How to get list of process ids that are in a given group id

Question

Going by the man page, I would expect this to work:

ps ah -o pid,pgrp -G 18322

But this shows exactly the same list as without the -G arguments. I want a saner way to produce this output:

ps ah -o pid,pgrp | perl -e 'while(<STDIN>){ my @ws = split " ", $_; if ($ws[1] eq $ARGV[0]) { print $ws[0]."\n" } }' 18322

(Thanks to mst on #perl for the perl-fu)

Here's a more traditional command line version, (thanks againt to mst), but still a bit awkward. $process_group needs to be set beforehand:

ps ah -o pgrp,pid | egrep '^'$process_group' ' | awk '{print $2}'

Answer 1

Use pgrep instead:

pgrep -g 18322

From man pgrep:

   -g, --pgroup pgrp,...
          Only  match  processes in the process group IDs listed.  Process
          group 0 is translated into pgrep's or pkill's own process group.

Alternatively, you can just parse the ps output in simpler ways:

ps xh -o pgrp,pid | awk '$1==18322{print $2}'

Or just simplify your (needlessly complex) original Perl approach:

ps xh -o pgrp,pid | perl -lane 'print $F[1] if $F[0] eq 5592'

Or just grep:

ps xh -o pgrp,pid | grep -Po '\s*5592\s*\K.+'