可能是一个简单的答案：如何在不导入外部模块的情况下进行转换？我已阅读CPAN但无法明确指出执行以下操作的方法：
Convert: 20080428 to 28-APR-08
有任何想法吗？
即使指导我阅读教程也会受到赞赏.
问候,
PIAS
解决方法
此代码在Y10k中失败,但这应该足够好了.正则表达式可能更严格,但如果日期已经过验证(或将以新表格验证),则无关紧要.
#!/usr/bin/perl use strict; use warnings; my @mon = qw/null JAN FEB MAR APR MAY JUN JUL AUG SEP OCT NOV DEC/; my $d = "20080428"; $d =~ s/..(..)(..)(..)/$3-$mon[$2]-$1/; print "date is now $d\n";
或者,如果你是疯了,想要在正则表达式中验证(需要Perl 5.10)：
#!/usr/bin/env perl5.10.0
use strict;
use warnings;
my @mon = qw/null JAN FEB MAR APR MAY JUN JUL AUG SEP OCT NOV DEC/;
my $d = join '',@ARGV;
# only validates between 1600 and 9999
# because of calendar weirdness prior to 1600
$d =~ s/
^
(?:
# non-leap years and everything but 29th of Feb in leap years
(?:
1[6-9] (?<y> [0-9]{2}) |
[2-9][0-9] (?<y> [0-9]{2})
)
(?: #any month 1st through 28th
(?: (?<m> 0[1-9] | 1[0-2]) (?<d> 0[0-9] | 1[0-9] | 2[0-8]) )
| #or 30th of any month but 2
(?: (?<m>0[13-9] | 1[0-2]) (?<d> 30) )
| # or 31st of 1,3,5,7,8,10,or 12
(?: (?<m>0[13578] | 1[02]) (?<d> 31) )
)
| # or 29th of Feb in leap years
(?:
(?: #centuries divisible by 4 minus the ones divisible by 100
16 |
[2468][048] |
[3579][26]
)
(?<y> 00)
| #or non-centuries divisible by 4
(?: 1[6-9] | [2-9][0-9] )
(?<y>
0[48] |
[2468][048] |
[13579][26]
)
)
(?<m> 02) (?<d> 29)
)
$
/$+{y}-$mon[$+{m}]-$+{d}/x or die "invalid date: $d";
print "date is now $d\n";