Find and sort by file size

Question

I have this command to find files larger than 2 KB and sort by size:

find . -size +2k -name *.log -printf "%p \t%k kb\n" | ls -lS

but the ls -lS gives the files that are less than 2 KB also.

How can I display the names and sizes of the files that are larger than 2 KB, sorted by size?

RudiC · Accepted Answer · 2018-08-22 06:46:46Z

3

Try

 find . -size +2k  -printf "%p \t%k kb\n" | sort -k2n

answered Aug 22 '18 at 6:46

RudiC

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what does k2n do? – Curious Aug 22 '18 at 6:47
1

Assuming none of the file names contain blanks or newline characters. – Stéphane Chazelas Aug 22 '18 at 8:13
Note that strictly speaking, that sorts by disk usage, not by size, though that's likely what the OP wants anyway. – Stéphane Chazelas Aug 22 '18 at 8:14
-k2n sets the sort key to the second field, to be interpreted numerically. – RudiC Aug 22 '18 at 8:41

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RalfFriedl · Accepted Answer · 2018-08-22 06:44:37Z

1

The ls command will list the current directory, not read something from stdin.

Use

find . -size +2k -name *.log -printf "%s %p \t%k kb\n" | sort -n | sed -e '/^[0-9]* //'

answered Aug 22 '18 at 6:44

RalfFriedl

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Stéphane Chazelas · Accepted Answer · 2019-03-18 14:53:49Z

1

With zsh:

zmodload zsh/stat
printf '%s\n' **/*.log(DLK+2oLe'[
  stat -A blk +block -- $REPLY && REPLY+=" $((blk / 2)) KiB"]')

That reports the disk usage of the log files whose size is greater that 2048, ordered by size.

answered Aug 22 '18 at 9:11

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3 Answers 3