LPM: Test / question on date stuff

[Mike Andrews]

On Wed, 9 Feb 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

I'd add 28 days worth of seconds to 'time' and have localtime just return
that:

   ($sec,$min,$hour,$mday,$mon,$year,$wday,$yday,$isdst)
       = localtime(time+2419200);

(60 secs * 60 mins * 24 hours = 86400 seconds, * 28 days = 2419200)


Mike Andrews (MA12) * mandrews at dcr.net * http://www.bit0.com/
VP, sysadmin, & network guy, Digital Crescent Inc, Frankfort KY
Internet services for Frankfort, Lawrenceburg, Owenton, Shelbyville
"Don't sweat the petty things, and don't pet the sweaty things."