# of letters in a string

Is there anyway to tell how many letters are in a string? More specifically, I have a userform where the user enters a string into a textbox. The textbox will only accept a total of 10 letters. When the string variable gets passed to my sub routine, I need to know how many letters are in that string.

If I can determine how many letters are in the string, if there are less than 10, I then want to add spaces to the end of that string to make it 10 characters long.

For example, if I enter:
"August" (6 characters) - it will then change it to "August____" (_ = " ")
"California" (10 characters) - it will not add any spaces


TIA for any help,

DejaVu

sStr = Left(sStr & " ",10)

--
Regards,
Tom Ogilvy


"DejaVu" <[email protected]> wrote in
message news:[email protected]...
>
> Is there anyway to tell how many letters are in a string? More
> specifically, I have a userform where the user enters a string into a
> textbox. The textbox will only accept a total of 10 letters. When the
> string variable gets passed to my sub routine, I need to know how many
> letters are in that string.
>
> If I can determine how many letters are in the string, if there are
> less than 10, I then want to add spaces to the end of that string to
> make it 10 characters long.
>
> For example, if I enter:
> "August" (6 characters) - it will then change it to "August____" (_ =
> " ")
> "California" (10 characters) - it will not add any spaces
>
>
> TIA for any help,
>
> DejaVu
>
>
> --
> DejaVu
> ------------------------------------------------------------------------
> DejaVu's Profile:
http://www.excelforum.com/member.php...o&userid=22629
> View this thread: http://www.excelforum.com/showthread...hreadid=393382
>

msgbox len(sStr) gives the number of characters in the string held in
variable sStr (just for completeness)

--
Regards,
Tom Ogilvy

"DejaVu" <[email protected]> wrote in
message news:[email protected]...
>
> Is there anyway to tell how many letters are in a string? More
> specifically, I have a userform where the user enters a string into a
> textbox. The textbox will only accept a total of 10 letters. When the
> string variable gets passed to my sub routine, I need to know how many
> letters are in that string.
>
> If I can determine how many letters are in the string, if there are
> less than 10, I then want to add spaces to the end of that string to
> make it 10 characters long.
>
> For example, if I enter:
> "August" (6 characters) - it will then change it to "August____" (_ =
> " ")
> "California" (10 characters) - it will not add any spaces
>
>
> TIA for any help,
>
> DejaVu
>
>
> --
> DejaVu
> ------------------------------------------------------------------------
> DejaVu's Profile:
http://www.excelforum.com/member.php...o&userid=22629
> View this thread: http://www.excelforum.com/showthread...hreadid=393382
>

Or you could do the following:

sStr = sStr & Space(10 - Len(sStr))

"Tom Ogilvy" wrote:

> sStr = Left(sStr & " ",10)
>
> --
> Regards,
> Tom Ogilvy
>
>
> "DejaVu" <[email protected]> wrote in
> message news:[email protected]...
> >
> > Is there anyway to tell how many letters are in a string? More
> > specifically, I have a userform where the user enters a string into a
> > textbox. The textbox will only accept a total of 10 letters. When the
> > string variable gets passed to my sub routine, I need to know how many
> > letters are in that string.
> >
> > If I can determine how many letters are in the string, if there are
> > less than 10, I then want to add spaces to the end of that string to
> > make it 10 characters long.
> >
> > For example, if I enter:
> > "August" (6 characters) - it will then change it to "August____" (_ =
> > " ")
> > "California" (10 characters) - it will not add any spaces
> >
> >
> > TIA for any help,
> >
> > DejaVu
> >
> >
> > --
> > DejaVu
> > ------------------------------------------------------------------------
> > DejaVu's Profile:
> http://www.excelforum.com/member.php...o&userid=22629
> > View this thread: http://www.excelforum.com/showthread...hreadid=393382
> >
>
>
>

Thanks to both of you... your help is much appreciated.

DejaVu

Str = Len(string)

Syntax LEN will provide the actual number of characters within a string
Example # = len(activecell)