Bash: Strip trailing linebreak from output

When I execute commands in Bash (or to be specific, wc -l < log.txt), the output contains a linebreak after it. How do I get rid of it?

If your expected output is a single line, you can simply remove all newline characters from the output. It would not be uncommon to pipe to the tr utility, or to Perl if preferred:

wc -l < log.txt | tr -d '\n'

wc -l < log.txt | perl -pe 'chomp'

You can also use command substitution to remove the trailing newline:

echo -n "$(wc -l < log.txt)"

printf "%s" "$(wc -l < log.txt)"

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If your expected output may contain multiple lines, you have another decision to make:

If you want to remove MULTIPLE newline characters from the end of the file, again use cmd substitution:

printf "%s" "$(< log.txt)"

If you want to strictly remove THE LAST newline character from a file, use Perl:

perl -pe 'chomp if eof' log.txt

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Note that if you are certain you have a trailing newline character you want to remove, you can use head from GNU coreutils to select everything except the last byte. This should be quite quick:

head -c -1 log.txt

Also, for completeness, you can quickly check where your newline (or other special) characters are in your file using cat and the 'show-all' flag -A. The dollar sign character will indicate the end of each line:

cat -A log.txt

One way:

wc -l < log.txt | xargs echo -n

If you want to remove only the last newline, pipe through:

sed -z '$ s/\n$//'

sed won't add a \0 to then end of the stream if the delimiter is set to NUL via -z, whereas to create a POSIX text file (defined to end in a \n), it will always output a final \n without -z.

Eg:

$ { echo foo; echo bar; } | sed -z '$ s/\n$//'; echo tender
foo
bartender

And to prove no NUL added:

$ { echo foo; echo bar; } | sed -z '$ s/\n$//' | xxd
00000000: 666f 6f0a 6261 72                        foo.bar

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To remove multiple trailing newlines, pipe through:

sed -Ez '$ s/\n+$//'

There is also direct support for white space removal in Bash variable substitution:

testvar=$(wc -l < log.txt)
trailing_space_removed=${testvar%%[[:space:]]}
leading_space_removed=${testvar##[[:space:]]}

If you assign its output to a variable, bash automatically strips whitespace:

linecount=`wc -l < log.txt`

If you want to print output of anything in Bash without end of line, you echo it with the -n switch.

If you have it in a variable already, then echo it with the trailing newline cropped:

$ testvar=$(wc -l < log.txt)
$ echo -n $testvar

Or you can do it in one line, instead:

$ echo -n $(wc -l < log.txt)

printf already crops the trailing newline for you:

$ printf '%s' $(wc -l < log.txt)

Detail:

• printf will print your content in place of the %s string place holder.
• If you do not tell it to print a newline (%s\n), it won't.

Adding this for my reference more than anything else ^_^

You can also strip a new line from the output using the bash expansion magic

VAR=$'helloworld\n'

CLEANED="${VAR%$'\n'}"

echo "${CLEANED}"

Using Awk:

awk -v ORS="" '1' log.txt 

Explanation:

1. -v assignment for ORS
2. ORS - output record separator set to blank. This will replace new line (Input record separator) with ""