Rename files in bash keeping the originals unchanged

I have a folder with a lot of images for a mult开发者_开发技巧ilingual site, the images are stored in the next format filename.lang_code.jpg, on the deploy script i want to pick the correct one for the site I'm deploying and copy it to filename.jpg so I can share the same css between the sites.

So, what I need is something like the command rename, but performing a copy, not a move, because I need to keep all files.

I'm doing it with the next code, but I find it overcomplicated and awful.

find -name "*.es.*" -print0 | xargs --null -I {} sh -c "echo {} | sed 's/.es//' | xargs -I }{ sh -c 'cp {} }{'"

This kind of copying/renaming is usually easier to perform in zsh, where you don't need to write a for loop or find command.

zmv -Ls '(**)/(*).es.(*)' '$1/$2.$3'

The -Ls is to create symbolic links; replace by -C for a copy, or nothing for a move. Other useful options are -i (ask for confirmation for each copy/move) and -n (just show what would happen but don't actually perform the copies/moves).

The $1, $2, $3 in the replacement text refer to the first, second and third parenthesized group in the pattern. In the pattern, **/ means any chain of directories.

You may need to first load the zmv command with autoload zmv. This can usefully go into your .zshrc, as well as alias zcp='zmv -L' and alias zln='zmv -L'.

find -name '*.es.*' -exec bash -c 'cp "$1" "${1/.es./.}"' modlang {} \;

bash 4

shopt -s globstar
for file in **/*.es*
do
  newfilename=${file/.es/}
  cp "$file" ""$newfilename"
done

of using find

find . -type f -iname "*.es.*" -print | sed 's/\(.*\)\.es\.\(.*\)/mv \1.es.\2 \1\2/' |bash

How about a for loop?

for x in *.es.* ; do
  xx=$(echo ${x} | sed 's/\.es//')
  cp ${x} ${xx}
done

Hmm.. you could just copy the whole lot of jpg's you want, put them in the directory you need them and just rename them from there?

Something like:

cp *.jpg /src /dst
cd /dst && rename 's/\.es//' *.jpg

The rename command is a wrapper for a perl script.