我试图计算从数据库中提取的两个日期之间的日历天数.我认为将日期转换为秒将是一个简单而正确的解决方案.
#!/usr/bin/perl
use warnings;
use strict;
use POSIX qw(strftime);
use Date::Parse;
my $minDate = "2016-03-27";
my $maxDate = "2016-06-15";
print "Format as extracted from db: mindate: $minDate and maxdate: $maxDate\n";
my ($dbYear,$dbMonth,$dbDay) = split ('-', $minDate);
my $datum = "$dbYear$dbMonth$dbDay";
my $minDateSec = str2time($datum);
($dbYear,$dbMonth,$dbDay) = split ('-', $maxDate);
$datum = "$dbYear$dbMonth$dbDay";
my $maxDateSec = str2time($datum);
my $numCalDaysSec = ($maxDateSec-$minDateSec)/86400;
print "Min date in Seconds: $minDateSec\n";
print "Max date in Seconds: $maxDateSec\n";
print "Num days: $numCalDaysSec\n";
Run Code Online (Sandbox Code Playgroud)
最初,我认为这种方法为我提供了可靠的结果:
bash-3.2$ ./testNumDays.pl
As extracted from db: mindate: 2016-06-14 and maxdate: 2016-06-15
Min date in Seconds: 1465855200
Max date in Seconds: 1465941600
Num days: 1
bash-3.2$ ./testNumDays.pl
As extracted from db: mindate: 2016-05-31 and maxdate: 2016-06-15
Min date in Seconds: 1464645600
Max date in Seconds: 1465941600
Num days: 15
bash-3.2$ ./testNumDays.pl
As extracted from db: mindate: 2016-03-28 and maxdate: 2016-06-15
Min date in Seconds: 1459116000
Max date in Seconds: 1465941600
Num days: 79
bash-3.2$ ./testNumDays.pl
As extracted from db: mindate: 2016-03-27 and maxdate: 2016-06-15
Min date in Seconds: 1459033200
Max date in Seconds: 1465941600
Num days: 79.9583333333333
bash-3.2$
Run Code Online (Sandbox Code Playgroud)
显然,日期之间的日历日期数应该是整数.嗯,我做错了什么?为什么转换为秒不可靠?
由于我是一个Perl新手,我可能会忽略这一点.因此,非常欢迎任何帮助.
一种方法是使用Time :: Piece Core模块:
use warnings;
use strict;
use Time::Piece qw();
my $t1 = Time::Piece->strptime('2016-03-27', '%Y-%m-%d');
my $t2 = Time::Piece->strptime('2016-06-15', '%Y-%m-%d');
my $seconds = $t2 - $t1;
print $seconds->days(), "\n";
__END__
80
Run Code Online (Sandbox Code Playgroud)
另一种选择是DateTime.它不在Core中,但它通常非常方便.因为它不解析日期,所以我们很快就制作了自己的解析函数.
use strict;
use warnings;
use DateTime;
my $minDate = "2016-03-27";
my $maxDate = "2016-06-15";
sub parse {
my $date = shift;
my ( $y, $m, $d ) = split /-/, $date;
return DateTime->new( year => $y, month => $m, day => $d );
}
my $days = parse($minDate)->delta_days(parse($maxDate))->in_units('days');
print $days;
__END__
80
Run Code Online (Sandbox Code Playgroud)
| 归档时间: |
|
| 查看次数: |
302 次 |
| 最近记录: |