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.)


9 Answers 9


You can use perl without chomp:

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

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

But why not use chomp itself:

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

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").


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
    Commented Jul 8, 2014 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. Commented Jul 8, 2014 at 12:01

I found bash was able to do what I wanted without any other tools. This isn't going to replicate chomp precisely but might help someone.

A command whose output ends with a newline will have it chomped during substitution:


It can be used as part of a pipeline with echo -n (quote the substitution to preserve the other newlines):

echo -n "$(my_command_with_a_newline)" |tail_or_whatever

Using printf "one\ntwo\n" like the OP:

$ echo -n "$(printf "one\ntwo\n")"; echo " done"
two done
  • Bash will remove all trailing newlines in command substitution, not just one, see for yourself: foobar="$(echo -e 'hello\n\n\n'); printf %s "$foobar"
    – Flimm
    Commented Aug 3, 2020 at 9:28

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 -0777pe '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 ([email protected])
# License: GPL
set -euf
sed -e :a -e '/^\n*$/{$d;N;ba' -e '}'


Print lines without newline, add a newline only if there is another line to print.

$ printf 'one\ntwo\n' | 

     awk '{ printf( "%s%s" , NR>1?"\n":"" , $0 ) }';   echo " done"

two done

#Other solutions If we were working with a file, we can just truncate one character from it (if it ends on a newline):

removeTrailNewline(){ [[ $(tail -c 1 "$1") ]] || truncate -s-1 "$1"; }

That is a fast solution as it needs to read only one character from the file and then remove it directly (truncate) without reading the whole file.

However, while working with data from stdin (an stream) the data must be read, all of it. And, it is "consumed" as soon as it is read. No backtrack (as with truncate). To find the end of an stream we need to read to the end of the stream. At that point, there is no way to move back on the input stream, the data has been already "consumed". This means that data must be stored in some form of buffer until we match the end of the stream and then do something with the data in the buffer.

The most obvious of solutions is to convert the stream into a file and process that file. But the question asks for some kind of filter of the stream. Not about the use of additional files.

###variable The naive solution would be to capture the whole input into a variable:

FilterOne(){ filecontents=$(cat; echo "x");        # capture the whole input
             filecontents=${filecontents%x};       # Remove the "x" added above.
             nl=$'\n';                             # use a variable for newline.
             printf '%s' "${filecontents%"$nl"}";  # Remove newline (if it exists).

printf 'one\ntwo'     | FilterOne ; echo 1done
printf 'one\ntwo\n'   | FilterOne ; echo 2done
printf 'one\ntwo\n\n' | FilterOne ; echo 3done

###memory It is possible to load a whole file in memory with sed. In sed it is impossible to avoid the trailing newline on the last line. GNU sed might avoid printing a trailing newline, but only if the source file is already missing it. So, no, simple sed can't help.

Except on GNU awk with the -z option:

sed -z 's/\(.*\)\n$/\1/'

With awk (any awk), slurp the whole stream, and printf it without the trailing newline.

awk '    { content = content $0 RS } 
     END { gsub( "\n$", "", content ); printf( "%s", content ) }

Loading a whole file into memory might not be a good idea, it may consume a lot of memory.

###Two lines in memory In awk, we can process two lines per loop by storing the previous line in a variable and printing the present one:

awk 'NR>1{print previous} {previous=$0} END {printf("%s",$0)}'

###Direct processing But we could we do better.

If we print the present line without a newline and print a newline only when a next line exists, we process one line at a time and the last line will not have a trailing newline:

awk 'NR==1{ printf("%s",$0);next }; { printf( "\n%s", $0 ) }'

Or, written in some other way:

awk 'NR>1{ print "" }; { printf( "%s", $0 ) }'


awk '{ printf( "%s%s" , NR>1?"\n":"" , $0 ) }'


$ printf 'one\ntwo\n' | awk '{ printf( "%s%s" , NR>1?"\n":"" , $0 ) }'; echo " done"
two done

Using Raku (formerly known as Perl_6)

~$ printf "one\ntwo\n" | raku -e 'print lines.join("\n")'; echo " done"
two done


~$ printf "one\ntwo" | raku -e 'print lines.join("\n")'; echo " done"
two done

In Raku, lines auto-chomps by default. the call to join("\n") puts the \n back in between the lines. Finally, print is used in Raku to output data because it doesn't add an extra \n at the end. [Should you need to output data with a trailing newline, use put].

Alternative form:

$  printf "one\ntwo\n" | raku -e 'lines.join("\n").print'; echo " done"
two done


~$  printf "one\ntwo" | raku -e 'lines.join("\n").print'; echo " done"
two done



The simplest I can think of is using tr.

echo string | tr '\n' ' '

That does the job for me.

  • 1
    This replaces all newlines with a space. The question was about replacing only the last newline.
    – Flimm
    Commented Jun 12, 2023 at 13:38

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