counting question marks and full stops.

[spazzirifick]

Below is my coding for the counting of sentences in a paragraph of inputted text. So far though i can only get it to count full stops (eg. however many full stops there are there are sentences.). What if i had a question mark though, because that would be the end of a sentence too???? so how do i include question marks in the code below to count question marks AND full stops. (eg. however many full stops AND question marks there are sentences).
hope u caught my drift!!

Private Sub cmdsentences_Click()
    Dim Sentence As String, Char As String
    Dim Count As Integer, I As Integer
    Count = 0
    Sentence = Trim(txtInput.Text)
    If Len(Sentence) > 0 Then
        For I = 1 To Len(Sentence)
            Char = Mid(Sentence, I, 1)
            If Char = "." Then
           
                Count = Count + 1
            End If
        Next I
            Count = Count
    End If
    lblCountsentences.Caption = Count
End Sub 

[dglienna]

If it's formatted correctly, you could check for two spaces, which should be between the period and the next sentence. Same from a question mark.

Otherwise, change it to this:

If Char = "." Or Char = "?"Then 

[spazzirifick]

THANKYOU!!!!

we have the "or" inbetween but i didnt realise that you had to have char= twice!!

thanks!

it works in other words..could you tell :P

[Hojo]

Don't forget sentences can finish with an exclamation point to!

And does it mater if people over puncuate??? (this would be 3 sentences to you code)

Just some things to consider.

[MartinLiss]

Here's another way.

Dim strSentences() As String
    Dim intCount As Integer
    
    strSentences = Split(Text1.Text, ".")
    intCount = intCount + UBound(strSentences)
    strSentences = Split(Text1.Text, "?")
    intCount = intCount + UBound(strSentences)
    strSentences = Split(Text1.Text, "!")
    intCount = intCount + UBound(strSentences)
    MsgBox intCount 

[MartinLiss]

Oh, wait, "THANK YOU!!!!" is a valid sentence!

[penagate]

Courtesy of yrwyddfa and I, with a few slight modifications:

Declare Sub RtlMoveMemory Lib "kernel32" ( _
    ByRef lpvDest As Any, _
    ByRef lpvSrc As Any, _
    ByVal cbLen As Long _
)

Function InStr2CharCount( _
    ByVal pszString As Long, _
    ByRef pszFind1 As String, _
    ByRef pszFind2 As String _
) As Long
Static chBuf(1024)  As Byte
Dim chSearchChar1   As Byte
Dim chSearchChar2   As Byte
Dim lStringLen      As Long
Dim i               As Long

    RtlMoveMemory lStringLen, ByVal (pszString - 4), 4&
    RtlMoveMemory chBuf(0), ByVal pszString, lStringLen

    chSearchChar1 = AscW(pszFind1)
    chSearchChar2 = AscW(pszFind2)
    For i = 0 To lStringLen Step 2
        If (chBuf(i) = chSearchChar1) Or (chBuf(i) = chSearchChar2) Then _
            InStr2CharCount = InStr2CharCount + 1
    Next i
End Function

' Usage:
MsgBox Instr2CharCount(StrPtr(Trim$(txtInput.Text)), ".", "?") 

[GreenEye]

Its not a very good way of counting sentences. But rather number of occurances of fullstops, question marks and exclamation marks. To increase accuracy try counting a trailing space. Like count for ". " and "? " etc. And dont forget to add 1 to count for the last sentence. This way confusions made by sentences like "Thank you!!!" will be eliminated. Actually its just more accurate way not foll-proof. More accurate count requires more sophisticated and complex code.