I'm looking for something that behaves like Perl's chomp. I'm looking for a command that simply prints its input, minus the last character if it's a newline:

$ printf "one\ntwo\n" | COMMAND_IM_LOOKING_FOR ; echo " done"
two done
$ printf "one\ntwo" | COMMAND_IM_LOOKING_FOR ; echo " done"
two done

(Command substitution in Bash and Zsh deletes all trailing new lines, but I'm looking for something that deletes one trailing new line at most.)


This should work:

printf "one\ntwo\n" | awk 'NR>1{print PREV} {PREV=$0} END{printf("%s",$0)}' ; echo " done"

The script always prints previous line instead of current, and the last line is treated differently.

What it does in more detail:

  1. NR>1{print PREV} Print previous line (except the first time).
  2. {PREV=$0} Stores current line in PREV variable.
  3. END{printf("%s",$0)} Finally, print last line withtout line break.

Also note this would remove at most one empty line at the end (no support for removing "one\ntwo\n\n\n").


You can use perl without chomp:

$ printf "one\ntwo\n" | perl -0 -pe 's/\n\Z//'; echo " done"
two done

$ printf "one\ntwo" | perl -0 -pe 's/\n\Z//'; echo " done"
two done

But why not use chomp itself:

$ printf "one\ntwo\n" | perl -pe 'chomp if eof'; echo " done"

If you want an exact equivalent to chomp, the first method that comes to my mind is the awk solution that LatinSuD already posted. I'll add some other methods that don't implement chomp but implement some common tasks that chomp is often used for.

When you stuff some text into a variable, all newlines at the end are stripped. So all these commands produce the same single-line output:

echo "$(printf 'one\ntwo') done"
echo "$(printf 'one\ntwo\n') done"
echo "$(printf 'one\ntwo\n\n') done"
echo "$(printf 'one\ntwo\n\n\n\n\n\n\n\n\n\n') done"

If you want to append some text to the last line of a file or of a command's output, sed can be convenient. With GNU sed and most other modern implementations, this works even if the input doesn't end in a newline¹; however, this won't add a newline if there wasn't one already.

sed '$ s/$/ done/'

¹ However this doesn't work with all sed implementations: sed is a text processing tool, and a file that isn't empty and doesn't end with a newline character is not a text file.

  • This is not exactly equivalent to chomp, as chomp only deletes at most one trailing newline. – Flimm Jul 8 '14 at 11:48
  • @Flimm Yes, the most obvious exact equivalent to chomp would be the awk solution that LatinSuD already posted. But in many cases chomp is just a tool to do a job, and I provide ways to do some common tasks. Let me update my answer to clarify this. – Gilles Jul 8 '14 at 12:01

Another perl approach. This one reads the entire input into memory so it might not be a good idea for large amounts of data (use cuonglm's or the awk approach for that):

$ printf "one\ntwo\n" | perl -00pe 's/\n$//'; echo " done"
two done

I snagged this from a github repo somewhere, but can't find where


# Delete all trailing blank lines.
# From http://sed.sourceforge.net/sed1line.txt
# Version: 1.3.0
# Created: 2011-01-02
# Updated: 2015-01-25
# Contact: Joel Parker Henderson (joel@joelparkerhenderson.com)
# License: GPL
set -euf
sed -e :a -e '/^\n*$/{$d;N;ba' -e '}'

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