Batch rename files with unknown names and unknown extensions

Question

I have a bunch of pictures, some are .jpg, some are .png. any given picture could have an entirely random name (possibly including spaces). in any given directory, how can I rename all of them, while keeping the extension as is? the input, for example:

001.jpg
02.png
05.png
zzzz.jpg

I would like to completely replace their names with four-digit numbers, starting with 0001, and going up to 9999 (if necessary). thus, the output might look like:

0001.jpg
0002.png
0003.png
0004.jpg

the order of these files is important to me, but they are already in sequential order if I were to sort them by filename alone.

I have checked out the perl rename, and its usage is a mystery to me in this example.

Answer 1

Perl Solution

perl -e '
    @files = sort glob "*.png *.jpg";
    for(@files){ 
        ($extension = $_) =~ s/.*\.(.*)/$1/;
        $new_name = sprintf("%04d",++$counter).".$extension";
        die "File $new_name already exists and would be clobbered\n" if -e $new_name;
        rename $_ => sprintf("%04d",++$counter).".$extension"
     }'

Bash Solution

export COUNT=0 
find . -maxdepth 1 -name "*.jpg" -o -name "*.png" -print0 | xargs -0 sort | while read file do\
extension=${file##*.}\
COUNT=$((COUNT+1))\
mv -- "$file" $(printf "%04d" $COUNT).$extension\
done

Note

• Both solutions are Bash solution is untested: try them it on dummy files first.
• The Bash solution may break if you have file names with newlines in them.
• Danger: As pointed out by Gilles in the comments below, if one of the original files has a name that follows the target naming convention, it will at best be renamed to another "number" and at worst be clobbered by another file. Which one happens will depend on how it sorts relative to other files. I have added a provision for this in the Perl solution since this is the one the OP seems to favor.
• The Perl solution (with the provision from the above point implemented) is non-idempotent.

Answer 2

(bash solution)

Using printf to format the new names and awk to pick out the filename extension...

$ ls
a.b  c.d  e.f  g.h  i.j  k.l  m.n  o.p  q.r  s.t  this file.y  u.v  w.x
$ COUNTER=0
$ for i in *; do COUNTER=$(($COUNTER + 1)); mv -v "$i" $(printf "%04g" $COUNTER).$(echo $i | awk -F'.' '{print $NF}'); done
`a.b' -> `0001.b'
`c.d' -> `0002.d'
`e.f' -> `0003.f'
`g.h' -> `0004.h'
`i.j' -> `0005.j'
`k.l' -> `0006.l'
`m.n' -> `0007.n'
`o.p' -> `0008.p'
`q.r' -> `0009.r'
`s.t' -> `0010.t'
`this file.y' -> `0011.y'
`u.v' -> `0012.v'
`w.x' -> `0013.x'

Remember to surround the input $i in quotes to escape spaces in filenames, as you have mentioned.

Answer 3

In TXR Lisp:

(let ((counter 0))
  (each ((name (sort (glob "*.{png,jpg}"))))
    (let ((new (pic "0###<" (inc counter) (short-suffix name))))
      (if (path-exists-p new)
        (error `@new already exists`))
      (rename-path name new))))

We could do some other things, like prepare the list of renames first, check for the coincidental existence of the right hand side names, and not perform any rename in that case.

Answer 4

With zsh:

autoload -Uz zmv
n=0; zmv -n '(**/)*(.*)(#q.)' '$1${(l[5][0])$((++n))}$2'

Remove the -n (dry run) if happy.

Beware that a file-1.2.tar.gz.sha1 for instance would be renamed to 00001.sha1, not 00001.tar.gz.sha1 as you might prefer.

To do it only for files whose extensions usually represents image files, you could get that list from the mime type database:

ext=( $(grep -Po '^image/\S+\s+\K\S.*' /etc/mime.types) )
n=0; zmv -n "(**/)*(.(#i)(${(j[|])${(@b)ext}})*)(#q.)" '$1${(l[5][0])$((++n))}$2'

Answer 5

Save this script:

#!/bin/bash

NUM=0
NUMBER_LENGTH=4

cd $1;

ls . -1 | sort | while read file; do
    NUM=$(( $NUM + 1 ));
    PREFIX=$NUM;
    while [ ${#PREFIX} != $NUMBER_LENGTH ]; do
        PREFIX="0${PREFIX}";
    done;
    NEW_NAME=$PREFIX.`echo $file | sed 's|[^.]\+\.\([\A-Za-z0-9]\+\)|\1|g'`;
    mv "$file" "$NEW_NAME";
done

and execute it as:

./_scriptname_ _dirWhereYourStuffIs_

CAVEAT: don't put the script in the same dir where your stuff is, because the script itself would be renamed.

If you're insane enough, you can do it all in a line (without using the script):

NUM=0; NUMBER_LENGTH=4; ls . -1 | sort | while read file; do NUM=$(( $NUM + 1 )); PREFIX=$NUM; while [ ${#PREFIX} != $NUMBER_LENGTH ]; do PREFIX="0${PREFIX}"; done; NEW_NAME=$PREFIX.`echo $file | sed 's|[^.]\+\.\([\A-Za-z0-9]\+\)|\1|g'`; mv "$file" "$NEW_NAME"; done

CAVEAT #2: for this single line, I'm assuming you're using BASH