Bash：从输出中去除尾随换行符
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Bash: Strip trailing linebreak from output
提问by hexacyanide
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?
当我在 Bash 中执行命令（或者更具体地说，wc -l < log.txt）时，输出后面包含一个换行符。我该如何摆脱它？
回答by Steve
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:
如果您的预期输出是单行，您可以简单地从输出中删除所有换行符。如果愿意，管道到“tr”实用程序或 Perl 并不少见：
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:
您还可以使用命令替换来删除尾随的换行符：
echo -n "$(wc -l < log.txt)"
printf "%s" "$(wc -l < log.txt)"
Please note that I disagree with the OP's decision to choose the accepted answer. I believe one should avoid using 'xargs' where possible. Yes, it's a cool toy. No, you do not need it here.
请注意，我不同意 OP 选择已接受答案的决定。我相信应该尽可能避免使用“xargs”。是的，这是一个很酷的玩具。不，你在这里不需要它。
If your expected output may contain multiple lines, you have another decision to make:
如果您的预期输出可能包含多行，您需要做出另一个决定：
If you want to remove MULTIPLE newline characters from the end of the file, again use cmd substitution:
如果要从文件末尾删除多个换行符，请再次使用 cmd 替换：
printf "%s" "$(< log.txt)"
If you want to strictly remove THE LAST newline character from a file, use Perl:
如果要从文件中严格删除最后一个换行符，请使用 Perl：
perl -pe 'chomp if eof' log.txt
Note that if you are certain you have a trailing newline character you want to remove, you can use 'head' from GNU coreutils to select everything except the last byte. This should be quite quick:
请注意，如果您确定要删除尾随换行符，则可以使用 GNU coreutils 中的“head”来选择除最后一个字节之外的所有内容。这应该很快：
head -c -1 log.txt
Also, for completeness, you can quickly check where your newline (or other special) characters are in your file using 'cat' and the 'show-all' flag. The dollar sign character will indicate the end of each line:
此外，为了完整性，您可以使用“cat”和“show-all”标志快速检查换行符（或其他特殊）字符在文件中的位置。美元符号字符将指示每一行的结尾：
cat -A log.txt
回答by Satya
One way:
单程：
wc -l < log.txt | xargs echo -n
回答by Andrey Taranov
There is also direct support for white space removal in Bash variable substitution:
在Bash 变量替换中也直接支持删除空格：
testvar=$(wc -l < log.txt)
trailing_space_removed=${testvar%%[[:space:]]}
leading_space_removed=${testvar##[[:space:]]}
回答by zero2cx
回答by nneonneo
If you assign its output to a variable, bashautomatically strips whitespace:
如果将其输出分配给变量，bash则会自动去除空格：
linecount=`wc -l < log.txt`
回答by Tom Hale
If you want to remove only the last newline, pipe through:
如果您只想删除最后一个换行符，请通过管道：
sed -z '$ s/\n$//'
sedwon't add a \0to then end of the stream if the delimiter is set to NULvia -z, whereas to create a POSIX text file (defined to end in a \n), it will always output a final \nwithout -z.
sed\0如果分隔符设置为NULvia -z，则不会在流的结尾添加 a ，而要创建 POSIX 文本文件（定义为以 a 结尾\n），它将始终输出\n不带-z.
Eg:
例如：
$ { echo foo; echo bar; } | sed -z '$ s/\n$//'; echo tender
foo
bartender
And to prove no NULadded:
并证明没有NUL添加：
$ { echo foo; echo bar; } | sed -z '$ s/\n$//' | xxd
00000000: 666f 6f0a 6261 72 foo.bar
To remove multiple trailing newlines, pipe through:
要删除多个尾随换行符，请通过管道：
sed -Ez '$ s/\n+$//'
回答by Rastislav Hasicek
If you want to print output of anything in Bash without end of line, you echo it with the -nswitch.
如果你想在没有行尾的情况下打印 Bash 中的任何输出，你可以用-nswitch回显它。
If you have it in a variable already, then echo it with the trailing newlinecropped:
如果您已经在一个变量中拥有它，那么用裁剪的尾随换行符来回显它：
$ testvar=$(wc -l < log.txt)
$ echo -n $testvar
Or you can do it in one line, instead:
或者你可以在一行中完成，而不是：
$ echo -n $(wc -l < log.txt)
