Comparing owners and permissions of content of two folders?

Question

How to to compare owners and permissions of content of two folders? Is there something like diff command which compare recursively two folders and display owner and permissions differences?

Answer 1

The solution, as with all things, is a perl script:

#!/usr/bin/perl

use File::Find;

my $directory1 = '/tmp/temp1';
my $directory2 = '/tmp/temp2';

find(\&hashfiles, $directory1);

sub hashfiles {
  my $file1 = $File::Find::name;
  (my $file2 = $file1) =~ s/^$directory1/$directory2/;

  my $mode1 = (stat($file1))[2] ;
  my $mode2 = (stat($file2))[2] ;

  my $uid1 = (stat($file1))[4] ;
  my $uid2 = (stat($file2))[4] ;

  print "Permissions for $file1 and $file2 are not the same\n" if ( $mode1 != $mode2 );
  print "Ownership for $file1 and $file2 are not the same\n" if ( $uid1 != $uid2 );
}

Look at http://perldoc.perl.org/functions/stat.html and http://perldoc.perl.org/File/Find.html for more info, particularly the stat one if you want to compare other file attributes.

If files don't exist in directory2 but exist in directory1, there will also be output because the stat will be different.

Answer 2

ls -al will show the permissions, if both of them are in the same folder you will get something like this:

drwxr-xr-x 4 root  root 4096 nov 28 20:48 temp
drwxr-xr-x 2 lucas 1002 4096 mrt 24 22:33 temp2

The 3rd column is the owner, the 4th is the group.