Counting open files per process

问题

I'm working on an application that monitors the processes' resources and gives a periodic report in Linux, but I faced a problem in extracting the open files count per process.

This takes quite a while if I take all of the files and group them according to their PID and count them.

How can I take the open files count for each process in Linux?

回答1:

Have a look at the /proc/ file system:

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

To do this for all processes, use this:

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

EDIT: 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 -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

回答2:

Try this:

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

回答3:

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

来源：https://stackoverflow.com/questions/21752067/counting-open-files-per-process

标签

Linux

bash