Linux中“ lsof”命令的替代方法

Question

I need a command in Linux(SUSE & RHEL) to find the list of open files as "lsof" command is taking time to give output.Even "lsof -n" command also takes a lot of time. Do we have any alternative command to "lsof"?

Thanks, Sanghamitra

Answer 1

You can read the open files from the proc file system.

for p in /proc/{0..9}*; do
  i=$(basename "$p")
  for f in "$p"/fd/*; do
    l=$(readlink -e "$f")
    if [ "$l" ]; then
      echo "$i: $l"
    fi
  done
done | sort -u | sort -n

Answer 2

In short, use lsof -n .

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lsof waits a lot because

1. It uses a lot of time to scan all the devices in /dev
2. It uses also a lot of time to get a reverse dns query from all the endpoints of the network sockets it found. Many of them doesn't have a valid revdns (like 0.0.0.0 ), and querying for them results timeout.
3. It does it as a single-threaded, not event-oriented app.

You can give it a significant acceleration by disabling the dns resolution part with a -n .

Anyways, extending @ceving 's answer, you could also watch not only the list of the opened file descriptors, but also the list of the files mapped into the address space of the processes ( cat /proc/<pid>/map ).

---

Typically, if you have a process running surprisingly slowly, a

strace -f -tt -o sux <your command line>

will dump into the file sux the kernel calls what the process did, with timestamps. In essence, so you will be able to track, what exactly this process did and how long .