For each item

[Jim Bufalini]

Think of each tab as an End Of Line char with an implied EOL char at the end
of string. Bottom line is you have 7 data elements. If it returned 9 and you
put this into a repeat loop with a count of 9, you would end up trying to
process two extra empty data elements at the end.

> -----Original Message-----
> From: use-revolution-bounces at lists.runrev.com
> [mailto:use-revolution-bounces at lists.runrev.com]On Behalf Of Hershel
> Fisch
> Sent: Thursday, August 11, 2005 12:23 PM
> To: How to use Revolution
> Subject: Re: For each item
>
>
> On 8/11/05 6:12 PM, "Alex Tweedly" <alex at tweedly.net> wrote:
>
> > Hershel Fisch wrote:
> >
> That's exactly my question, why at the letter "a" you combine it with the
> first tab and an the last "c" is not combined with the previous tab,and
> In-between every tab is an item?
> Now if you say that every second tab is a delimiter and every first tab is
> an item I understand and again it shouldn't be 7 ?
> Thanks, Hershel
> > I count it to be 7.
> >
> > put "a" & tab & tab & tab & tab & "b" & tab  & tab & "c"  into tV
> >      11111111   222   333   444   555555555    666   777
> >
> >
> > (Hope this comes through as fixed-width font :-)
> >
> >
>
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