J West - revisit counting words in string?

The answer supplied by Jonathan West to me previous post is copied below:

>> I have used Len(strVariable) to get the length of my text string in
>> characters. Can I also get the length in "words", or get a count of how
>> many words are contained in that string?

>Yes, you can, but first you need to define what you mean by a "word". For
>instance is "apple-pie" one word or two?
>
>If you want to take the simplest case of words being separated by spaces,
>then the following line of code will do the trick
>
>numWords = UBound(Split(stringWithWords, " ")) + 1

Unfortunately, when used with
Selection.MoveRight Unit:=wdWord, Count:=numWords
I get the wrong answer, because Word stops at every punctuation mark along
the way as well as spaces. So I didn't know what I was asking then - but
know a bit more now. Is there a way to count the words in a string as Word
would using Ctrl+Right Arrow?

Ed

Jonathan
Wed Jan 07 14:55:42 CST 2004

I did say that it depends on what you mean by a word :-)

If you want to be sure that you are getting the same method as Word does,
then use Word's method. Insert the text into a document, select it, and get
the value of Selection.Words.Count.

--
Regards
Jonathan West - Word MVP
http://www.multilinker.com
Please reply to the newsgroup

"Ed" <ed_millis@removethis.hotmail.com> wrote in message
news:e8$aDtV1DHA.2636@TK2MSFTNGP09.phx.gbl...
> The answer supplied by Jonathan West to me previous post is copied below:
>
> >> I have used Len(strVariable) to get the length of my text string in
> >> characters. Can I also get the length in "words", or get a count of
how
> >> many words are contained in that string?
>
> >Yes, you can, but first you need to define what you mean by a "word". For
> >instance is "apple-pie" one word or two?
> >
> >If you want to take the simplest case of words being separated by spaces,
> >then the following line of code will do the trick
> >
> >numWords = UBound(Split(stringWithWords, " ")) + 1
>
> Unfortunately, when used with
> Selection.MoveRight Unit:=wdWord, Count:=numWords
> I get the wrong answer, because Word stops at every punctuation mark along
> the way as well as spaces. So I didn't know what I was asking then - but
> know a bit more now. Is there a way to count the words in a string as
Word
> would using Ctrl+Right Arrow?
>
> Ed
>
>

Klaus
Wed Jan 07 15:46:36 CST 2004

Hi Ed,

Would have been easier to remain in the old thread ;-)

> >If you want to take the simplest case of words being separated by spaces,
> >then the following line of code will do the trick
> >
> >numWords = UBound(Split(stringWithWords, " ")) + 1

> Unfortunately, when used with
> Selection.MoveRight Unit:=wdWord, Count:=numWords
> I get the wrong answer, because Word stops at every punctuation mark along
> the way as well as spaces. So I didn't know what I was asking then - but
> know a bit more now. Is there a way to count the words in a string as
Word
> would using Ctrl+Right Arrow?

If you (can) stay with the way Word counts words, you shouldn't run into
problems:

Dim numWords As Long
numWords = Selection.Words.Count
Selection.Collapse (wdCollapseStart)
Selection.MoveRight Unit:=wdWord, Count:=numWords

That would be a rather clumsy way to move the IP to the end of a
selection...

If you could post what you are trying to do, we might come up with more
specific tips.

Regards,
Klaus

Ed
Wed Jan 07 16:05:00 CST 2004

Sorry, I guess I should be more clear. I have an InputBox; the user's input
is set to a String variable. I then find that string in the document and
select it. I then have to extend the selection by a number of "words" (as
counted using Ctrl+Right Arrow), which is predetermined by another InputBox.
I figured the easiest way was to Find and Select the String, Collapse to the
start, then select the number of "words" in the string plus the other
amount. Am I doing this the hard way?

Ed

"Klaus Linke" <info@fotosatz-kaufmann.de.no.junk> wrote in message
news:es7ZAfW1DHA.2544@TK2MSFTNGP10.phx.gbl...
> Hi Ed,
>
> Would have been easier to remain in the old thread ;-)
>
> > >If you want to take the simplest case of words being separated by
spaces,
> > >then the following line of code will do the trick
> > >
> > >numWords = UBound(Split(stringWithWords, " ")) + 1
>
> > Unfortunately, when used with
> > Selection.MoveRight Unit:=wdWord, Count:=numWords
> > I get the wrong answer, because Word stops at every punctuation mark
along
> > the way as well as spaces. So I didn't know what I was asking then -
but
> > know a bit more now. Is there a way to count the words in a string as
> Word
> > would using Ctrl+Right Arrow?
>
> If you (can) stay with the way Word counts words, you shouldn't run into
> problems:
>
> Dim numWords As Long
> numWords = Selection.Words.Count
> Selection.Collapse (wdCollapseStart)
> Selection.MoveRight Unit:=wdWord, Count:=numWords
>
> That would be a rather clumsy way to move the IP to the end of a
> selection...
>
> If you could post what you are trying to do, we might come up with more
> specific tips.
>
> Regards,
> Klaus
>
>

sestyd
Wed Jan 07 16:05:43 CST 2004

JWest & Ed,

I though this was pretty interesting, and since I'm trying
to figure out all the methods (after actually finding the
documentation on the darned things), I wrote this sub
which prints out the count from the count method.

When I counted this paragraph by hand I got 58. The count
method counts 59. when I actually print the words out,
word 59 seems to be a carriage return or maybe it's empty,
not sure yet. Word behaves the same way whether I add this
text to a paragraph, section, or just to the document
content. Does anyone know why that is?

Anyway, try this sub (unwrap the lines). It probably will
work consistantly if you subtract 1 from the count.

Vera

Sub countWords()

Dim strText As String
Dim i As Integer

strText = "I get the wrong answer, because Word stops at
every punctuation mark along the way as well as spaces.
So I didn't know what I was asking then - but know a bit
more now. Is there a way to count the words in a string
as Word would using Ctrl+Right Arrow?"

With ActiveDocument
.Paragraphs.Add 'add the introductory paragraph
.Paragraphs(1).Range.Text = strText 'add text to it

intWords = .Paragraphs(1).Range.Words.Count

Debug.Print "Word count: " & intWords
For i = 1 To intWords
Debug.Print "word " & i & ": " & .Paragraphs
(1).Range.Words.Item(i).Text
Next

End With

End Sub



>-----Original Message-----
>I did say that it depends on what you mean by a word :-)
>
>If you want to be sure that you are getting the same
method as Word does,
>then use Word's method. Insert the text into a document,
select it, and get
>the value of Selection.Words.Count.
>
>--
>Regards
>Jonathan West - Word MVP
>http://www.multilinker.com
>Please reply to the newsgroup
>
>"Ed" <ed_millis@removethis.hotmail.com> wrote in message
>news:e8$aDtV1DHA.2636@TK2MSFTNGP09.phx.gbl...
>> The answer supplied by Jonathan West to me previous
post is copied below:
>>
>> >> I have used Len(strVariable) to get the length of my
text string in
>> >> characters. Can I also get the length in "words",
or get a count of
>how
>> >> many words are contained in that string?
>>
>> >Yes, you can, but first you need to define what you
mean by a "word". For
>> >instance is "apple-pie" one word or two?
>> >
>> >If you want to take the simplest case of words being
separated by spaces,
>> >then the following line of code will do the trick
>> >
>> >numWords = UBound(Split(stringWithWords, " ")) + 1
>>
>> Unfortunately, when used with
>> Selection.MoveRight Unit:=wdWord, Count:=numWords
>> I get the wrong answer, because Word stops at every
punctuation mark along
>> the way as well as spaces. So I didn't know what I was
asking then - but
>> know a bit more now. Is there a way to count the words
in a string as
>Word
>> would using Ctrl+Right Arrow?
>>
>> Ed
>>
>>
>
>.
>

Klaus
Wed Jan 07 16:37:17 CST 2004

"Ed" <ed_millis@removethis.hotmail.com> wrote:
> I have an InputBox; the user's input is set to a String variable.
> I then find that string in the document and select it.
> I then have to extend the selection by a number of "words" (as
> counted using Ctrl+Right Arrow), which is predetermined by
> another InputBox.
> I figured the easiest way was to Find and Select the String, Collapse to
the
> start, then select the number of "words" in the string plus the other
> amount. Am I doing this the hard way?

You wouldn't need to collapse the selection. Just move the end by the number
of words specified in your second InputBox:

Selection.MoveEnd Unit:=wdWord, Count:=numWords

But Ctrl+Right Arrow also counts punctuation marks as words.
If a user enters a number in the input box, he might not be aware of that.

If you *don't* want to count punctuation marks/paragraph marks... as words,
you would need to write a loop:
-- Do
-- move the end of the selection one "word",
-- get the last "word" (Selection.Words.Last.Text),
-- if the text of that "word" only contains white space and punctuation
marks, don't increment the counter,
-- else, increment the counter by one,
-- Loop until the counter is equal to numWords.

Regards,
Klaus

Ed
Wed Jan 07 16:41:00 CST 2004

Vera, that was great! I was hoping there was a method that would let me do
this "inside" VBA, without having to print the string into the document, but
I can sneak it in, get what I need, and delete it back out before anyone
notices. 8>)

(BTW, I have a book which (so it says) lists out every object, property, and
method in Word VBA. It's "Word 2000 Developer's Handbook" (ISBN
0-7821-2329-5). Halfpricecomputerbooks.com has one for 24.99. It's worked
okay for me so far - as long as I know enough to know what to look up!)

Ed

"sestyd@yahoo.com" <anonymous@discussions.microsoft.com> wrote in message
news:038101c3d56a$654360b0$a501280a@phx.gbl...
> JWest & Ed,
>
> I though this was pretty interesting, and since I'm trying
> to figure out all the methods (after actually finding the
> documentation on the darned things), I wrote this sub
> which prints out the count from the count method.
>
> When I counted this paragraph by hand I got 58. The count
> method counts 59. when I actually print the words out,
> word 59 seems to be a carriage return or maybe it's empty,
> not sure yet. Word behaves the same way whether I add this
> text to a paragraph, section, or just to the document
> content. Does anyone know why that is?
>
> Anyway, try this sub (unwrap the lines). It probably will
> work consistantly if you subtract 1 from the count.
>
> Vera
>
> Sub countWords()
>
> Dim strText As String
> Dim i As Integer
>
> strText = "I get the wrong answer, because Word stops at
> every punctuation mark along the way as well as spaces.
> So I didn't know what I was asking then - but know a bit
> more now. Is there a way to count the words in a string
> as Word would using Ctrl+Right Arrow?"
>
> With ActiveDocument
> .Paragraphs.Add 'add the introductory paragraph
> .Paragraphs(1).Range.Text = strText 'add text to it
>
> intWords = .Paragraphs(1).Range.Words.Count
>
> Debug.Print "Word count: " & intWords
> For i = 1 To intWords
> Debug.Print "word " & i & ": " & .Paragraphs
> (1).Range.Words.Item(i).Text
> Next
>
> End With
>
> End Sub
>
>
>
> >-----Original Message-----
> >I did say that it depends on what you mean by a word :-)
> >
> >If you want to be sure that you are getting the same
> method as Word does,
> >then use Word's method. Insert the text into a document,
> select it, and get
> >the value of Selection.Words.Count.
> >
> >--
> >Regards
> >Jonathan West - Word MVP
> >http://www.multilinker.com
> >Please reply to the newsgroup
> >
> >"Ed" <ed_millis@removethis.hotmail.com> wrote in message
> >news:e8$aDtV1DHA.2636@TK2MSFTNGP09.phx.gbl...
> >> The answer supplied by Jonathan West to me previous
> post is copied below:
> >>
> >> >> I have used Len(strVariable) to get the length of my
> text string in
> >> >> characters. Can I also get the length in "words",
> or get a count of
> >how
> >> >> many words are contained in that string?
> >>
> >> >Yes, you can, but first you need to define what you
> mean by a "word". For
> >> >instance is "apple-pie" one word or two?
> >> >
> >> >If you want to take the simplest case of words being
> separated by spaces,
> >> >then the following line of code will do the trick
> >> >
> >> >numWords = UBound(Split(stringWithWords, " ")) + 1
> >>
> >> Unfortunately, when used with
> >> Selection.MoveRight Unit:=wdWord, Count:=numWords
> >> I get the wrong answer, because Word stops at every
> punctuation mark along
> >> the way as well as spaces. So I didn't know what I was
> asking then - but
> >> know a bit more now. Is there a way to count the words
> in a string as
> >Word
> >> would using Ctrl+Right Arrow?
> >>
> >> Ed
> >>
> >>
> >
> >.
> >

anonymous
Wed Jan 07 17:00:48 CST 2004

Ed,

Actually, you should just be able to pass your string to
the sub in a string variable.

sub countWords(ByVal strText as string)
or better yet

function countWords(ByVal strText as string) as integer
code, code, code



end function


so I think you can probably pass the selection in, not
sure about that

if you use the function then it will return the count to
you




>-----Original Message-----
>Vera, that was great! I was hoping there was a method
that would let me do
>this "inside" VBA, without having to print the string
into the document, but
>I can sneak it in, get what I need, and delete it back
out before anyone
>notices. 8>)
>
>(BTW, I have a book which (so it says) lists out every
object, property, and
>method in Word VBA. It's "Word 2000 Developer's
Handbook" (ISBN
>0-7821-2329-5). Halfpricecomputerbooks.com has one for
24.99. It's worked
>okay for me so far - as long as I know enough to know
what to look up!)
>
>Ed
>
>"sestyd@yahoo.com" <anonymous@discussions.microsoft.com>
wrote in message
>news:038101c3d56a$654360b0$a501280a@phx.gbl...
>> JWest & Ed,
>>
>> I though this was pretty interesting, and since I'm
trying
>> to figure out all the methods (after actually finding
the
>> documentation on the darned things), I wrote this sub
>> which prints out the count from the count method.
>>
>> When I counted this paragraph by hand I got 58. The
count
>> method counts 59. when I actually print the words out,
>> word 59 seems to be a carriage return or maybe it's
empty,
>> not sure yet. Word behaves the same way whether I add
this
>> text to a paragraph, section, or just to the document
>> content. Does anyone know why that is?
>>
>> Anyway, try this sub (unwrap the lines). It probably
will
>> work consistantly if you subtract 1 from the count.
>>
>> Vera
>>
>> Sub countWords()
>>
>> Dim strText As String
>> Dim i As Integer
>>
>> strText = "I get the wrong answer, because Word stops at
>> every punctuation mark along the way as well as spaces.
>> So I didn't know what I was asking then - but know a bit
>> more now. Is there a way to count the words in a string
>> as Word would using Ctrl+Right Arrow?"
>>
>> With ActiveDocument
>> .Paragraphs.Add 'add the introductory paragraph
>> .Paragraphs(1).Range.Text = strText 'add text
to it
>>
>> intWords = .Paragraphs(1).Range.Words.Count
>>
>> Debug.Print "Word count: " & intWords
>> For i = 1 To intWords
>> Debug.Print "word " & i & ": " & .Paragraphs
>> (1).Range.Words.Item(i).Text
>> Next
>>
>> End With
>>
>> End Sub
>>
>>
>>
>> >-----Original Message-----
>> >I did say that it depends on what you mean by a word :-
)
>> >
>> >If you want to be sure that you are getting the same
>> method as Word does,
>> >then use Word's method. Insert the text into a
document,
>> select it, and get
>> >the value of Selection.Words.Count.
>> >
>> >--
>> >Regards
>> >Jonathan West - Word MVP
>> >http://www.multilinker.com
>> >Please reply to the newsgroup
>> >
>> >"Ed" <ed_millis@removethis.hotmail.com> wrote in
message
>> >news:e8$aDtV1DHA.2636@TK2MSFTNGP09.phx.gbl...
>> >> The answer supplied by Jonathan West to me previous
>> post is copied below:
>> >>
>> >> >> I have used Len(strVariable) to get the length of
my
>> text string in
>> >> >> characters. Can I also get the length in "words",
>> or get a count of
>> >how
>> >> >> many words are contained in that string?
>> >>
>> >> >Yes, you can, but first you need to define what you
>> mean by a "word". For
>> >> >instance is "apple-pie" one word or two?
>> >> >
>> >> >If you want to take the simplest case of words being
>> separated by spaces,
>> >> >then the following line of code will do the trick
>> >> >
>> >> >numWords = UBound(Split(stringWithWords, " ")) + 1
>> >>
>> >> Unfortunately, when used with
>> >> Selection.MoveRight Unit:=wdWord, Count:=numWords
>> >> I get the wrong answer, because Word stops at every
>> punctuation mark along
>> >> the way as well as spaces. So I didn't know what I
was
>> asking then - but
>> >> know a bit more now. Is there a way to count the
words
>> in a string as
>> >Word
>> >> would using Ctrl+Right Arrow?
>> >>
>> >> Ed
>> >>
>> >>
>> >
>> >.
>> >
>
>
>.
>

Ed
Wed Jan 07 17:08:24 CST 2004

Thanks, Klaus. I think I've got enough to play with for now - almost more
than I can juggle without dropping! I'll work this through and yell for
help if need be. I appreciate all your input.

Ed

"Klaus Linke" <info@fotosatz-kaufmann.de.no.junk> wrote in message
news:uA3Lh7W1DHA.2388@TK2MSFTNGP09.phx.gbl...
> "Ed" <ed_millis@removethis.hotmail.com> wrote:
> > I have an InputBox; the user's input is set to a String variable.
> > I then find that string in the document and select it.
> > I then have to extend the selection by a number of "words" (as
> > counted using Ctrl+Right Arrow), which is predetermined by
> > another InputBox.
> > I figured the easiest way was to Find and Select the String, Collapse to
> the
> > start, then select the number of "words" in the string plus the other
> > amount. Am I doing this the hard way?
>
> You wouldn't need to collapse the selection. Just move the end by the
number
> of words specified in your second InputBox:
>
> Selection.MoveEnd Unit:=wdWord, Count:=numWords
>
> But Ctrl+Right Arrow also counts punctuation marks as words.
> If a user enters a number in the input box, he might not be aware of that.
>
> If you *don't* want to count punctuation marks/paragraph marks... as
words,
> you would need to write a loop:
> -- Do
> -- move the end of the selection one "word",
> -- get the last "word" (Selection.Words.Last.Text),
> -- if the text of that "word" only contains white space and punctuation
> marks, don't increment the counter,
> -- else, increment the counter by one,
> -- Loop until the counter is equal to numWords.
>
> Regards,
> Klaus
>
>

vera
Wed Jan 07 17:08:45 CST 2004

Ed,

Whoops! Hit the wrong button!
I'll work out the function if you get stuck. Thanks for
the book tip!

Vera


>-----Original Message-----
>Ed,
>
>Actually, you should just be able to pass your string to
>the sub in a string variable.
>
>sub countWords(ByVal strText as string)
>or better yet
>
>function countWords(ByVal strText as string) as integer
> code, code, code
>
>
>
>end function
>
>
>so I think you can probably pass the selection in, not
>sure about that
>
>if you use the function then it will return the count to
>you
>
>
>
>
>>-----Original Message-----
>>Vera, that was great! I was hoping there was a method
>that would let me do
>>this "inside" VBA, without having to print the string
>into the document, but
>>I can sneak it in, get what I need, and delete it back
>out before anyone
>>notices. 8>)
>>
>>(BTW, I have a book which (so it says) lists out every
>object, property, and
>>method in Word VBA. It's "Word 2000 Developer's
>Handbook" (ISBN
>>0-7821-2329-5). Halfpricecomputerbooks.com has one for
>24.99. It's worked
>>okay for me so far - as long as I know enough to know
>what to look up!)
>>
>>Ed
>>
>>"sestyd@yahoo.com" <anonymous@discussions.microsoft.com>
>wrote in message
>>news:038101c3d56a$654360b0$a501280a@phx.gbl...
>>> JWest & Ed,
>>>
>>> I though this was pretty interesting, and since I'm
>trying
>>> to figure out all the methods (after actually finding
>the
>>> documentation on the darned things), I wrote this sub
>>> which prints out the count from the count method.
>>>
>>> When I counted this paragraph by hand I got 58. The
>count
>>> method counts 59. when I actually print the words out,
>>> word 59 seems to be a carriage return or maybe it's
>empty,
>>> not sure yet. Word behaves the same way whether I add
>this
>>> text to a paragraph, section, or just to the document
>>> content. Does anyone know why that is?
>>>
>>> Anyway, try this sub (unwrap the lines). It probably
>will
>>> work consistantly if you subtract 1 from the count.
>>>
>>> Vera
>>>
>>> Sub countWords()
>>>
>>> Dim strText As String
>>> Dim i As Integer
>>>
>>> strText = "I get the wrong answer, because Word stops
at
>>> every punctuation mark along the way as well as spaces.
>>> So I didn't know what I was asking then - but know a
bit
>>> more now. Is there a way to count the words in a
string
>>> as Word would using Ctrl+Right Arrow?"
>>>
>>> With ActiveDocument
>>> .Paragraphs.Add 'add the introductory paragraph
>>> .Paragraphs(1).Range.Text = strText 'add text
>to it
>>>
>>> intWords = .Paragraphs(1).Range.Words.Count
>>>
>>> Debug.Print "Word count: " & intWords
>>> For i = 1 To intWords
>>> Debug.Print "word " & i & ": " & .Paragraphs
>>> (1).Range.Words.Item(i).Text
>>> Next
>>>
>>> End With
>>>
>>> End Sub
>>>
>>>
>>>
>>> >-----Original Message-----
>>> >I did say that it depends on what you mean by a
word :-
>)
>>> >
>>> >If you want to be sure that you are getting the same
>>> method as Word does,
>>> >then use Word's method. Insert the text into a
>document,
>>> select it, and get
>>> >the value of Selection.Words.Count.
>>> >
>>> >--
>>> >Regards
>>> >Jonathan West - Word MVP
>>> >http://www.multilinker.com
>>> >Please reply to the newsgroup
>>> >
>>> >"Ed" <ed_millis@removethis.hotmail.com> wrote in
>message
>>> >news:e8$aDtV1DHA.2636@TK2MSFTNGP09.phx.gbl...
>>> >> The answer supplied by Jonathan West to me previous
>>> post is copied below:
>>> >>
>>> >> >> I have used Len(strVariable) to get the length
of
>my
>>> text string in
>>> >> >> characters. Can I also get the length
in "words",
>>> or get a count of
>>> >how
>>> >> >> many words are contained in that string?
>>> >>
>>> >> >Yes, you can, but first you need to define what you
>>> mean by a "word". For
>>> >> >instance is "apple-pie" one word or two?
>>> >> >
>>> >> >If you want to take the simplest case of words
being
>>> separated by spaces,
>>> >> >then the following line of code will do the trick
>>> >> >
>>> >> >numWords = UBound(Split(stringWithWords, " ")) + 1
>>> >>
>>> >> Unfortunately, when used with
>>> >> Selection.MoveRight Unit:=wdWord, Count:=numWords
>>> >> I get the wrong answer, because Word stops at every
>>> punctuation mark along
>>> >> the way as well as spaces. So I didn't know what I
>was
>>> asking then - but
>>> >> know a bit more now. Is there a way to count the
>words
>>> in a string as
>>> >Word
>>> >> would using Ctrl+Right Arrow?
>>> >>
>>> >> Ed
>>> >>
>>> >>
>>> >
>>> >.
>>> >
>>
>>
>>.
>>
>.
>

vera
Wed Jan 07 17:25:25 CST 2004

Ed,

Oh, heck, I just had to try something else. Try this by
creating a new word doc, putting some text on it,
selecting the text, then run the sub testWordCount in your
immediate window. And don't forget to unwrap the lines.

Vera


Function countWords() As Integer

Dim i As Integer

intWords = selection.Words.Count - 1

Debug.Print "Word count: " & intWords
For i = 1 To intWords
Debug.Print "word " & i & ": " &
selection.Words.Item(i).Text
Next

countWords = intWords

End Function

Sub testCountWords()

Dim intWordCount As Integer
intWordCount = countWords()
Debug.Print "Here is the word count: " & intWordCount

End Sub


>-----Original Message-----
>Ed,
>
>Whoops! Hit the wrong button!
>I'll work out the function if you get stuck. Thanks for
>the book tip!
>
>Vera
>
>
>>-----Original Message-----
>>Ed,
>>
>>Actually, you should just be able to pass your string to
>>the sub in a string variable.
>>
>>sub countWords(ByVal strText as string)
>>or better yet
>>
>>function countWords(ByVal strText as string) as integer
>> code, code, code
>>
>>
>>
>>end function
>>
>>
>>so I think you can probably pass the selection in, not
>>sure about that
>>
>>if you use the function then it will return the count to
>>you
>>
>>
>>
>>
>>>-----Original Message-----
>>>Vera, that was great! I was hoping there was a method
>>that would let me do
>>>this "inside" VBA, without having to print the string
>>into the document, but
>>>I can sneak it in, get what I need, and delete it back
>>out before anyone
>>>notices. 8>)
>>>
>>>(BTW, I have a book which (so it says) lists out every
>>object, property, and
>>>method in Word VBA. It's "Word 2000 Developer's
>>Handbook" (ISBN
>>>0-7821-2329-5). Halfpricecomputerbooks.com has one for
>>24.99. It's worked
>>>okay for me so far - as long as I know enough to know
>>what to look up!)
>>>
>>>Ed
>>>
>>>"sestyd@yahoo.com"
<anonymous@discussions.microsoft.com>
>>wrote in message
>>>news:038101c3d56a$654360b0$a501280a@phx.gbl...
>>>> JWest & Ed,
>>>>
>>>> I though this was pretty interesting, and since I'm
>>trying
>>>> to figure out all the methods (after actually finding
>>the
>>>> documentation on the darned things), I wrote this sub
>>>> which prints out the count from the count method.
>>>>
>>>> When I counted this paragraph by hand I got 58. The
>>count
>>>> method counts 59. when I actually print the words out,
>>>> word 59 seems to be a carriage return or maybe it's
>>empty,
>>>> not sure yet. Word behaves the same way whether I add
>>this
>>>> text to a paragraph, section, or just to the document
>>>> content. Does anyone know why that is?
>>>>
>>>> Anyway, try this sub (unwrap the lines). It probably
>>will
>>>> work consistantly if you subtract 1 from the count.
>>>>
>>>> Vera
>>>>
>>>> Sub countWords()
>>>>
>>>> Dim strText As String
>>>> Dim i As Integer
>>>>
>>>> strText = "I get the wrong answer, because Word stops
>at
>>>> every punctuation mark along the way as well as
spaces.
>>>> So I didn't know what I was asking then - but know a
>bit
>>>> more now. Is there a way to count the words in a
>string
>>>> as Word would using Ctrl+Right Arrow?"
>>>>
>>>> With ActiveDocument
>>>> .Paragraphs.Add 'add the introductory
paragraph
>>>> .Paragraphs(1).Range.Text = strText 'add text
>>to it
>>>>
>>>> intWords = .Paragraphs(1).Range.Words.Count
>>>>
>>>> Debug.Print "Word count: " & intWords
>>>> For i = 1 To intWords
>>>> Debug.Print "word " & i & ": " & .Paragraphs
>>>> (1).Range.Words.Item(i).Text
>>>> Next
>>>>
>>>> End With
>>>>
>>>> End Sub
>>>>
>>>>
>>>>
>>>> >-----Original Message-----
>>>> >I did say that it depends on what you mean by a
>word :-
>>)
>>>> >
>>>> >If you want to be sure that you are getting the same
>>>> method as Word does,
>>>> >then use Word's method. Insert the text into a
>>document,
>>>> select it, and get
>>>> >the value of Selection.Words.Count.
>>>> >
>>>> >--
>>>> >Regards
>>>> >Jonathan West - Word MVP
>>>> >http://www.multilinker.com
>>>> >Please reply to the newsgroup
>>>> >
>>>> >"Ed" <ed_millis@removethis.hotmail.com> wrote in
>>message
>>>> >news:e8$aDtV1DHA.2636@TK2MSFTNGP09.phx.gbl...
>>>> >> The answer supplied by Jonathan West to me previous
>>>> post is copied below:
>>>> >>
>>>> >> >> I have used Len(strVariable) to get the length
>of
>>my
>>>> text string in
>>>> >> >> characters. Can I also get the length
>in "words",
>>>> or get a count of
>>>> >how
>>>> >> >> many words are contained in that string?
>>>> >>
>>>> >> >Yes, you can, but first you need to define what
you
>>>> mean by a "word". For
>>>> >> >instance is "apple-pie" one word or two?
>>>> >> >
>>>> >> >If you want to take the simplest case of words
>being
>>>> separated by spaces,
>>>> >> >then the following line of code will do the trick
>>>> >> >
>>>> >> >numWords = UBound(Split(stringWithWords, " ")) + 1
>>>> >>
>>>> >> Unfortunately, when used with
>>>> >> Selection.MoveRight Unit:=wdWord, Count:=numWords
>>>> >> I get the wrong answer, because Word stops at every
>>>> punctuation mark along
>>>> >> the way as well as spaces. So I didn't know what
I
>>was
>>>> asking then - but
>>>> >> know a bit more now. Is there a way to count the
>>words
>>>> in a string as
>>>> >Word
>>>> >> would using Ctrl+Right Arrow?
>>>> >>
>>>> >> Ed
>>>> >>
>>>> >>
>>>> >
>>>> >.
>>>> >
>>>
>>>
>>>.
>>>
>>.
>>
>.
>

Jay
Wed Jan 07 21:29:51 CST 2004

Hi, Ed,

As a completely different alternative, have a look at the
Selection.Extend method.

"Ed" <ed_millis@removethis.hotmail.com> wrote:

>Sorry, I guess I should be more clear. I have an InputBox; the user's input
>is set to a String variable. I then find that string in the document and
>select it. I then have to extend the selection by a number of "words" (as
>counted using Ctrl+Right Arrow), which is predetermined by another InputBox.
>I figured the easiest way was to Find and Select the String, Collapse to the
>start, then select the number of "words" in the string plus the other
>amount. Am I doing this the hard way?
>
>Ed
>
>"Klaus Linke" <info@fotosatz-kaufmann.de.no.junk> wrote in message
>news:es7ZAfW1DHA.2544@TK2MSFTNGP10.phx.gbl...
>> Hi Ed,
>>
>> Would have been easier to remain in the old thread ;-)
>>
>> > >If you want to take the simplest case of words being separated by
>spaces,
>> > >then the following line of code will do the trick
>> > >
>> > >numWords = UBound(Split(stringWithWords, " ")) + 1
>>
>> > Unfortunately, when used with
>> > Selection.MoveRight Unit:=wdWord, Count:=numWords
>> > I get the wrong answer, because Word stops at every punctuation mark
>along
>> > the way as well as spaces. So I didn't know what I was asking then -
>but
>> > know a bit more now. Is there a way to count the words in a string as
>> Word
>> > would using Ctrl+Right Arrow?
>>
>> If you (can) stay with the way Word counts words, you shouldn't run into
>> problems:
>>
>> Dim numWords As Long
>> numWords = Selection.Words.Count
>> Selection.Collapse (wdCollapseStart)
>> Selection.MoveRight Unit:=wdWord, Count:=numWords
>>
>> That would be a rather clumsy way to move the IP to the end of a
>> selection...
>>
>> If you could post what you are trying to do, we might come up with more
>> specific tips.
>>
>> Regards,
>> Klaus



--
Regards,
Jay Freedman
Microsoft Word MVP FAQ: http://www.mvps.org/word

EmpyreanEmpathy
Thu Jan 08 01:16:06 CST 2004

Hi, gang

This is a homely old sleighdog of a macro but I'd be curious to see
how else an accurate word count could be arrived at for the following
sentence:

"One + two - three = four," five, 6, seven; "eight nine ten!!!"
Eleven twelve>>thirteen<< fourteen? '"` ~15||| . . .
'sixteen'\.^seventeen``= (eighteen@nineteen%):::::


$twenty???

Code:

(Place cursor before first word.)

Dim R As Range, WordCnt As Integer
Set R = Selection.Range
Do
R.MoveEndWhile vbCr & vbTab & Chr(160) & "
.,;:'()[]!@#$%^&*-_+-={}~`\/?<>|"""""
R.MoveEnd wdWord
R.MoveEndWhile vbCr & vbTab & Chr(160) & "
.,;:'()[]!@#$%^&*-_+={}~`\/?<>|"""""
WordCnt = WordCnt + 1
R.Select
MsgBox WordCnt, , "WordCnt"
Loop Until WordCnt = 20
Selection.Collapse

P.S. Won't work with "smart" quotes or em dashes.

- Bruce

Jay Freedman <jay.freedman@verizon.net> wrote in message news:<1ijpvvoe1co7bsunfgg18vsdm3rskd7s6k@4ax.com>...
> Hi, Ed,
>
> As a completely different alternative, have a look at the
> Selection.Extend method.
>
> "Ed" <ed_millis@removethis.hotmail.com> wrote:
>
> >Sorry, I guess I should be more clear. I have an InputBox; the user's input
> >is set to a String variable. I then find that string in the document and
> >select it. I then have to extend the selection by a number of "words" (as
> >counted using Ctrl+Right Arrow), which is predetermined by another InputBox.
> >I figured the easiest way was to Find and Select the String, Collapse to the
> >start, then select the number of "words" in the string plus the other
> >amount. Am I doing this the hard way?
> >
> >Ed
> >
> >"Klaus Linke" <info@fotosatz-kaufmann.de.no.junk> wrote in message
> >news:es7ZAfW1DHA.2544@TK2MSFTNGP10.phx.gbl...
> >> Hi Ed,
> >>
> >> Would have been easier to remain in the old thread ;-)
> >>
> >> > >If you want to take the simplest case of words being separated by
> spaces,
> >> > >then the following line of code will do the trick
> >> > >
> >> > >numWords = UBound(Split(stringWithWords, " ")) + 1
>
> >> > Unfortunately, when used with
> >> > Selection.MoveRight Unit:=wdWord, Count:=numWords
> >> > I get the wrong answer, because Word stops at every punctuation mark
> along
> >> > the way as well as spaces. So I didn't know what I was asking then -
> but
> >> > know a bit more now. Is there a way to count the words in a string as
> Word
> >> > would using Ctrl+Right Arrow?
> >>
> >> If you (can) stay with the way Word counts words, you shouldn't run into
> >> problems:
> >>
> >> Dim numWords As Long
> >> numWords = Selection.Words.Count
> >> Selection.Collapse (wdCollapseStart)
> >> Selection.MoveRight Unit:=wdWord, Count:=numWords
> >>
> >> That would be a rather clumsy way to move the IP to the end of a
> >> selection...
> >>
> >> If you could post what you are trying to do, we might come up with more
> >> specific tips.
> >>
> >> Regards,
> >> Klaus

Ed
Thu Jan 08 07:50:52 CST 2004

That worked "Vera" good! 8>) Thanks for all your help.

Ed

"vera" <sestyd@yahoo.com> wrote in message
news:048301c3d575$8776ddf0$a401280a@phx.gbl...
> Ed,
>
> Oh, heck, I just had to try something else. Try this by
> creating a new word doc, putting some text on it,
> selecting the text, then run the sub testWordCount in your
> immediate window. And don't forget to unwrap the lines.
>
> Vera
>
>
> Function countWords() As Integer
>
> Dim i As Integer
>
> intWords = selection.Words.Count - 1
>
> Debug.Print "Word count: " & intWords
> For i = 1 To intWords
> Debug.Print "word " & i & ": " &
> selection.Words.Item(i).Text
> Next
>
> countWords = intWords
>
> End Function
>
> Sub testCountWords()
>
> Dim intWordCount As Integer
> intWordCount = countWords()
> Debug.Print "Here is the word count: " & intWordCount
>
> End Sub
>
>
> >-----Original Message-----
> >Ed,
> >
> >Whoops! Hit the wrong button!
> >I'll work out the function if you get stuck. Thanks for
> >the book tip!
> >
> >Vera
> >
> >
> >>-----Original Message-----
> >>Ed,
> >>
> >>Actually, you should just be able to pass your string to
> >>the sub in a string variable.
> >>
> >>sub countWords(ByVal strText as string)
> >>or better yet
> >>
> >>function countWords(ByVal strText as string) as integer
> >> code, code, code
> >>
> >>
> >>
> >>end function
> >>
> >>
> >>so I think you can probably pass the selection in, not
> >>sure about that
> >>
> >>if you use the function then it will return the count to
> >>you
> >>
> >>
> >>
> >>
> >>>-----Original Message-----
> >>>Vera, that was great! I was hoping there was a method
> >>that would let me do
> >>>this "inside" VBA, without having to print the string
> >>into the document, but
> >>>I can sneak it in, get what I need, and delete it back
> >>out before anyone
> >>>notices. 8>)
> >>>
> >>>(BTW, I have a book which (so it says) lists out every
> >>object, property, and
> >>>method in Word VBA. It's "Word 2000 Developer's
> >>Handbook" (ISBN
> >>>0-7821-2329-5). Halfpricecomputerbooks.com has one for
> >>24.99. It's worked
> >>>okay for me so far - as long as I know enough to know
> >>what to look up!)
> >>>
> >>>Ed
> >>>
> >>>"sestyd@yahoo.com"
> <anonymous@discussions.microsoft.com>
> >>wrote in message
> >>>news:038101c3d56a$654360b0$a501280a@phx.gbl...
> >>>> JWest & Ed,
> >>>>
> >>>> I though this was pretty interesting, and since I'm
> >>trying
> >>>> to figure out all the methods (after actually finding
> >>the
> >>>> documentation on the darned things), I wrote this sub
> >>>> which prints out the count from the count method.
> >>>>
> >>>> When I counted this paragraph by hand I got 58. The
> >>count
> >>>> method counts 59. when I actually print the words out,
> >>>> word 59 seems to be a carriage return or maybe it's
> >>empty,
> >>>> not sure yet. Word behaves the same way whether I add
> >>this
> >>>> text to a paragraph, section, or just to the document
> >>>> content. Does anyone know why that is?
> >>>>
> >>>> Anyway, try this sub (unwrap the lines). It probably
> >>will
> >>>> work consistantly if you subtract 1 from the count.
> >>>>
> >>>> Vera
> >>>>
> >>>> Sub countWords()
> >>>>
> >>>> Dim strText As String
> >>>> Dim i As Integer
> >>>>
> >>>> strText = "I get the wrong answer, because Word stops
> >at
> >>>> every punctuation mark along the way as well as
> spaces.
> >>>> So I didn't know what I was asking then - but know a
> >bit
> >>>> more now. Is there a way to count the words in a
> >string
> >>>> as Word would using Ctrl+Right Arrow?"
> >>>>
> >>>> With ActiveDocument
> >>>> .Paragraphs.Add 'add the introductory
> paragraph
> >>>> .Paragraphs(1).Range.Text = strText 'add text
> >>to it
> >>>>
> >>>> intWords = .Paragraphs(1).Range.Words.Count
> >>>>
> >>>> Debug.Print "Word count: " & intWords
> >>>> For i = 1 To intWords
> >>>> Debug.Print "word " & i & ": " & .Paragraphs
> >>>> (1).Range.Words.Item(i).Text
> >>>> Next
> >>>>
> >>>> End With
> >>>>
> >>>> End Sub
> >>>>
> >>>>
> >>>>
> >>>> >-----Original Message-----
> >>>> >I did say that it depends on what you mean by a
> >word :-
> >>)
> >>>> >
> >>>> >If you want to be sure that you are getting the same
> >>>> method as Word does,
> >>>> >then use Word's method. Insert the text into a
> >>document,
> >>>> select it, and get
> >>>> >the value of Selection.Words.Count.
> >>>> >
> >>>> >--
> >>>> >Regards
> >>>> >Jonathan West - Word MVP
> >>>> >http://www.multilinker.com
> >>>> >Please reply to the newsgroup
> >>>> >
> >>>> >"Ed" <ed_millis@removethis.hotmail.com> wrote in
> >>message
> >>>> >news:e8$aDtV1DHA.2636@TK2MSFTNGP09.phx.gbl...
> >>>> >> The answer supplied by Jonathan West to me previous
> >>>> post is copied below:
> >>>> >>
> >>>> >> >> I have used Len(strVariable) to get the length
> >of
> >>my
> >>>> text string in
> >>>> >> >> characters. Can I also get the length
> >in "words",
> >>>> or get a count of
> >>>> >how
> >>>> >> >> many words are contained in that string?
> >>>> >>
> >>>> >> >Yes, you can, but first you need to define what
> you
> >>>> mean by a "word". For
> >>>> >> >instance is "apple-pie" one word or two?
> >>>> >> >
> >>>> >> >If you want to take the simplest case of words
> >being
> >>>> separated by spaces,
> >>>> >> >then the following line of code will do the trick
> >>>> >> >
> >>>> >> >numWords = UBound(Split(stringWithWords, " ")) + 1
> >>>> >>
> >>>> >> Unfortunately, when used with
> >>>> >> Selection.MoveRight Unit:=wdWord, Count:=numWords
> >>>> >> I get the wrong answer, because Word stops at every
> >>>> punctuation mark along
> >>>> >> the way as well as spaces. So I didn't know what
> I
> >>was
> >>>> asking then - but
> >>>> >> know a bit more now. Is there a way to count the
> >>words
> >>>> in a string as
> >>>> >Word
> >>>> >> would using Ctrl+Right Arrow?
> >>>> >>
> >>>> >> Ed
> >>>> >>
> >>>> >>
> >>>> >
> >>>> >.
> >>>> >
> >>>
> >>>
> >>>.
> >>>
> >>.
> >>
> >.
> >

EmpyreanEmpathy
Thu Jan 08 10:37:09 CST 2004

Here are three more word-counting sleighdogs, slightly more refined
but still slobbering. The first one's a little more fun than the
others. This time they skip "smart" quotes too.

The third one, count-all-words-in-doc, took about 5 seconds to find
over 26,000 words and about 10 seconds to find over 45,000 words.

Add characters to skip, as needed, within the MoveEndWhile statement.


'========================================================================
Sub CountUserSpecifiedNumberOfWords()
'COUNT USER-SPECIFIED NUMBER OF WORDS, STARTING AT INSERTION POINT
'SKIP ANYTHING OTHER THAN WORDS OR NUMBERS
If Selection.Type <> wdSelectionIP Then
MsgBox "Quitting.", , "Text Can't Be Selected"
Exit Sub
End If
Dim R As Range, X As Range, WordCnt As Long, UserNumber As Long
UserNumber = InputBox("Number of words:", "Number of Words to Count",
10)
Set R = Selection.Range
Do
R.MoveEndWhile vbCr & vbTab & Chr(160) & Chr(147) & Chr(148) & "
.,;:'()[]!@#$%^&*-_+-={}~`\/?<>|"""""
Set X = R.Duplicate
X.Start = X.End
X.MoveEnd wdWord
X.Select
WordCnt = WordCnt + 1
MsgBox X.Text, , WordCnt
R.MoveEnd wdWord
R.MoveEndWhile vbCr & vbTab & Chr(160) & Chr(147) & Chr(148) & "
.,;:'()[]!@#$%^&*-_+-={}~`\/?<>|"""""
Loop Until WordCnt = UserNumber
Selection.Collapse
End Sub


'========================================================================
Sub CountWordsInSelectedText()
'COUNT WORDS IN BLOCK OF SELECTED TEXT
'SKIP ANYTHING OTHER THAN WORDS OR NUMBERS
If Selection.Type = wdSelectionIP Then
MsgBox "Quitting.", , "Nothing Selected"
Exit Sub
End If
Dim R As Range, WordCnt As Long
Set R = Selection.Range.Duplicate
R.End = R.Start
Do While R.End < Selection.Range.End
R.MoveEndWhile vbCr & vbTab & Chr(160) & Chr(147) & Chr(148) & "
.,;:'()[]!@#$%^&*-_+-={}~`\/?<>|"""""
R.MoveEnd wdWord
WordCnt = WordCnt + 1
R.MoveEndWhile vbCr & vbTab & Chr(160) & Chr(147) & Chr(148) & "
.,;:'()[]!@#$%^&*-_+-={}~`\/?<>|"""""
Loop
MsgBox WordCnt
End Sub


'========================================================================
Sub CountAllWordsInWholeDoc()
'COUNT ALL WORDS IN MAIN STORY RANGE OF DOCUMENT
'SKIP ANYTHING OTHER THAN WORDS OR NUMBERS
Dim R As Range, X As Range, WordCnt As Long
Set X = ActiveDocument.Range.Duplicate
X.End = X.End - 1
Set R = Selection.Range
Do While R.End < X.End
R.MoveEndWhile vbCr & vbTab & Chr(160) & Chr(147) & Chr(148) & "
.,;:'()[]!@#$%^&*-_+-={}~`\/?<>|"""""
R.MoveEnd wdWord
WordCnt = WordCnt + 1
R.MoveEndWhile vbCr & vbTab & Chr(160) & Chr(147) & Chr(148) & "
.,;:'()[]!@#$%^&*-_+-={}~`\/?<>|"""""
Loop
MsgBox WordCnt, , "WordCnt for Whole Doc"
End Sub


EmpyreanEmpathy@aol.com (Bruce Brown) wrote in message news:<1d2f086c.0401072316.5c112583@posting.google.com>...
> Hi, gang
>
> This is a homely old sleighdog of a macro but I'd be curious to see
> how else an accurate word count could be arrived at for the following
> sentence:
>
> "One + two - three = four," five, 6, seven; "eight nine ten!!!"
> Eleven twelve>>thirteen<< fourteen? '"` ~15||| . . .
> 'sixteen'\.^seventeen``= (eighteen@nineteen%):::::
>
>
> $twenty???
>
> Code:
>
> (Place cursor before first word.)
>
> Dim R As Range, WordCnt As Integer
> Set R = Selection.Range
> Do
> R.MoveEndWhile vbCr & vbTab & Chr(160) & "
> .,;:'()[]!@#$%^&*-_+-={}~`\/?<>|"""""
> R.MoveEnd wdWord
> R.MoveEndWhile vbCr & vbTab & Chr(160) & "
> .,;:'()[]!@#$%^&*-_+={}~`\/?<>|"""""
> WordCnt = WordCnt + 1
> R.Select
> MsgBox WordCnt, , "WordCnt"
> Loop Until WordCnt = 20
> Selection.Collapse
>
> P.S. Won't work with "smart" quotes or em dashes.
>
> - Bruce
>
> Jay Freedman <jay.freedman@verizon.net> wrote in message news:<1ijpvvoe1co7bsunfgg18vsdm3rskd7s6k@4ax.com>...
> > Hi, Ed,
> >
> > As a completely different alternative, have a look at the
> > Selection.Extend method.
> >
> > "Ed" <ed_millis@removethis.hotmail.com> wrote:
> >
> > >Sorry, I guess I should be more clear. I have an InputBox; the user's input
> > >is set to a String variable. I then find that string in the document and
> > >select it. I then have to extend the selection by a number of "words" (as
> > >counted using Ctrl+Right Arrow), which is predetermined by another InputBox.
> > >I figured the easiest way was to Find and Select the String, Collapse to the
> > >start, then select the number of "words" in the string plus the other
> > >amount. Am I doing this the hard way?
> > >
> > >Ed
> > >
> > >"Klaus Linke" <info@fotosatz-kaufmann.de.no.junk> wrote in message
> > >news:es7ZAfW1DHA.2544@TK2MSFTNGP10.phx.gbl...
> > >> Hi Ed,
> > >>
> > >> Would have been easier to remain in the old thread ;-)
> > >>
> > >> > >If you want to take the simplest case of words being separated by
> spaces,
> > >> > >then the following line of code will do the trick
> > >> > >
> > >> > >numWords = UBound(Split(stringWithWords, " ")) + 1
>
> > >> > Unfortunately, when used with
> > >> > Selection.MoveRight Unit:=wdWord, Count:=numWords
> > >> > I get the wrong answer, because Word stops at every punctuation mark
> along
> > >> > the way as well as spaces. So I didn't know what I was asking then -
> but
> > >> > know a bit more now. Is there a way to count the words in a string as
> Word
> > >> > would using Ctrl+Right Arrow?
> > >>
> > >> If you (can) stay