Count words in word vba

asked Feb 3, 2012 at 0:04

Here’s another regex solution:

Dim WordCount = New Regex("w+").Matches(s).Count

answered Feb 3, 2012 at 1:45

You need to remove the «return» key

Add these lines:

str = Replace(str, "chr(10)", "")  'Replaces  line feed
str = Replace(str, "chr(13)", "")  'Replaces carriage return

before

  str = LTrim(str) 'removes blank spaces at the beginning of text
  str = RTrim(str) ' removes blank spaces at the end of text 

answered Feb 3, 2012 at 1:35

Don’t use

str = Replace(str, "chr(10)", "")  'Replaces  line feed
str = Replace(str, "chr(13)", "")  'Replaces carriage return

Better

str = Replace(str, vbCrLf, " ")

If you would like an easy word count without using Regex

Dim s As String = " Count    the   words "

Dim iWords As Integer = 0
Dim bNonPrintable As Integer = 1

Dim len As Integer = s.Length - 1
Dim i As Integer

For i = 0 To len
    If s(i) = " "c Or s(i) = Chr(10) Or s(i) = Chr(13) Then
        If bNonPrintable = 0 Then
            iWords += 1

            bNonPrintable = i
        End If
    Else
        If bNonPrintable > 0 Then bNonPrintable = 0
    End If
Next
If bNonPrintable = 0 Then iWords += 1

iWords value will be 3 in this case (no need to use trim or replaces)

answered Oct 24, 2017 at 9:21

I have seen a lot of examples where people count empty spaces and think these are equivalent to the count of words. Well this is not correct and you should not use this method. There are several cases I can think of where this does not work:

1. When an empty space is not followed by a word, the number of spaces is +1 the number of words.
2. You can have double spaces, they will count as +1 word as well.
3. Most importantly, you can have a word followed by a newline character(s) and then followed by another word. There are no spaces here, so the whole thing will be wrongfully considered one word instead of two.

Now here is what I have come up with. Add an empty space to the beginning of an existing string. Then replace all newline occurrences with empty space characters. Then check one by one the characters from the start of the string, if the current character is an empty space and character following it is not, then you got a word. Simple as that! Works with consecutive spaces, newline characters and pretty much anything I could throw at it.

Implementation in VB6:

Private Function countwords(cstring As String) As Long
cstring = " " & cstring
cstring = Replace(cstring, vbNewLine, " ")
countwords = 0
    For i = 1 To Len(cstring) - 1
        If Mid(cstring, i, 1) = " " And Mid(cstring, i + 1, 1) <> " " Then
            countwords = countwords + 1
        End If
    Next
End Function

answered Feb 11, 2021 at 21:52