如何检查 unix 时间提前 1 天

Question

How to check the UNIX time is 1 day older

for ex: UNIX time is 1308715355

I want to find that that time 1 day older or not with now

Answer 1

If you consider 2011-06-27T01:05 to be less than one day older than 2011-06-28T01:00,

use DateTime;

my $epoch = '1308715355';

my $dt_ref = DateTime->from_epoch(epoch => $epoch, time_zone => 'local');
my $dt_now = DateTime->now(time_zone => 'local');

$dt_ref->add(days => 1);
if ($dt_ref <= $dt_now) {
   # At least one day old.
}

If you consider 2011-06-27 to be one day older than 2011-06-28 regardless of times of day,

use DateTime;

my $epoch = '1308715355';

my $dt_ref = DateTime->from_epoch(epoch => $epoch, time_zone => 'local')
   ->truncate( to => 'days' );
my $dt_now = DateTime->today(time_zone => 'local');

$dt_ref->add(days => 1);
if ($dt_ref <= $dt_now) {
   # At least one day old.
}

Upd : Grammar error fix: s/older/older than/

Answer 2

> date -dtomorrow +%s; date +%s
1309366474
1309280074

This should handle, BTW, the DST problem that evan mentioned.

You didn't mention your desired solution space, so hopefully this would suffice.

Answer 3

Comparing with the number of seconds for a day is generally adequate (60 * 60 * 24 = 86400). However, if this always needs to work exactly, you need to take into account when you switch to and from DST. Those days are 23 & 25 hours.

Most people tend to overlook daylight savings time. This can cause a lot of issues if you need precision when dealing with time.

Answer 4

use strict;
use warnings;
use DateTime;

my $epoch = '1308715355';

my $dt = DateTime->from_epoch(epoch => $epoch, time_zone => 'local')
                 ->add(days => 1);
print $dt; # '2011-06-22T23:02:35' -- for *my* time zone (CDT)
print $dt->epoch; # '1308801755'