Hi Seona,

that is it! Thanks.

 * count
 * count + divisor
 * count + (divisor * 2)


On 4/5/07, Seona Bellamy <[EMAIL PROTECTED]> wrote:
>
> 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.
>
> >
>


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