rearranging dates

campbell has asked for the wisdom of the Perl Monks concerning the following question:

Replies are listed 'Best First'.
Re: rearranging dates by Zaxo (Archbishop) on Jun 19, 2003 at 12:58 UTC
Here goes, as an exercise call perldoc -f on each keyword: `my $new = join '', reverse split /\//, $date;` This wasn't homework was it? After Compline, Zaxo	[reply] [d/l]
Re: rearranging dates by davorg (Chancellor) on Jun 19, 2003 at 13:01 UTC
`my $date = 'dd/mm/yyyy'; if (my @date = split /\//, $date) { $date = join '', @date[2, 1, 0]; }` [download] -- <http://www.dave.org.uk> "The first rule of Perl club is you do not talk about Perl club." -- Chip Salzenberg	[reply] [d/l]
Re: rearranging dates by vek (Prior) on Jun 19, 2003 at 13:08 UTC
Just for grins you could also use `substr`: `my $DDMMYYYY = '16/06/2003'; my $YYYYMMDD = substr($DDMMYYYY, 6, 4) . substr($DDMMYYYY, 3, 2) . substr($DDMMYYYY, 0, 2);` [download] -- vek --	[reply] [d/l] [select]
Re: rearranging dates by nite_man (Deacon) on Jun 19, 2003 at 13:18 UTC
You can do it at this way: `my $date = '23/06/2003'; $date =~ m!([0-9]{2})/([0-9]{2})/([0-9]{4})!;` [download] or `$date =~ m!(\d{2})/(\d{2})/(\d{4})!; print "$3-$2-$1\n"; __DATE__ 2003-06-23` [download] Updated! `my $date = '23/06/2003'; $date =~ s!(\d{2})/(\d{2})/(\d{4})!$3-$2-$1!; print $date; __DATE__ 2003-06-23` [download] -------------------------------- SV* sv_bless(SV* sv, HV* stash);	[reply] [d/l] [select]
Re: rearranging dates by ant9000 (Monk) on Jun 19, 2003 at 13:32 UTC
It's been largely answered, but here's my attempt: `$_="DDMMYYYY"; print pack("A4A2A2",reverse(unpack("A2A2A4",$_))),$/;` [download]	[reply] [d/l]
Re: rearranging dates by hmerrill (Friar) on Jun 19, 2003 at 14:12 UTC
Here's my shot - slightly different than other options so far: `my $ddsmmsyyyy = "19/06/2003"; my($dd, $mm, $yyyy) = split /\//, $ddsmmsyyyy; my $yyyymmdd = $yyyy.$mm.$dd;` [download]	[reply] [d/l]