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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