whole, it's better to work with a Range object. You can have only one Selection; you can have as many Ranges as you need.
Because your code uses the Selection object it's not 100% clear what the code does. Based on my best guess, I put together the following example which you can tweak if it's not exactly right.
At the beginning, I check whether there's something in the selection, or it's a blinking insertion point. If no text is selected, the macro ends. This is better than invoking Error handling, then not handling anything: If other problems crop up in your code, you wouldn't know about them.
A Range object is instantiated for the selection - there's no need to "cut" it, as you'll see further along. Based on this, the entire sentence is also assigned to a Range object. The text of the sentence is picked up, then the sentence's Range is "collapsed" to its starting point. (Think of this like pressing the left arrow on the keyboard.)
Now the sentence's text is checked for the character Chr(145). If it's not there, the original selection's text (including formatting) is added at the beginning of the sentence. If it's there, then it's added four times.
Finally, the original selection is deleted.
Sub MoveToBeginningSentence()
Application.ScreenUpdating = False
Dim selectedText As String
Dim punctuation As String
punctuation = Chr(145) ' ‘ "smart" apostrophe
Dim selRange As word.Range
Dim curSentence As word.Range
Dim i As Long
' Cancel macro when there's no text selected
If Selection.Type = wdSelectionIP Then Exit Sub
Set selRange = Selection.Range
Set curSentence = selRange.Sentences(1)
selectedText = curSentence.Text
curSentence.Collapse wdCollapseStart
If InStr(selectedText, punctuation) = 0 Then
curSentence.FormattedText = selRange.FormattedText
Else
For i = 1 To 4
curSentence.FormattedText = selRange.FormattedText
curSentence.Collapse wdCollapseEnd
Next
End If
selRange.Delete
End Sub

Answer 2

Please check out this code.
Sub MoveToBeginningSentence()
' 19 Jan 2018
Dim Rng As Range
Dim SelText As String
Dim Repeats As Integer
Dim i As Integer
With Selection.Range
SelText = .Text ' copy the selected text
Set Rng = .Sentences(1) ' identify the current sentence
End With
If Len(SelText) Then ' Skip when no text is selected
With Rng
Application.ScreenUpdating = False
Selection.Range.Text = "" ' delete the selected text
Repeats = IIf(IsQuote(.Text), 4, 1)
If Repeats = 4 Then .MoveStart wdCharacter, 1
For i = 1 To Repeats
.Text = SelText & .Text
Next i
Application.ScreenUpdating = True
End With
Else
MsgBox "Please select some text.", _
vbExclamation, "Selection is empty"
End If
End Sub
Private Function IsQuote(Txt As String) As Boolean
' 19 Jan 2018
Dim Quotes
Dim Ch As Long
Dim i As Long
Quotes = Array(34, 147, 148, -24143, -24144)
Ch = Asc(Txt)
' Debug.Print Ch ' read ASCII code of first character
For i = 0 To UBound(Quotes)
If Ch = Quotes(i) Then Exit For
Next i
IsQuote = (i <= UBound(Quotes))
End Function
The approach taken is to identify the first character of the selected sentence using the ASC() function. For a normal quotation mark that would be 34. In my test I came up with -24143 and -24144 (opening and closing). I couldn't identify Chr(145) but found MS stating that curly quotation marks are Chr(147) and Chr(148) respectively. Therefore I added a function that checks all of them. If you enable the line Debug.Print Ch in the function the character code actually found will be printed to the immediate window. You might add more character codes to the array Quotes.
The code itself doesn't consider spaces between words. Perhaps Word will take care of that, and perhaps you don't need it.

Answer 3

You need to supply InStr with the starting position as a first parameter:
If InStr(1, selectedText, punctuation) = 0 Then
Also
Const punctuation As String = " & Chr(145) & "
is going to search for space-ampersand-space-Chr(145)-space-ampersand-space. If you want to search for the smart quote character then use
Const punctuation As String = Chr(145)
Hope that helps.