How to display 4/25/10 to 2010-25-04?

Question

my$str= \'4/25/10\';
my$sr = join(\' \',split (/\\//,$str));
#my$s = sprintf \'%3$d %2$d %1$d\',$srt;
print$sr,\"\\n\";

output:

<

Answer 1

Well, a braindead solution might be:

my @date = split( /\//,$str)
printf("%04d-%02d-%02d", $date[2] + 2000, $date[1], $date[0]);

You could write something a little more self-documenting by highlighting what you expect to be year, month and day like so:

my ($day, $month, $year) = split /\//, $str;
printf("%04d-%02d-%02d", $year + 2000, $month, $day);

Answer 2

use DateTime::Format::Strptime;
my $strp = DateTime::Format::Strptime->new(
    pattern   => '%m/%d/%y',
    locale    => 'en_US'
);
my $dt = $strp->parse_datetime('4/25/10');
print $dt->strftime('%Y-%d-%m'), "\n";

Gives:

2010-25-04

Answer 3

You're not that far off.

Instead of splitting and joining in a single operation, you can keep individual variables to handle the data better:

my ($d,$m,$y) = split /\//, $str;

Then you can format it in most any way you please, for example:

printf "20%02d-%02d-%02d\n", $y, $d, $m;

A few notes, though:

• I won't comment about the source format, but the format you're converting to doesn't make a lot of sense. You'd probably be better using ISO-8601: 2010-04-25.
• Obviously, this way of doing only works up to year 2099.
• For anything more serious, you'd be better off delegating this kind of work to date handling modules. See for example this question for parsing and this question for formatting

Answer 4

No need to do to DateTime for this. Time::Piece has been included with the Perl core distribution for years.

#!/usr/bin/perl

use strict;
use warnings;
use 5.010;

use Time::Piece;

my $date = '4/25/10';

my $date_tp = Time::Piece->strptime($date, '%m/%d/%y');

say $date_tp->strftime('%Y-%m-%d');