On 11/16/06 7:10 PM, "John Craig" <[EMAIL PROTECTED]> wrote:
> Say we have a 10 digit number called tInt - 1234567890 > > repeat with i = length(tInt) - 3 to 1 step -3 I got it, a reverse repeat by 3. I was wondering for a while how to do a reverse repeat, I got it. Thanks. Hershel > > We start at length(tInt) - 3 (position 7) and put a comma after the > character at position 7 > This gives us 1234567,890 > > The next iteration (-3) takes us to position 4 - put in another comma; > Now we have 1234,567,890 > > Same again, takes us to position 1; > Now we have 1,234,567,890 > > The last iteration takes us to position -2. If we added a comma here, > it would appear near the end of the string! > So we add the condition; > if i > 0 then put "," after char i of tInt > > the condition takes care of any strings which aren't exact multiples of 3. > > :-) > JC > >> it did, very nice code. >> Just don't understand the repeat header. >> "to 1 step -3" >> Thanks, Hershel >> > > > _______________________________________________ > 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
