top 3 users in a directory

[Rosli Ahmad]

Hi,
I wonder if there's a quick way to list top 3 users consuming the most diskspaces in a certain directory.

Thanks.

[MarkSyder]

Do these users have their own sub-directories within the directory? If so, then du -k will help you:

du -k|sort -k1,1nr|pg

If not, I don't know of a way.

Mark Syder (like the drink but spelt different)

[Henk Geurts]

du -sk * |sort -n
shows the disk usage of the directories beneath the point you run this command from.
not exactly the answer to your question but works for /home
regards.

[Kent Ostby]

Rosli --

If you don't care about subdirectories then you can use this:

Create a file called "useme.awk" which contains:

{addit[$3]+=$5}
END {for (idx1 in addit) {printf ("%8s %10d\n",idx1,addit[idx1]);}}

Then do this:

ll | grep -v -e total -e ^d | awk -f useme.awk

Best regards,

Kent M. Ostby

[Noel Miranda]

You can use the "quot" command which give the space taken up by users in blocks and sort on the block field.

[Peter Godron]

Rosli,
not the quickest way I guess, but it seems to work ok.
Regards

[Hein van den Heuvel]

And in perl, much similar to kent's solution but with sort build in:

while (<>){
($mode,$links,$owner,$group,$size)=split;
$total{$owner} +=$size;
}
foreach $owner (sort {$total{$b} <=> $total{$a}} keys %total) {
printf ("%12d %s\n",$total{$owner},$owner) if ($i++ < 3);
}


Hein.