How to find a file and move it into the directory it was found in

Question

I am trying to rename files in sub directories using xargs by running the following command:

find . -name oldfile.txt -print0 | xargs -n 1 -0 -I % mv % newfile.txt

The problem is that the file gets moved into the directory I run the command in instead of the directory that oldfile.txt is found in. I cant seem the solution for this, any ideas?

I'm thinking that perhaps I can get the directory from the found file's path and then append it to the front of the second paramater of mv but how would that work?

Answer 1

Instead of relying on external commands, where you lose the information about the current directory, you can use find's -execdir option to have it execute commands in the directory a file was found in.

find . -name oldfile.txt -execdir mv -- {} newfile.txt \;

This is actually a more secure version of -exec that is preferred in most situations. The only drawback is that it's not specified by POSIX, so you won't find it in all find variants – but at least in GNU find and BSD find you can.

If you want to be extra safe, you should call mv like mv -- source target so that a file starting with a dash isn't incorrectly interpreted as an argument to mv.

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If for some reason you don't want to use -execdir or have other complicated things to do, or just want to learn a little about Bash, here's how to get the folder and move the file based on that:

find . -name oldfile.txt -exec bash -c 'mv -- "${0}" "$(dirname $0)"/newfile.txt' {} \;

Here, {} passes the full path to bash as $0. We get its directory name through command substitution ($(…)) and then append the new name. Both the original argument and the directory name have to be double quoted to prevent whitespace from breaking the command – otherwise mv would think you're trying to move two files called foo and bar when the file is actually called foo bar, for example.