Bash: Strip trailing linebreak from output

One way:

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

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

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+$//'