CIs there a way to do a word count on a text field?

[jimk100189]

In Word 2013, I can do a word count on a document under REVIEW.
Does such a thing exist in ACCESS 2013?

[ranman256]

youd have to write your own, or get one from this forum.

[mike60smart]

Hi

Use this Function

Function WordCount(ByRef Str As String) As Long
 
    Dim cnt As Long
    Dim Words As Long
 
    Words = 0
    WordCount = 0
    If Len(Str) = 0 Then
        Exit Function
    End If
    For cnt = 1 To Len(Str)
        Select Case Mid(Str, cnt, 1)
            Case " "
                Words = Words + 1
            Case "."
                Words = Words + 1
            Case ","
                Words = Words + 1
            Case ";"
                Words = Words + 1
            Case ":"
                Words = Words + 1
            Case "-"
                Words = Words + 1
        End Select
    Next cnt
    If Words = 0 Then Words = 1
    WordCount = Words
End Function

[mike60smart]

Then in a query you can use this:-

Count: Wordcount(Nz([Item_Notes],""))

[Welshgasman]

Hi

Use this Function

Function WordCount(ByRef Str As String) As Long
 
    Dim cnt As Long
    Dim Words As Long
 
    Words = 0
    WordCount = 0
    If Len(Str) = 0 Then
        Exit Function
    End If
    For cnt = 1 To Len(Str)
        Select Case Mid(Str, cnt, 1)
            Case " "
                Words = Words + 1
            Case "."
                Words = Words + 1
            Case ","
                Words = Words + 1
            Case ";"
                Words = Words + 1
            Case ":"
                Words = Words + 1
            Case "-"
                Words = Words + 1
        End Select
    Next cnt
    If Words = 0 Then Words = 1
    WordCount = Words
End Function

@mike60smart
I had always thought hyphenated words counted as 1?.
Hyphenated words in Word do count as 1?

[mike60smart]

Hi Welshgasman

The function used in a query produced the following results:-

[Welshgasman]

They are not hyphenated though?, there is a space each side of the "-" character?

Plus
Additional detail – WW ROHS shows up as 4 words in Word?

Additional detail – WW ROHS, 1.5x batt = 3600mAh shows up as 8 words ?

I cannot really see if your , in the last entry has a space after it, but would make it worse if it does not?

[mike60smart]

Yes The function appears to count the Hyphens and the Comma as a word so not 100% accurate.

I would imagine the Function needs tweaking slightly to ignore Hyphens & Commas. I would not have a clue as to how it can be tweeked as
my VB Skills are minimal.

[Welshgasman]

I was thinking of using the Split() function with a space?
If a word ends with a , it is still one word?, same for other characters.?

If for some reason multiple delimiters are needed then this link shows a way.
https://stackoverflow.com/questions/...s-in-vba-excel

Only the O/P knows what he needs, but I was going on what Word produces as a word count.

Thorough testing would be required of course?

[Micron]

Word must have its own definition. I just tested a sql statement in Word and the result was 48. IMO that is only about 50 words short of what I would count.

[orange]

I used Micron's response

Word must have its own definition. I just tested a sql statement in Word and the result was 48. IMO that is only about 50 words short of what I would count.

Just replace each space with space--> gives me 32 which seems to match the word count.
So counting spaces may be quick and dirty method. How exact must your count be?

[Micron]

@orange, what is the input for that comment? I don't understand. Surely it's not the record that's counted as 10?
What I've not seen yet is an input and expected result from OP and until we get that I'm going to sit on the sidelines.

WW
ROHS
1.5x
batt
3600mAh

IMO, that field is 5 words but the exact words are unknown because of the lack of info. If = is not a word, then which of these is the word: 1.5 or 1.5x since x is also a mathematical symbol, just like = ? Perhaps the count is more important so it doesn't matter, but I dare say there will be cases where lack of tight rules could be an issue. At best, I think an approximation is all that could be hoped for.

[orange]

micron,
I quoted your response post #10. And used that.
I copied the 2 sentences, pasted in Notepad++ and replaced space with space. 32 replacements.

Now if you have user inputted text, I would suggest you remove all double (or more) contiguous spaces to single space, then do the replace. Seems it could be a fairly simple function, but accuracy and specific instruction/requirement details needed before any further suggestions.

A function:

' ----------------------------------------------------------------
' Procedure Name: GetWordCount
' Purpose: Get a word count from a string based on number of spaces in the input
' Procedure Kind: Function
' Procedure Access: Public
' Parameter s (String): the string to be parsed for word count
' Return Type: Integer
' Author: Jack
' Date: 01-Jul-21
' ----------------------------------------------------------------
Function GetWordCount(s As String) As Integer

    Dim StringIn As Variant
    If Len(s) = 0 Then
        GetWordCount = 0
    Else
        StringIn = Split(s, " ")
        GetWordCount = UBound(StringIn) + 1 'must account for array element(0)
    End If
End Function

[jimk100189]

I was thinking of using the Split() function with a space?
If a word ends with a , it is still one word?, same for other characters.?

If for some reason multiple delimiters are needed then this link shows a way.
https://stackoverflow.com/questions/...s-in-vba-excel

Only the O/P knows what he needs, but I was going on what Word produces as a word count.

Thorough testing would be required of course?

YES : Yes, word count is what I'm trying to get. In MS Word REVIEW tab the WORD FOUNT function provides
Pages
Words
Characters (With spaces)
Characters (Without spaces)
Paragragraphs, and
Lines

[pbaldy]

Post 14 was moderated, I'm posting to trigger email notifications.