substract from actual date 7 days and printout in month.days.hours.min.year

nicopelle has asked for the wisdom of the Perl Monks concerning the following question:

Hi to all, I need to substract from actual date/time 7 days. The format could be for example #031810002015 (march 18 10:00 of year 2015):

#!/usr/bin/perl -w
#output as 031810002015 (march 18 10:00 del 2015)
($sec,$min,$hour,$mday,$mon,$year,$wday,$yday,$isdst) = localtime(time
+);
my $now = sprintf("%02d%02d%02d%02d%04d", $mon+1, $mday,$hour, $min,$y
+ear+1900);
print "$now\n";
[download]

Thanks for your time as usuals, Nick.

Comment on substract from actual date 7 days and printout in month.days.hours.min.year Download Code

Replies are listed 'Best First'.
Re: substract from actual date 7 days and printout in month.days.hours.min.year by Anonymous Monk on Mar 23, 2015 at 10:29 UTC
Use one of the modules available on CPAN, some examples are Date::Calc, Date::Manip, or DateTime. `use DateTime; print DateTime->now->add(days=>-7) ->strftime('%m%d%H%M%Y'), "\n"; __END__ 031614272015` [download]	[reply] [d/l]
Re^2: substract from actual date 7 days and printout in month.days.hours.min.year by nicopelle (Acolyte) on Mar 23, 2015 at 10:36 UTC
Thanks to everyone has replied this post. Thanks to all, Nick.	[reply]
Re: substract from actual date 7 days .. by Discipulus (Canon) on Mar 23, 2015 at 10:29 UTC
Hello nicopelle, You have many options to do this: see this node to review them (production code must use serious modules..) HtH L* There are no rules, there are no thumbs.. Reinvent the wheel, then learn The Wheel; may be one day you reinvent one of THE WHEELS.	[reply] [d/l]
Re: substract from actual date 7 days and printout in month.days.hours.min.year by MidLifeXis (Monsignor) on Mar 23, 2015 at 13:34 UTC
Do you mean 7 days, 72460*60 seconds, or something else? Think daylight savings time boundaries, leap seconds, etc. --MidLifeXis	[reply]