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find the biggest files recursively, no matter how many

Terminal - find the biggest files recursively, no matter how many
find . -type f -printf '%20s %p\n' | sort -n | cut -b22- | tr '\n' '\000' | xargs -0 ls -laSr
2009-08-13 13:13:33
User: fsilveira
Functions: cut find ls sort tr xargs
10
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find the biggest files recursively, no matter how many

This command will find the biggest files recursively under a certain directory, no matter if they are too many. If you try the regular commands ("find -type f -exec ls -laSr {} +" or "find -type f -print0 | xargs -0 ls -laSr") the sorting won't be correct because of command line arguments limit.

This command won't use command line arguments to sort the files and will display the sorted list correctly.

Show sample output | Add to favourites | Report as malicious

Alternatives

There are 6 alternatives - vote for the best!

Terminal - Alternatives
find . -type f|perl -lne '@x=sort {$b->[0]<=>$a->[0]}[(stat($_))[7],$_],@x;splice(@x,11);print "@{$x[0]}";END{for(@x){print "@$_"}'
2012-01-08 14:43:43
User: bazzargh
Functions: find perl
Tags: sort perl find
-2
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find the biggest files recursively, no matter how many

A different approach to the problem - maintain a small sorted list, print the largest as we go, then the top 10 at the end. I often find that the find and sort take a long time, and the large file might appear near the start of the find. By printing as we go, I get better feedback. The sort used in this will be much slower on perls older than 5.8.

Show sample output | Comments (0) | Add to favourites | Report as malicious

Know a better way?

If you can do better, submit your command here.

What others think

the sort manpage is a little cryptic, but you can sort on fields other than the beginning of the line (similar to cut):

find . -type f -ls | sort -n --key=7

Pipe that to "cut -b68-" to get only the filenames.

Comment by bwoodacre 197 weeks and 2 days ago

I thought about "find -ls" before but it is bad because the file name isn't always at the same position, depending on the owner/group/size/time/date strings length. The time/date length changes for some specific locales.

Comment by fsilveira 197 weeks and 2 days ago

agreed! Another shoddy way is to pipe into "awk '{print $11}' " or some such to get the filenames, but they still have to be quoted or null-separated. However, I think you can get rid of cut and tr if you modify the -printf format string:

find . -type f -printf '%s\t"%p"\n' | sort -n | cut -f2 | xargs ls -laS

the tab is for cut and the quotes for xargs. Another option: on a single fs, use inode numbers to avoid messy filenames.

Comment by bwoodacre 197 weeks and 1 day ago

The 'cut' could not be avoided in that command and the 'tr' is to handle filenames with spaces correctly, although filenames with '\n' (which are *much* harder to be found) will be missed.

Comment by fsilveira 197 weeks and 1 day ago

I see, here's maybe even a shorter version. Have you seen the -print0 option to find?

find -type f -print0 | xargs -0 ls -la | sort -nr --key=5

this handles ANY type of filename even those with newlines and spaces.

Comment by bwoodacre 196 weeks and 4 days ago

Your point of view

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