Help!

[spazzirifick]

I have until the end of today to find a way to stop my word counter counting (word)..(word). as 2 sentences when it isnt! i want it to count one full stop as a sentence but not two or three!! HELP!

Private Sub cmdsentences_Click()
    Dim Sentence As String, Char As String 'setting the data types
    Dim Count As Integer, I As Integer 'setting the data types
    Count = 0 'initialise the count to 0
    Sentence = Trim(txtInput.Text) 'set the end of the sentence.
    If Len(Sentence) > 0 Then 'if there is no text, count stays at zero
        For I = 1 To Len(Sentence) 'when it does have characters then execute instructions
            Char = Mid(Sentence, I, 1) 'characters in sentence
            If Char = "." Or Char = "?" Or Char = "!" Then 'characters which show end of sentence(loop begins)
                Count = Count + 1 'add one to count
            End If
        Next I
            Count = Count 'count remains the same
    End If
    lblCountsentences.Caption = Count 'where to display the result
End Sub

[lintz]

Do your sentances only end in ".", "!" and "?" ?

[spazzirifick]

yeh they do

[pnish]

Try something along these lines:

Sentence = Replace$(Sentence, "?", ".")
Sentence = Replace$(Sentence, "!", ".")
Do Until Instr(Sentence, "..") = 0
    Sentence = Replace$(Sentence, "..", ".")
Loop

Add this code just before your For/Next loop starts. Good luck.

[spazzirifick]

can i leave it exactly as you put it or do i need to change it?

[BrianHawley]

Try this.


Basically I've used a flag to show if we are inside a sentence or not.

If we are in a sentence, then a .! or ? is the end of that sentence, so we count it and then we are NOT in a sentence, so if we get any more we don't add to the sentence count.

We detect that we are back in a sentence by checkign the character to see if it is a letter or number. (You may want to add other characters to that line, depending on how you want to define a sentence.)

Private Sub cmdsentences_Click()
    
    Dim InsideASentence As Boolean
    
    Dim Sentence As String, Char As String 'setting the data types
    Dim Count As Integer, I As Integer 'setting the data types
    Count = 0 'initialise the count to 0
    Sentence = Trim(txtInput.Text) 'set the end of the sentence.
    If Len(Sentence) > 0 Then 'if there is no text, count stays at zero
        For I = 1 To Len(Sentence) 'when it does have characters then execute instructions
            Char = Mid(Sentence, I, 1) 'characters in sentence
            
            If InStr("abcdefghijklmnopqrstuvwxyz1234567890", LCase$(Char)) > 0 Then
              InsideASentence = True
            End If
            
            If Char = "." Or Char = "?" Or Char = "!" Then 'characters which show end of sentence(loop begins)
                
                If InsideASentence Then
                  Count = Count + 1 'add one to count
                  InsideASentence = False
                End If
            
            End If
        Next I
            Count = Count 'count remains the same
    End If
    lblCountsentences.Caption = Count 'where to display the result
End Sub

You can take out count=count if you like. It does nothing.

[spazzirifick]

it doesnt like the replace$ thing....o_O what should i do?? "sub or function not defined" <--thats what it says. its a compile error.

[pnish]

It should work exactly as it is

Private Sub cmdsentences_Click()
    Dim Sentence As String, Char As String 'setting the data types
    Dim Count As Integer, I As Integer 'setting the data types
    Count = 0 'initialise the count to 0
    Sentence = Trim(txtInput.Text) 'set the end of the sentence.
    If Len(Sentence) > 0 Then 'if there is no text, count stays at zero
 
        Sentence = Replace$(Sentence, "?", ".")
        Sentence = Replace$(Sentence, "!", ".")
        Do Until Instr(Sentence, "..") = 0
            Sentence = Replace$(Sentence, "..", ".")
        Loop
 
        For I = 1 To Len(Sentence) 'when it does have characters then execute instructions
            Char = Mid(Sentence, I, 1) 'characters in sentence
            If Char = "." Or Char = "?" Or Char = "!" Then 'characters which show end of sentence(loop begins)
                Count = Count + 1 'add one to count
            End If
        Next I
 [b]Don't need to do this=>[/b] Count = Count 'count remains the same
    End If
    lblCountsentences.Caption = Count 'where to display the result
End Sub

[spazzirifick]

haha, seems like it should work but the insideasentence comes up with "variable not found" should i be replacing it with something?

[pnish]

it doesnt like the replace$ thing....o_O what should i do?? "sub or function not defined" <--thats what it says. its a compile error.

What version of VB are you using??

[spazzirifick]

i put that in, but it doesnt like the replace$ thing. i'm not sure why?

[spazzirifick]

i'm using vb 5

[pnish]

I've got a feeling that VB 5 doesn't have the Replace$() function

[spazzirifick]

nooo it doesnt!! oh no!! what should i put :O

[BrianHawley]

haha, seems like it should work but the insideasentence comes up with "variable not found" should i be replacing it with something?

Did you Dim it?

Anyway pnishes solution is better (unless you don't have replace).

[spazzirifick]

i dont have a replace...so what should i do with dim??????

[BrianHawley]

Private Sub cmdsentences_Click()
    
    Dim InsideASentence As Boolean
    
    Dim Sentence As String, Char As String 'setting the data types
    Dim Count As Integer, I As Integer 'setting the data types
    Count = 0 'initialise the count to 0
    Sentence = Trim(txtInput.Text) 'set the end of the sentence.
    If Len(Sentence) > 0 Then 'if there is no text, count stays at zero
        For I = 1 To Len(Sentence) 'when it does have characters then execute instructions
            Char = Mid(Sentence, I, 1) 'characters in sentence
            
            If InStr("abcdefghijklmnopqrstuvwxyz1234567890", LCase$(Char)) > 0 Then
              InsideASentence = True
            End If
            
            If Char = "." Or Char = "?" Or Char = "!" Then 'characters which show end of sentence(loop begins)
                
                If InsideASentence Then
                  Count = Count + 1 'add one to count
                  InsideASentence = False
                End If
            
            End If
        Next I
            Count = Count 'count remains the same
    End If
    lblCountsentences.Caption = Count 'where to display the result
End Sub

This works. I've just tested it.

If in doubt, delete everything in your click event and post this instead.

The 'replace' method is better, but won't work if you don't have 'replace'. (Could write your own replace of course)

[pnish]

I deleted the Replace code I posted as it was pretty buggy. Here's some code I found on the VB-Helper website. Hope this helps:

' Replace all occurrances of from_str with to_str.
Public Function ReplaceText(ByVal txt As String, ByVal _
    from_str As String, ByVal to_str As String) As String
Dim result As String
Dim from_len As Integer
Dim pos As Integer
 
    from_len = Len(from_str)
    Do While Len(txt) > 0
        ' Find from_str.
        pos = InStr(txt, from_str)
        If pos = 0 Then
            ' No more occurrences.
            result = result & txt
            txt = ""
        Else
            ' Make the replacement.
            result = result & Left$(txt, pos - 1) & to_str
            txt = Mid$(txt, pos + from_len)
        End If
    Loop
 
    ReplaceText = result
End Function

[spazzirifick]

yep, its works!! thanks so much!!

[spazzirifick]

hey, that code works but when i hve two dots between something it counts it as the end of a sentence. eg. i have...a fish.
it would count this as having two sentences, which is better then what i had before saying "18 sentences" etc..but it still has that error. is this fixable? to say only count the sentence if there is one full stop by itself.

[dglienna]

Post your code. I don't know which code you are using.

[spazzirifick]

Private Sub cmdsentences_Click()
    
    Dim InsideASentence As Boolean
    
    Dim Sentence As String, Char As String 'setting the data types
    Dim Count As Integer, I As Integer 'setting the data types
    Count = 0 'initialise the count to 0
    Sentence = Trim(txtInput.Text) 'set the end of the sentence.
    If Len(Sentence) > 0 Then 'if there is no text, count stays at zero
        For I = 1 To Len(Sentence) 'when it does have characters then execute instructions
            Char = Mid(Sentence, I, 1) 'characters in sentence
            
            If InStr("abcdefghijklmnopqrstuvwxyz1234567890", LCase$(Char)) > 0 Then
              InsideASentence = True
            End If
            
            If Char = "." Or Char = "?" Or Char = "!" Then 'characters which show end of sentence(loop begins)
                
                If InsideASentence Then
                  Count = Count + 1 'add one to count
                  InsideASentence = False
                End If
            
            End If
        Next I
            Count = Count 'count remains the same
    End If
    lblCountsentences.Caption = Count 'where to display the result
End Sub

[pnish]

Hmm. That's going to be tricky to try and handle every conceivable type of sentence. In strict grammatical terms, 'I have...a fish.' is not a proper sentence anyway, so perhaps you needn't worry about that.

You could have something like 'I.have.many.big.fish.' which your program would count as 5 sentences. I'm not sure how best to handle that situation.

[spazzirifick]

Yeh, its a tad cnfusing. hmhmmhhm
maybe i will just assume that its correct. how would i create an error message if its not enetered correctly..i think that might be a better way to go!

[pnish]

Probably the best thing to do would be to validate the sentence as it's being entered by the user. eg

if the last char entered was a '.' or a '!' or a '?' the don't allow them to enter another '.' or '!' or '?'
don't allow a '.' or a '!' or a '?' to be entered if the sentence doesn't contain any spaces

Things like that.

[dglienna]

You could check for an occurrence of '..", but I would think that you could check for a space (actually two spaces) after the end of a sentence.

[BrianHawley]

Change

If InStr("abcdefghijklmnopqrstuvwxyz1234567890", LCase$(Char)) > 0 Then
              InsideASentence = True
            End If

to

If InStr("ABCDEFGHIJKLMNOPQRSTUVWXYZ1234567890", Char) > 0 Then
              InsideASentence = True
            End If

That way it will not detect a new sentence unless it starts with a capital letter. Should nail most things, unless you have ... followed by a name.