On Fri, Dec 15, 2017 at 12:52 PM, Galen Seitz wrote:
> find / -path /tmp -prune -o -path /sys -prune -o -path /proc -prune -o
> -printf %u:%g\\n | awk '{usergroup[$0]++}; END {for(key in usergroup) printf
> "%-20s %d\n",key,usergroup[key]}'
You could also limit find to the filesystems that you
On 12/15/17 11:55, Galen Seitz wrote:
> Hi,
>
> Can anyone suggest a quick way to get a count of the number files owned
> by each uid, or alternatively, print each unique uid that exists in the
> filesystem? Bonus points for doing the same thing for gid's in the same
> pass.
Thanks for the other
Depending on your definition of quick:
ls -lR | grep ^[d-] | perl -ane '$user{$F[2]}++; $group{$F[3]}++; END{
print " Users\n"; for $s (keys %user) {print "$s $user{$s}\n"} print "
Groups\n"; for $s (keys %group) { print "$s $group{$s}\n"}}'
On 2017-12-15 11:55, Galen Seitz wrote:
Hi,
On Fri, 15 Dec 2017, Galen Seitz wrote:
Hi,
Can anyone suggest a quick way to get a count of the number files
owned by each uid, or alternatively, print each unique uid that
exists in the filesystem? Bonus points for doing the same thing for
gid's in the same pass.
Quickly? Probably not.
First knee jerk response is find. Do you just want normal files? or
other things too, like directories, devices, links, et al?
On Fri, Dec 15, 2017 at 11:55 AM, Galen Seitz wrote:
> Hi,
>
> Can anyone suggest a quick way to get a count of the number files owned
> by each uid, or alternatively,
Hi,
Can anyone suggest a quick way to get a count of the number files owned
by each uid, or alternatively, print each unique uid that exists in the
filesystem? Bonus points for doing the same thing for gid's in the same
pass.
thanks,
galen
--
Galen Seitz
gal...@seitzassoc.com
__
