Find the owner of a directory or file, but only return that and nothing else

Question

I am looking for a command that will return the owner of a directory and only that--such as a regex parsing the ls -lat command or something similar? I want to use the result in another script.

Answer 1

stat from GNU coreutils can do this:

stat -c '%U' /path/of/file/or/directory

Unfortunately, there are a number of versions of stat, and there's not a lot of consistency in their syntax. For example, on FreeBSD, it would be

stat -f '%Su' /path/of/file/or/directory

If portability is a concern, you're probably better off using Gilles's suggestion of combining ls and awk. It has to start two processes instead of one, but it has the advantage of using only POSIX-standard functionality:

ls -ld /path/of/file/or/directory | awk '{print $3}'

Answer 2

Parsing the output of ls is rarely a good idea, but obtaining the first few fields is an exception, it actually works.

ls -ld /path/to/directory | awk '{print $3}'

Another option is to use a stat command, but the problem with stat from the shell is that there are multiple commands with different syntax, so stat in a shell script is unportable (even across Linux installations).

Note that testing whether a given user is the owner is a different proposition.

if [ -n "$(find . -user "$username" -print -prune -o -prune)" ]; then
  echo "The current directory is owned by $username."
fi
if [ -n "$(find . -user "$(id -u)" -print -prune -o -prune)" ]; then
  echo "The current directory is owned by the current user."
fi

Answer 3

One can also do this with GNU find:

find $directoryname -maxdepth 0 -printf '%u\n'

This isn't portable outside of the GNU system, but I'd be surprised to find a Linux distribution where it doesn't work.