LPM: Test / question on date stuff
Rich Bowen
rbowen at rcbowen.com
Wed Feb 9 09:45:15 CST 2000
Carl Rose wrote:
>
...
> ($sec,$min,$hour,$mday,$mon,$year,$wday,$yday,$isdst)
> = localtime(time);
> $mon += 1;
> $year = $year +1900;
>
> print "$mon/$mday/$year \n";
>
> It of course returns the date just like I want it
> (02/09/2000). So what code do I issue to get 28 days
> from now added onto the day and get the correct
> rollaround. (ie when you add 28 above you get Feb 37th
> -ugg instead of the early part of March). Any help
> would be appreciated and go toward good karma.
How about:
$now = time;
$later = $now + (28 * 24 * 60 * 60);
($sec,$min,$hour,$mday,$mon,$year,$wday,$yday,$isdst)
= localtime($later);
$mon += 1;
$year = $year +1900;
print "$mon/$mday/$year \n";
Enjoy, and welcome to the list.
Rich
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