Counting open files per process

Have a look at the /proc/ file system:

ls /proc/$pid/fd/ | wc -l

To do this for all processes, use this:

cd /procfor pid in [0-9]*do    echo "PID = $pid with $(ls /proc/$pid/fd/ | wc -l) file descriptors"done

As a one-liner (filter by appending | grep -v "0 FDs"):

for pid in /proc/[0-9]*; do printf "PID %6d has %4d FDs\n" $(basename $pid) $(ls $pid/fd | wc -l); done

As a one-liner including the command name, sorted by file descriptor count in descending order (limit the results by appending | head -10):

for pid in /proc/[0-9]*; do p=$(basename $pid); printf "%4d FDs for PID %6d; command=%s\n" $(ls $pid/fd | wc -l) $p "$(ps -p $p -o comm=)"; done | sort -nr

Credit to @Boban for this addendum:

You can pipe the output of the script above into the following script to see the ten processes (and their names) which have the most file descriptors open:

  ...done | sort -rn -k5 | head | while read -r _ _ pid _ fdcount _do  command=$(ps -o cmd -p "$pid" -hc)  printf "pid = %5d with %4d fds: %s\n" "$pid" "$fdcount" "$command"done

Here's another approach to list the top-ten processes with the most open fds, probably less readable, so I don't put it in front:

find /proc -maxdepth 1 -type d -name '[0-9]*' \     -exec bash -c "ls {}/fd/ | wc -l | tr '\n' ' '" \; \     -printf "fds (PID = %P), command: " \     -exec bash -c "tr '\0' ' ' < {}/cmdline" \; \     -exec echo \; | sort -rn | head

Try this:

ps aux | sed 1d | awk '{print "fd_count=$(lsof -p " $2 " | wc -l) && echo " $2 " $fd_count"}' | xargs -I {} bash -c {}

I used this to find top filehandler-consuming processes for a given user (username) where dont have lsof or root access:

for pid in `ps -o pid -u username` ; do echo "$(ls /proc/$pid/fd/ 2>/dev/null | wc -l ) for PID: $pid" ; done  | sort -n | tail