Regex to rename Folder from "dd.mm.yy" to "yy.mm.dd"?

Question

I want to rename all folders from "dd.mm.yy" to "yy.mm.dd". How to do that in the shell?

I recommend you go the extra step and use the standard date format YYYY-mm-dd — glenn jackman, Oct 21 '19 at 14:03
Related: https://unix.stackexchange.com/questions/534033/how-to-restructure-a-date-variable-in-bash — Jim L., Oct 22 '19 at 18:17

score 5 · Answer 1 · answered Oct 19 '19 at 12:58

5

A bash solution renaming the directories in your current directory:

for f in [0-3][0-9].[01][0-9].[0-9][0-9]; do
  [ -d "$f" ] && mv -v "$f" "${f:6}.${f:3:2}.${f:0:2}"
done

answered Oct 19 '19 at 12:58

Freddy

25,565

score 3 · Answer 2 · answered Oct 19 '19 at 08:43

Your question could use a little more information, therefore I am assuming:

by all folders you mean all folders in the current directory
while you mention regex, you neglected to mention OS, so I'm working on the basis that your rename version supports them

rename 's/^([0-9]{2})\.([0-9]{2})\.([0-9]{2})\//$3.$2.$1/' */

If you don't have rename, or you do and it doesn't support regex, this is a more portable approach (albeit less elegant):

find . -mindepth 1 -maxdepth 1 -type d \
  | sed 's ^./  ' \
  | grep -E "^([0-9]{2}\.){2}[0-9]{2}$" \
  | while IFS=. read dd mm yy ; do mv $dd.$mm.$yy $yy.$mm.$dd ; done

Hermann · Answer 3 · 2019-10-21T16:28:47.787

2

I recommend not using regexes, but rather resort to date tools when it comes to handling dates, e.g.

for dir in */; do echo mv $dir $(LC_ALL=C busybox date -uD '%d.%m.%Y' -d "$dir" "+%Y.%m.%d") ; done

(inspired by Force date to read Day/Month/Year.)
I included an echo for "preview" mode.

edited Oct 21 '19 at 16:28

answered Oct 19 '19 at 09:38

Hermann

6,148

score 2 · Answer 4 · answered Oct 21 '19 at 14:35

This is longer than a one-liner, but it's not making any assumptions about the date:

$ mkdir 12.12.12 01.02.99

$ perl -MTime::Piece -Mautodie -E '
  opendir my $dh, ".";
  while (my $f = readdir $dh) {
    if (-d $f and $f =~ /^\d\d.\d\d.\d\d$/) {
      my $t = Time::Piece->strptime($f, "%d.%m.%y");
      rename $f, $t->ymd;
    }
  }
'

$ ls
1999-02-01/  2012-12-12/

score 1 · Accepted Answer · edited Oct 22 '19 at 18:04

1

A very easy way to do it, since already folders created.

ls -1d */|cut -f1 -d/ |awk -F"." '{print "mv " $0 " " $3"."$2"."$1"/"}' | bash

edited Oct 22 '19 at 18:04

Chris Davies

116,213
16
160
287

answered Oct 21 '19 at 20:47

Pradeep G

22

JJoao · Answer 6 · 2019-10-22T19:00:23.163

0

I know this is ugly (and fails in some situations...) but...

reg='^(..)\.(..)\.(..)/'
for a in */ ; do
  [[ $a =~ $reg ]] && mv $a ${BASH_REMATCH[3]}.${BASH_REMATCH[2]}.${BASH_REMATCH[1]}
done

edited Oct 22 '19 at 19:00

answered Oct 22 '19 at 18:54

JJoao

12,170
1
23
45

Regex to rename Folder from "dd.mm.yy" to "yy.mm.dd"?

6 Answers6