Get the owner of a directory

frayja · 08-31-2007, 12:59 PM

Hello

I am writing a script that needs to know who the owner is of a file or directory.

Code:

#!/bin/bash
for D in /home/*; do
   OWNER=`gettheownersomethow $D`
   echo "The owner of $D is $OWNER"
done

The problem is, of course, the command to get the file or directory owner (in the code example `gettheownersomehow`).

Is there a command or trick that I could use to get this?

Regards,
Frayja

pixellany · 08-31-2007, 01:13 PM

Code:

ls -l | awk '{print $3}'

Prints the owner of every file or directory in the current directory.

The AWK syntax is simply "print the 3rd field in every line of data"

To get the owner of only one file (filename):

Code:

ls -l | grep filename | awk '{print $3}'

unSpawn · 08-31-2007, 01:37 PM

If you want everything at once then try 'find' with "-printf "%U %f\n" ". If you want per file specs then use 'stat' ('stat -c %u'). For both 'find' and 'stat' goes you only need one binary, don't need to pipe output and munge it and whatever you want to print (numerical ID's, resolved names etc, etc) is configurable.

frayja · 08-31-2007, 01:49 PM

Great! Thanks for the fast reply, this solves my problem completely!

Now... back to my little script