Timothy S. Nelson wrote:
> I've been working on an extension to Time::Piece which I call
> Time::Period. Then you can subtract one Time::Period from another, and it
> gives a Time::Seconds which is the length of the first time period minus
> the length of the *overlapping* second time period.
>
> It'd be nice if we had an easy way to shoot this problem in the head
> :).
'Time::Period' is DateTime::Span.
'Subtract' is ->complement().
'Overlapping' is ->intersection.
So while you ask in the above for the remaining portion of period 1, I
wonder if you want the remaining portion of the two periods. Of course if
period 2 is entirely within period 1, the two are the same.
#!/usr/bin/perl
use DateTime;
use DateTime::Span;
use Data::Dumper;
$date1 = DateTime->new( year => 2003, month => 4, day => 11 );
$date2 = DateTime->new( year => 2003, month => 4, day => 13 );
$date3 = DateTime->new( year => 2003, month => 4, day => 12 );
$date4 = DateTime->new( year => 2003, month => 4, day => 14 );
$time_period_1 = DateTime::Span->from_datetimes(
start => $date1,
end =>$date2
);
$time_period_2 = DateTime::Span->from_datetimes(
start => $date3,
end =>$date4
);
# VERSION 1
# When you subtract 1 from 2, you want the length of the
# remaining portion of 1:
# Period 1: 111111111111111111111111
# Period 2: 222222222222222222222222
# 1 Comp 2: CCCCCCCCCCCC
# Result: ^^^^^^^^^^^^
$version_1 = ($time_period_1->complement($time_period_2))->duration;
print Dumper( { $version_1->deltas } );
# 1 day ( 2002-04-11 )
# VERSION 2
# You want the length of the remaining portion once the
# intersection is removed from the union (a 'nand'):
# Period 1: 111111111111111111111111
# Period 2: 222222222222222222222222
# Union: UUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUU
# Intersct: IIIIIIIIIIII
# U Comp I: CCCCCCCCCCCC CCCCCCCCCCCC
# Result: ^^^^^^^^^^^^ ^^^^^^^^^^^^
$union = $time_period_1->union($time_period_2);
$intersection = $time_period_1->intersection($time_period_2);
$version_2 = ($union->complement($intersection))->duration;
print Dumper( { $version_2->deltas } );
# 2 days ( 2002-04-11 & 2002-04-14 )
[end]
Of course, as I said, you'll get exactly the same results for the above if
period 1 ->contains period 2. That's because the union of the two periods
is the same as period 1 while the intersection is the same as period 2:
VERSION 1:
# Period 1: 111111111111111111111111111111111111
# Period 2: 222222222222
# 1 Comp 2: CCCCCCCCCCCC CCCCCCCCCCCC
# Result: ^^^^^^^^^^^^ ^^^^^^^^^^^^
VERSION 2:
# Period 1: 111111111111111111111111111111111111
# Period 2: 222222222222
# Union: UUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUU
# Intersct: IIIIIIIIIIII
# U Comp I: CCCCCCCCCCCC CCCCCCCCCCCC
# Result: ^^^^^^^^^^^^ ^^^^^^^^^^^^
--
Obviously the reply-to is a fake. Just change the 'spam-' to 'i' so that the
result sounds like why you go to an optometerist.
