# Visual basic 6.0

Can you send me a code to my program..
this is my problem:
"enter a sentence and determine the number of words present in a particular sentence."
thank you..i hope you can send me the code i need,
thank you...god bless,,,,,,,,

Posted by on

• K-ann Feb 29, 2008

Good morning!

can you send me a code in visual basic 6.0 ?

this is my program:

" enter a sentence and determine how many word are there in a particular sentence"

Thank you very much. Hope that you can send me a code in my problem

God bless you........

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• Contributor

Below is the code....you can skip the isWord check if you want to count numbers as well. Hope this helps...

Function countWords(str As String) As Long
Dim words() As String, i As Long, numWords As Long
words = Split(str, " ")
For i = LBound(words) To UBound(words)
If isWord(words(i)) Then
numWords = numWords + 1
End If
Next
countWords = numWords
End Function
Function isWord(str As String) As Boolean
Dim ret As Boolean
If str = Null Then
ret = False
ElseIf Len(str) = 0 Then
ret = False
ElseIf Mid(str, 1, 1) >= "A" And Mid(str, 1, 1) <= "Z" Then
ret = True
ElseIf Mid(str, 1, 1) >= "a" And Mid(str, 1, 1) <= "z" Then
ret = True
Else
ret = False
End If
isWord = ret
End Function

Posted on Mar 19, 2008

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• Expert

Use a compare (instr) command to search the sentence for punctution and spaces. Increment a variable for each word found, resetting the search string and retaining the start position of each find.

Posted on Feb 27, 2008

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Posted on Jan 02, 2017

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try to this topic. HOW TO CONVERT NUMBERS INTO WORDS. you can search though internet the free program for converting numbers into words.

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### Visual basic 6.0

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### Source code for word count

Haha, i wrote one of thee a while ago for a competition. You need to us e the char at function mate :D
Inside a for loop
Ill havea look for you though :D

Sep 09, 2008 | ArcMedia JavaScript Source Code 3000 Pro...

### Program

Sub aa()
'
' aa Macro
'
'
Dim a As String
a = "how are you this is dharani"
Dim leng, cnt As Integer
leng = 1
cnt = 0
While leng <= Len(RTrim(a))
If Mid(a, leng, 1) = " " Then
cnt = cnt + 1
End If
leng = leng + 1
Wend
If Len(RTrim(a)) > 0 Then
cnt = cnt + 1
End If
MsgBox Str(cnt) + " Words"
End Sub

Feb 27, 2008 | Computers & Internet

### Program

I wrote this as a vbscript. It's Visual Basic code in windows script. If you have XP, just copy the text into notepad and save it as "vowel.vbs". Now double click the newly created script file.

' vowel.vbs
'
' This VBScript is used to count the vowels and consonants in a string.
' Designed and Tested on Windows XP Pro SP2
'
' Version 1.0.0 - 02.20.2008
'
' This code may be freely distributed or modified.
' -----------------------------------------------------------------'

Option Explicit
Dim str
Dim vowels
Dim consonants
Dim i

vowels = 0
consonants = 0
str = "hello"

for i = 1 to len(str)
select case mid(str,i,1)
case "a"
vowels = vowels + 1
case "e"
vowels = vowels + 1
case "i"
vowels = vowels + 1
case "o"
vowels = vowels + 1
case "u"
vowels = vowels + 1
case else
consonants = consonants + 1
end select
next

msgbox "Vowels = " & vowels
msgbox "Consonants = " & consonants
'End of script.

Feb 21, 2008 | Computers & Internet

### Help me

Sub CountCharacters(Text As String)
Dim iVowel As Integer
Dim iCons As Integer
Dim iNum As Integer
Dim iSpace As Integer
Dim iOther As Integer

Dim cnt As Integer
Dim sChar As String

For cnt = 1 To Len(Text)
sChar = LCase(Mid(Text, cnt, 1))

Select Case sChar
Case "a", "e", "i", "o", "u" ' And sometimes "y"?
' Vowel
iVowel = iVowel + 1
Case "a" To "z"
' Includes the first case, but VB matches the first case and stops
iCons = iCons + 1
Case "0" To "9"
' Numbers
iNum = iNum + 1
Case " ", vbTab, vbLf
' Space characters
' Not including vbCr as this will likely be dealing with Windows text
' and I only want to match one character for a newline
iSpace = iSpace + 1
Case Else
If sChar <> vbCr Then iOther = iOther + 1
End Select
Next cnt

Debug.Print "Text: " & Text
Debug.Print "Vowels: " & CStr(iVowel)
Debug.Print "Consonants: " & CStr(iCons)
Debug.Print "Numbers: " & CStr(iNum)
Debug.Print "Space characters: " & CStr(iSpace)
Debug.Print "Other: " & CStr(iOther)
End Sub

Feb 21, 2008 | Computers & Internet

### Visual basic code

'Text1 is a textbox or a string you can use

Private Sub Command1_Click()
Dim a
Dim vow, con As Integer
vow = 0
con = 0
For i = 1 To Len(Text1)
a = Mid(Text1, i, 1)
If (a = "a" Or a = "e" Or a = "i" Or a = "o" Or a = "u") Then
vow = vow + 1
Else
con = con + 1
End If
Next
MsgBox "Vowels: " & vow & vbNewLine & "Consonants : " & con

End Sub

Feb 19, 2008 | Computers & Internet

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