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.

5 Answers 5


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}'
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    stat -c %U /path, if brevity is a bonus. Commented Feb 20, 2011 at 23:18
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    That assumes GNU stat, which is not the case on older Linux systems (even on newer systems I'd be wary, there might be a different stat (a site-wide standard) in /usr/local/bin or somewhere in the user's home), and is rarely available on other unices. Commented Feb 20, 2011 at 23:18
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    stat -c %U has the advantage of also working with BusyBox, if the stat command is compiled in. Commented Feb 21, 2011 at 0:23
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    Nice, the last example (ls) works both on Unix/OSX and Linux
    – kenorb
    Commented Nov 28, 2013 at 15:52
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    Note that ls will print the uid if there's no local user matching the owner (i.e., on a network share), but stat -c '%U' DIR will print UNKNOWN, which is less helpful or more appropriate, depending on how you look at it.
    – basic6
    Commented May 14, 2014 at 8:59

Parsing the output of ls is rarely a good idea, but obtaining the first few fields is an exception, it actually works on all “traditional” unices (it doesn't work on platforms such as some Windows implementations that allow spaces in user names).

ls -ld /path/to/directory | awk 'NR==1 {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 $username."
if [ -n "$(find . -user "$(id -u)" -print -prune -o -prune)" ]; then
  echo "The current directory is owned by the current user."
  • There are some caveats with the ls | awk approach too unfortunately, as I've noted here. I haven't yet come up with a solution for the, "target file/dir is a symlink with a different name," issue I mentioned in my recent comment.
    – beporter
    Commented Jul 15, 2014 at 17:16
  • Instead of find . -user "$username" -print -prune -o -prune you could just do find . -maxdepth 0 -user "$username" Commented Dec 9, 2016 at 12:18
  • @Gilles Is there any reason you use awk 'NR==1 {print $3}' instead of just awk '{print $3}'? I'm not sure why the NR==1 is necessary here. Commented Sep 10, 2019 at 21:08
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    @HaroldFischer Only in an edge case where the path contains a newline. It's rarely necessary, but never harmful. Commented Sep 10, 2019 at 21:19
  • @Gilles Very good to know!! Just curious, are you aware of an implementation of ls where a newline in a path causes the line to be split in two (which I believe is the edge case you are trying to cover)? On GNU ls (newish version), BusyBox ls and FreeBSD ls a newline is returned as $'\n', ? and ?, respectively. Commented Sep 10, 2019 at 21:48

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.


In pure bash you can convert the output of ls to an array and index into it.

# (lrwxr-xr-x, 1, myuser, staff, 36, Oct, 21, 16:36, /path/to/file)    
file_meta=($(ls -ld /path/to/file))
file_owner="${file_meta[2]}" # myuser

It's not as elegant as using stat, find, or awk, but could work in a pinch.

ls -ld /path/to/file/or/directory | cut -d' ' -f3

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