Re: 2 errors I can not understand

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.