How to find files containing exactly 16 lines? - bash

Question

I have to find files that containing exactly 16 lines in Bash.
My idea is:
find -type f | grep '/^...$/'
Does anyone know how to utilise find + grep or maybe find + awk?
Then,
Move the matching files another directory.
Deleting all non-matching files.

Answer 1

I would just do:
wc -l **/* 2>/dev/null | awk '$1=="16"'

Answer 2

Keep it simple:
find . -type f |
while IFS= read -r file
do
size=$(wc -l < "$file")
if (( size == 16 ))
then
mv -- "$file" /wherever/you/like
else
rm -f -- "$file"
fi
done
If your file names can contain newlines then google for the find and read options to handle that.

Answer 3

You should use grep instead of wc because wc counts newline characters \n and will not count if the last line doesn't ends with a newline.
e.g.
grep -cH '' * 2>/dev/null | awk -F: '$2==16'
for more correct approach (without error messages, and without argument list too long error) you should combine it with the find and xargs commands, like
find . -type f -print0 | xargs -0 grep -cH '' | awk -F: '$2==16'
if you don't want count empty lines (so only lines what contains at least one character), you can replace the '' with the '.'. And instead of awk, you can use second grep, like:
find . -type f -print0 | xargs -0 grep -cH '.' | grep ':16$'
this will find all files what are contains 16 non-empty lines... and so on..

Answer 4

GNU sed
sed -E '/^.{16}$/!d' file

Answer 5

A pure bash version:
#!/usr/bin/bash
for f in *; do # Look for files in the present dir
[ ! -f "$f" ] && continue # Skip not simple files
cnt=0
# Count the first 17 lines
while ((cnt<17)) && read x; do ((++cnt)); done<"$f"
if [ $cnt == 16 ] ; then echo "Move '$f'"
else echo "Delete '$f'"
fi
done

Answer 6

This snippet will do the work:
find . -type f -readable -exec bash -c \
'if(( $(grep -m 17 -c "" "$0")==16 )); then echo "file $0 has 16 lines"; else echo "file $0 doesn'"'"'t have 16 lines"; fi' {} \;
Hence, if you need to delete the files that are not 16 lines long, and move those who are 16 lines long to folder /my/folder, this will do:
find . -type f -readable -exec bash -c \
'if(( $(grep -m 17 -c "" "$0")==16 )); then mv -nv "$0" /my/folder; else rm -v "$0"; fi' {} \;
Observe the quoting for "$0" so that it's safe regarding any file name with funny symbols in it (spaces, ...).
I'm using the -v option so that rm and mv are verbose (I like to know what's happening). The -n option to mv is no-clobber: a security to not overwrite an existing file; this option might not be available if you have an old system.
The good thing about this method. It's really safe regarding any filename containing funny symbols.
The bad thing(s). It forks a bash and a grep and an mv or rm for each file found. This can be quite slow. This can be fixed using trickier stuff (while still remaining safe regarding funny symbols in filenames). If you really need it, I can give you a possible answer. It will also break if file can't be (re)moved.
Remark. I'm using the -readable option to find, so that it only considers files that are readable. If you have this option, use it, you'll have a more robust command!

Answer 7

I would go with
find . -type f | while read f ; do
[[ "${f##*/}" =~ ^.{16}$ ]] && mv "${f}" <any_directory> || rm -f "${f}"
done
or
find . -type f | while read f ; do
[[ $(echo -n "${f##*/}" | wc -c) -eq 16 ]] && mv "${f}" <any_directory> || rm -f "${f}"
done
Replace <any_directory> with the directory you actually want to move the files to.
BTW, find command will go sub-directories. if you don't want this, then you should change the find command to fit your need.