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I am looking for a command that would list me number of open files for each user, sorted descending.

For example lsof -u postgres | wc - would list them only for the postgres user, but I do want to see on which user I may loose handlers.

Note: some question may apply to PIDs instead of user names, but still, I am looking for a one liner.

sorin
  • 12,189

2 Answers2

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I ran Warren Lavallee's script, and found it difficult to interpret.

Instead, I extended the original script to step through all the users to give the following:-

for f in $(sed</etc/passwd 's/:.*$//g'); do ( echo -n $f ' '; lsof -u $f 2>/dev/null | wc -l ); done | grep -v ' 0$'

This works fine on Ubuntu 14.04. The final grep removes the zero entries, as a lot of the users will not have active processes. You can also add | sort -rnk 2 to the end of the command to sort into descending file count order.

You will need to be in a root shell to make sure you can see the files from all users.

AFH
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If you are open to pipe's, how about this:

lsof | perl -pe 's/\s\s+/ /g' | cut -d' ' -f3 | grep -v ^USER | sort | uniq -c | sort -rn

I tested this on MacOS 10.10.2 (Yosemite), OEL 6.6, and Ubuntu 10.04.01.

Warren Lavallee
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