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( [id://3333]=superdoc: print w/replies, xml ) Need Help??

I don't know what your other requiremensts are, but here's a DateTime script to do it. Notice that you can plug in your own formatter class. :)

#!/usr/bin/perl use DateTime; use DateTime::Format::Strptime; foreach my $time ( map { time - $_ * 86400 } -3 .. 3 ) { my $dt = DateTime->from_epoch( epoch => $time, formatter => DateTime::Format::Strptime->new( pattern => "%B % +e, %Y" ) ); my $string = do { my $duration = $dt - DateTime->now; my $D = $duration->years + $duration->months; my $d_days = $duration->delta_days; if( $D or $d_days < -1 or $d_days > 1 ) { "$dt" } elsif( $d_days < 0 ) { "Yesterday" } elsif( $d_days > 0 ) { "Tomorrow" } else { "Today" } }; print "$time: ", $dt->ymd, " $string\n"; }

And here's the output:

1196530263: 2007-12-01 December 1, 2007 1196443863: 2007-11-30 November 30, 2007 1196357463: 2007-11-29 Tomorrow 1196271063: 2007-11-28 Today 1196184663: 2007-11-27 Yesterday 1196098263: 2007-11-26 November 26, 2007 1196011863: 2007-11-25 November 25, 2007
--
brian d foy <brian@stonehenge.com>
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In reply to Re: Formatting Datetime objects nicely by brian_d_foy
in thread Formatting Datetime objects nicely by dragonchild

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