On Tue, 2007-03-13 at 19:04 -0400, Jonathan Kahan wrote:
> This did fix the problem but I am amazed that
> 
> $s%$d=0 would be interpereted as a statement assigning d to 0 since there is 
> some other stuff in front of d... I would think that would produce an error 
> at compile time since $s%$d is an illegal variable name. Normally when my 
> php script errors at compile time nothing will display to the screen.

Nothing wrong with $s%$d=0. What you have is the following:

    $s % ($d = 0)

Probably what was intended was:

    ($s % $d) == 0

Moral of the story? Don't be sloppy. Take pride in writing readable
code. Anyone can produce gibberish.

Cheers,
Rob.
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