Robert Cummings wrote:
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 % $d) == 0
Moral of the story? Don't be sloppy. Take pride in writing readable
code. Anyone can produce gibberish.
Cheers,
Rob.
another suggestion would be to have it written this way
0 == ($s % $d)
if you by chance did this
0 = ($s % $d)
it will give you an error, because you cannot assign a value to a
literal value.
Just a thought.
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