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: [EMAIL PROTECTED] > [mailto:[EMAIL PROTECTED] 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" <[EMAIL PROTECTED]> 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 :-) > > > > > > _______________________________________________ > use-revolution mailing list > [email protected] > Please visit this url to subscribe, unsubscribe and manage your > subscription preferences: > http://lists.runrev.com/mailman/listinfo/use-revolution > > _______________________________________________ use-revolution mailing list [email protected] Please visit this url to subscribe, unsubscribe and manage your subscription preferences: http://lists.runrev.com/mailman/listinfo/use-revolution
