How do I create a copy of a file in multiple directories?

Question

I have a folder structure

conf/
    widget1/
        env1.cfg
        env2.cfg
    widget2/
        env1.cfg
        env2.cfg
    widget3/
        env1.cfg
        env2.cfg
    widget4/
        env1.cfg
        env2.cfg
    ...

What I want to do is find all the files named env2.cfg, and create a copy of them in their directory named env3.cfg.

I've tried:

find -name env2.cfg -exec cp '{}' "env3.cfg" ";"

but that just created env3.cfg in the conf/ directory.

Any thoughts?

For clarification I'm doing this on Mac OSX, but ideally the same command should work on CentOS if possible.

Answer 1

Try

find . -name env2.cfg -execdir cp {} env3.cfg \;

By replacing exec to execdir. It will mean that your 'action' (cp) will take place inside the folder where the file (env2.cfg) have been found.

Answer 2

Your example copies all env2.cfg files to the current working directory, which is ".". You need to give the file found a new name, including its directory. This one worked for me:

for file in $(find . -name env2.cfg); do backup=$(echo $file | sed 's/env2.cfg/env3.cfg/'); cp $file $backup; done

edit: Or the even more elegant way, so you don't have to bother with multiple commands:

for file in $(find . -name env2.cfg); do cp $file ${file/env2/env3}; done

Answer 3

Since it's a static structure you can directly address the filenames and let the shell copy them:

for f in conf/widget*/env2.cfg
do
    cp "${f}" "${f/env2/env3}"
done