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

FFPID=$(pidof firefox-bin) && lsof -p $FFPID | awk '{ if($7>0) print ($7/1024/1024)" MB -- "$9; }' | grep ".mozilla" | sort -rn
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.
Sample Output
29.8359 MB -- /home/user/.mozilla/firefox/l0ad8wrj.default/urlclassifier3.sqlite
4.91797 MB -- /home/user/.mozilla/firefox/l0ad8wrj.default/places.sqlite
4.11752 MB -- /home/user/.mozilla/firefox/l0ad8wrj.default/Cache/_CACHE_003_
4.0147 MB -- /home/user/.mozilla/firefox/l0ad8wrj.default/XUL.mfasl
2.73011 MB -- /home/user/.mozilla/firefox/l0ad8wrj.default/extensions/{000a9d1c-beef-4f90-9363-039d445309b8}/lib/ff35/
2.25571 MB -- /home/user/.mozilla/firefox/l0ad8wrj.default/extensions/{c45c406e-ab73-11d8-be73-000a95be3b12}/chrome/webdeveloper.jar
2.03915 MB -- /home/user/.mozilla/firefox/l0ad8wrj.default/Cache/_CACHE_001_
2.03215 MB -- /home/user/.mozilla/firefox/l0ad8wrj.default/Cache/_CACHE_002_
1.25815 MB -- /home/user/.mozilla/firefox/l0ad8wrj.default/extensions/{DDC359D1-844A-42a7-9AA1-88A850A938A8}/chrome/chrome.jar
0.382812 MB -- /home/user/.mozilla/firefox/l0ad8wrj.default/cookies.sqlite
0.191917 MB -- /home/user/.mozilla/firefox/l0ad8wrj.default/extensions/{a7c6cf7f-112c-4500-a7ea-39801a327e5f}/chrome/fireftp.jar
0.185547 MB -- /home/user/.mozilla/firefox/l0ad8wrj.default/signons.sqlite
0.166023 MB -- /home/user/.mozilla/firefox/l0ad8wrj.default/extensions/{582195F5-92E7-40a0-A127-DB71295901D7}/chrome/gmanager.jar
0.149414 MB -- /home/user/.mozilla/firefox/l0ad8wrj.default/ubiquity_ann.sqlite

By: josue
2009-08-16 08:58:22

1 Alternatives + Submit Alt

  • 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. Show Sample Output

    lsof -p $(pidof firefox) | awk '/.mozilla/ { s = int($7/(2^20)); if(s>0) print (s)" MB -- "$9 | "sort -rn" }'
    tzk · 2010-01-13 22:45:53 1

What Others Think

you can also get the open files list with from /proc/$(pidof firefox-bin)/fd under linux. So you could use find to follow all the links-to-files and then ls them sorted by size: find /proc/$(pidof firefox-bin)/fd -xtype f -exec readlink -f {} \; | xargs ls -lS
bwoodacre · 666 weeks and 3 days ago
Another way to look at this is to use the pmap utility which looks at /proc/$PID/maps: pmap $(pidof firefox-bin) | sort -rn --key=2 | grep -v anon The amounts given are upper limits for the memory consumed, like the "VIRT" field in top. Read more at:
bwoodacre · 666 weeks and 3 days ago
Thanks, in general we can list the largest open files by: lsof -s | awk '{ if($7>0) print ($7/1024/1024)" MB -- "$9; }' | sort -nr
vdchuyen · 658 weeks and 5 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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