>Yes. There are 2 hours each year when this formula is wrong.
I assume you're thinking of the switch from DST to CST and vice-versa,
yes? But since that change occurs at oh-dark-hundred (2:00AM), it's
already "today," and "yesterday" would be the same, wouldn't it? Or is it
a different pair of hours you're referring to?
More seriously, isn't this code subtracting a day twice?
>>>> my ($mday, $mon, $year) = (localtime(time() - 60 * 60 * 24) )[
3,4,5];
>>>> printf("%d/%02d/%02d\n", $year + 1900, $mon + 1, $mday - 1);
I would think that the first, subtracting from time() would be the correct
one--the one that would overcome changes in month, year, and time, while
the second, from $mday's value, is not just redundant, but at this point,
wrong, no? Not only that, it would introduce such curiosities as the
zero-th day of each month--the day before the first of a month.
On a picky note, wouldn't it be easier to just subtract 86400?
Deane Rothenmaier
Programmer/Analyst
Walgreens Corp.
224-542-5150
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