help with word and character counter

[spazzirifick]

hihi

i have to write some visual basic to make a word counter, character counter and count the commas and full stops too.
but i only know how to do the word counter!! can anyone help with the character counter (how many characters altogether in a paragraph) or the other two!! thanks

[dglienna]

msg len(text1.text)

will count the characters.

[Devion]

Public Function fCountOccurance(strData as string, strChar as string) as Long
     
     Dim CharCount as long
 
     CharCount = 0 ' reset counter
    
     For X = 1 To Len(StrData) 'loop through word
          If Mid$(Str, X, 1) = strChar Then 
               CharCount = CharCount + 1 'increase the counter by one
          End If
     Next
 
    fCountOccurance = CharCount
End function

Occurance counter The Easy Way(tm)(c) .. The one did without having any coffee yet -.-

[vbasicgirl]

I have seen this tested for speed somewhere and if i remember rightly, InStr() came out on top.
I use to use this

Private Function CountChar(MainString As String, TheChar As String) As Long
  CountChar = Len(MainString) - Len(Replace(MainString, TheChar, ""))
End Function

But i believe this is faster

Private Function CountChar(MainString As String, TheChar As String) As Long
Dim pos As Long
 pos = InStr(1, MainString, TheChar)
 
  Do While pos
   CountChar = CountChar + 1
   pos = InStr(pos + 1, MainString, TheChar)
  Loop
End Function

casey.

[MartinLiss]

Word counter...

Dim strWords() As String
Dim bReplaced As Boolean
Dim strTestData As String
 
strTestData = "This is  some test     data with extra spaces in it" _
            & " and it contains fifteen words."
            
' Get rid of the double, triple, etc spaces.
bReplaced = True
Do Until Not bReplaced
    bReplaced = False
    ' Look for double spaces
    If InStr(strTestData, "  ") Then
        ' Replace the double spaces with single spaces
        strTestData = Replace(strTestData, "  ", " ")
        bReplaced = True
    End If
Loop
 
' Now see how many words there are
strWords = Split(strTestData, " ")
' We add one because strWords(0) is the 1st word
MsgBox "There are " & UBound(strWords) + 1 & " words"

[Devion]

Martin.. your code would actually count dashes and any other loose character

[MartinLiss]

Martin.. your code would actually count dashes and any other loose character

You're right I hadn't thought of that. Interestingly I just tried Word to see what it does and it says "This - is a test" has 4 words but "This -- is a test" has 5, so if I wanted to modify my code to work like Word I guess I would look for single characters in the strWord() array that weren't letters and ignore them.

[Pradeep1210]

Dim str1 As String
str1 = Text1.Text
Do While InStr(str1, "  ") > 0
    str1 = Replace(str1, "  ", " ")
Loop
'To count no. of words
MsgBox Len(str1) - Len(Replace(str1, " ", "")) + 1
 
'To count no. of characters (with spaces)
MsgBox Len(str1)
 
'To count no. of characters (without spaces)
MsgBox Len(Replace(str1, " ", ""))
 
'To count no. of semi-colons or any other character
MsgBox Len(str1) - Len(Replace(str1, ";", ""))

[MartinLiss]

I not sure this is rigorous enough but you get the idea.

Dim strWords() As String
Dim bReplaced As Boolean
Dim strTestData As String
Dim lngIndex As Long
Dim intCount As Integer
 
strTestData = "This is  a test  -  string with extra spaces in it" _
            & " and it contains fifteen words."
            
' Get rid of the double, triple, etc spaces.
bReplaced = True
Do Until Not bReplaced
    bReplaced = False
    ' Look for double spaces
    If InStr(strTestData, "  ") Then
        ' Replace the double spaces with single spaces
        strTestData = Replace(strTestData, "  ", " ")
        bReplaced = True
    End If
Loop
 
' Now see how many words there are
strWords = Split(strTestData, " ")
 
' We add one because strWords(0) is the 1st word
intCount = UBound(strWords) + 1
        
 
For lngIndex = 0 To UBound(strWords)
    If Len(strWords(lngIndex)) = 1 Then
        If InStr(1, "ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789", UCase(strWords(lngIndex))) Then
           ' The above are OK
        Else
            intCount = intCount - 1
        End If
    End If
Next
           
MsgBox "There are " & intCount & " words"

[spazzirifick]

THANKS!! that was a little simple but combined with the other replies i think i've got it! cheers

[spazzirifick]

for this you are using a test piece of string right??

if i wanted to make that a paragraph of text i have inputted what would i do??

thanks for your help so far

[MartinLiss]

Public Function CountWords(strText As String) As Integer
 
Dim strWords() As String
Dim bReplaced As Boolean
Dim lngIndex As Long
            
' Get rid of the double, triple, etc spaces.
bReplaced = True
Do Until Not bReplaced
    bReplaced = False
    ' Look for double spaces
    If InStr(strText, "  ") Then
        ' Replace the double spaces with single spaces
        strText = Replace(strText, "  ", " ")
        bReplaced = True
    End If
Loop
 
' Now see how many words there are
strWords = Split(strText, " ")
 
' We add one because strWords(0) is the 1st word
CountWords = UBound(strWords) + 1
 
For lngIndex = 0 To UBound(strWords)
    If Len(strWords(lngIndex)) = 1 Then
        If InStr(1, "ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789", UCase(strWords(lngIndex))) Then
           ' The above are OK
        Else
            CountWords = CountWords - 1
        End If
    End If
Next
 
End Function

Usage

MsgBox CountWords(<Your string or variable name goes here>)

BTW, welcome to VB Forums!

[spazzirifick]

i dont know much about VB but you code looks easy enough to understand

however, i dont know what i am replacing with what..eg. my text boxs and command buttons have names but where should i be putting the names of each into my code??

[spazzirifick]

and if i put that code in straight away will it work? because it doesnt seem to..i think i need to chnage something to fit into mine..i just dont know what (i am not very experienced AT all with VB).

[dglienna]

In the click event of the button that you want to count words, place something like this:

MsgBox CountWords(myTextbox.text) & " words in document"

or to place them in a variable,

dim x as integer
x = CountWords(myTextbox.text) 
msgbox("There are " & x & " words in this document")