linux bash - Parse date in custom format

Question

I have a date in a the %c format (could be any other) and I need to use it in the date command. %c is NOT the American format. It is the G

Answer 1

You may use libdatetime-format-flexible-perl.

#!/usr/bin/perl
use DateTime::Format::Flexible;
my $date_str = "So 22 Dez 2013 07:29:35 CET";
$parser = DateTime::Format::Flexible->new;
my $date = $parser->parse_datetime($date_str);
print $date

Default output will be 2013-12-22T07:29:35, but since $date is not a regular string but object, you can do something like this:

printf '%02d.%02d.%d', $date->day, $date->month, $date->year;

Also date behavior probably should be considered as a bug. I think so, because date in the same format but in russian is parsed correctly.

$ export LC_TIME=ru_RU.UTF-8
$ NOW="$(date "+%c")"
$ date --date="$NOW" '+%d.%m.%Y'
22.12.2013