I'm trying to copy certain files from one directory to another. Using this command
find "$HOME" -name '*.txt' -type f -print0 | xargs -0 cp -t $HOME/newdir I get an warning message saying
cp: '/home/me/newdir/logfile.txt' and '/home/me/newdir/logfile.txt' are the same file
How can I avoid this warning message?
The problem is that you try to copy a file to itself. You can avoid it by excluding the destination directory from the results of the find command like this:
find "$HOME" -name '*.txt' -type f -not -path "$HOME/newdir/*" -print0 | xargs -0 cp -t "$HOME/newdir" 
Try using install instead. This replaces by removing the file first.
install -v target/release/dynnsd-client target/ Output:
removed 'target/dynnsd-client' 'target/release/dynnsd-client' -> 'target/dynnsd-client' And then remove the source files.
Make it unique in the process. But this require sorting
find "$HOME" -name '*.txt' -type f -print0 | sort -zu | xargs -0 cp -t "$HOME/newdir" Or if it's not about the generated files, try to use the -u option of cp.
find "$HOME" -name '*.txt' -type f -print0 | xargs -0 cp -ut "$HOME/newdir" -u copy only when the SOURCE file is newer than the destination file or when the destination file is missing
Install
It worked perfectly in a Makefile context with Docker:
copy: @echo '' bash -c 'install -v ./docker/shell .' bash -c 'install -v ./docker/docker-compose.yml .' bash -c 'install -v ./docker/statoshi .' bash -c 'install -v ./docker/gui .' bash -c 'install -v ./docker/$(DOCKERFILE) .' bash -c 'install -v ./docker/$(DOCKERFILE_SLIM) .' bash -c 'install -v ./docker/$(DOCKERFILE_GUI) .' bash -c 'install -v ./docker/$(DOCKERFILE_EXTRACT) .' @echo '' build-shell: copy docker-compose build shell 
your_command 2>/dev/null=)