Check which files are opened by Firefox then sort by largest size.

lsof -p $(pidof firefox) | awk '/.mozilla/ { s = int($7/(2^20)); if(s>0) print (s)" MB -- "$9 | "sort -rn" }'
Just refining last proposal for this check, showing awk power to make more complex math (instead /1024/1024, 2^20). We don't need declare variable before run lsof, because $(command) returns his output. Also, awk can perform filtering by regexp instead to call grep. I changed the 0.0000xxxx messy output, with a more readable form purging all fractional numbers and files less than 1 MB.
Sample Output
41 MB -- /home/tzk/.mozilla/firefox/default/urlclassifier3.sqlite
6 MB -- /home/tzk/.mozilla/firefox/default/places.sqlite
3 MB -- /home/tzk/.mozilla/firefox/default/XPC.mfasl
2 MB -- /home/tzk/.mozilla/firefox/default/extensions/writearea@writearea.com/chrome/writearea.jar
2 MB -- /home/tzk/.mozilla/firefox/default/extensions/writearea@writearea.com/chrome/writearea.jar
2 MB -- /home/tzk/.mozilla/firefox/default/extensions/{4BBDD651-70CF-4821-84F8-2B918CF89CA3}-trash/chrome/febe.jar

10
By: tzk
2010-01-13 22:45:53

1 Alternatives + Submit Alt

  • Check which files are opened by Firefox then sort by largest size (in MB). You can see all files opened by just replacing grep to "/". Useful if you'd like to debug and check which extensions or files are taking too much memory resources in Firefox. Show Sample Output


    6
    FFPID=$(pidof firefox-bin) && lsof -p $FFPID | awk '{ if($7>0) print ($7/1024/1024)" MB -- "$9; }' | grep ".mozilla" | sort -rn
    josue · 2009-08-16 08:58:22 3

What Others Think

Nicely done, I would use "lsof -c firefox" in place of "lsof -p $(pidof firefox)" to tune it a little more.
zlemini · 527 weeks and 2 days ago

What do you think?

Any thoughts on this command? Does it work on your machine? Can you do the same thing with only 14 characters?

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