On 05/04/07, Dale Fraser <[EMAIL PROTECTED]> wrote: > > So are you saying a fixed number of rows 6 > > Or a fixed number of columns 3? >
I believe the line in the original email was: I need to output a recordset in a table, infinite rows, and maximum 3 > columns in width. > I'm more interested in knowing if there was a reason for the number of empty cells in the third column. Because I'm pretty sure that the columns could have been closer to even in length, and that would be easier to do I think. The maths is making better sense in my head, anyway. Then again, knowing my history with maths, I'm not sure that's saying much... Basically, what I'm thinking is something along these lines: * Divide your number of records by 3 and round it off to an integer (I'll call this divisor). This is the number of records you will have in each of the first two (complete) columns. The third column may or may not have this many records. * Loop over your recordset this many times, with a counter of some sort to count the iterations. * In the loop, get the three cells as follows: * count * count + divisor * count + (divisor * 2) I haven't tested this, so it's purely theoretical and may or may not work. Oh, and you'd probably want to test if the value exists in the third column because that one may not be full. Cheers, Seona. --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "cfaussie" group. To post to this group, send email to cfaussie@googlegroups.com To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/cfaussie?hl=en -~----------~----~----~----~------~----~------~--~---
