Bash: Strip trailing linebreak from output - bash

Question

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?

Answer 1

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)"
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
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

Answer 2

One way:
wc -l < log.txt | xargs echo -n

Answer 3

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$//' | xxd
00000000: 666f 6f0a 6261 72 foo.bar
To remove multiple trailing newlines, pipe through:
sed -Ez '$ s/\n+$//'

Answer 4

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:]]}

Answer 5

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)

Answer 6

If you assign its output to a variable, bash automatically strips whitespace:
linecount=`wc -l < log.txt`

Answer 7

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.

Answer 8

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}"

Answer 9

Using Awk:
awk -v ORS="" '1' log.txt
Explanation:
-v assignment for ORS
ORS - output record separator set to blank. This will replace new line (Input record separator) with ""