我有一个shell脚本与这段代码:

var=`hg st -R "$path"`
if [ -n "$var" ]; then
    echo $var
fi

但是条件代码总是执行,因为hg st总是打印至少一个换行符。

是否有一个简单的方法从$var中剥离空白(如PHP中的trim())?

or

有没有处理这个问题的标准方法?

我可以使用sed或AWK,但我认为有一个更优雅的解决方案来解决这个问题。


当前回答

答案有很多,但我仍然认为我刚刚写的剧本值得一提,因为:

it was successfully tested in the shells bash/dash/busybox shell it is extremely small it doesn't depend on external commands and doesn't need to fork (->fast and low resource usage) it works as expected: it strips all spaces and tabs from beginning and end, but not more important: it doesn't remove anything from the middle of the string (many other answers do), even newlines will remain special: the "$*" joins multiple arguments using one space. if you want to trim & output only the first argument, use "$1" instead if doesn't have any problems with matching file name patterns etc

脚本:

trim() {
  local s2 s="$*"
  until s2="${s#[[:space:]]}"; [ "$s2" = "$s" ]; do s="$s2"; done
  until s2="${s%[[:space:]]}"; [ "$s2" = "$s" ]; do s="$s2"; done
  echo "$s"
}

用法:

mystring="   here     is
    something    "
mystring=$(trim "$mystring")
echo ">$mystring<"

输出:

>here     is
    something<

其他回答

Read已经修整了空白,所以在bash中你可以这样做:

$ read foo <<< "   foo  bar   two spaces follow   "
$ echo ".$foo."
.foo  bar   two spaces follow.

POSIX兼容的版本稍长一些

$ read foo << END
   foo  bar   two spaces follow   
END
$ echo ".$foo."
.foo  bar   two spaces follow.

我发现我需要从混乱的sdiff输出中添加一些代码来清理它:

sdiff -s column1.txt column2.txt | grep -F '<' | cut -f1 -d"<" > c12diff.txt 
sed -n 1'p' c12diff.txt | sed 's/ *$//g' | tr -d '\n' | tr -d '\t'

这将删除尾随空格和其他不可见字符。

我将简单地使用sed:

function trim
{
    echo "$1" | sed -n '1h;1!H;${;g;s/^[ \t]*//g;s/[ \t]*$//g;p;}'
}

a)单行字符串使用示例

string='    wordA wordB  wordC   wordD    '
trimmed=$( trim "$string" )

echo "GIVEN STRING: |$string|"
echo "TRIMMED STRING: |$trimmed|"

输出:

GIVEN STRING: |    wordA wordB  wordC   wordD    |
TRIMMED STRING: |wordA wordB  wordC   wordD|

b)多行字符串使用示例

string='    wordA
   >wordB<
wordC    '
trimmed=$( trim "$string" )

echo -e "GIVEN STRING: |$string|\n"
echo "TRIMMED STRING: |$trimmed|"

输出:

GIVEN STRING: |    wordAA
   >wordB<
wordC    |

TRIMMED STRING: |wordAA
   >wordB<
wordC|

c)最后说明: 如果你不喜欢使用函数,对于单行字符串,你可以简单地使用“更容易记住”的命令,比如:

echo "$string" | sed -e 's/^[ \t]*//' | sed -e 's/[ \t]*$//'

例子:

echo "   wordA wordB wordC   " | sed -e 's/^[ \t]*//' | sed -e 's/[ \t]*$//'

输出:

wordA wordB wordC

在多行字符串上使用上述方法也可以,但请注意,它也会切断任何尾随/前导内部多行空间,就像GuruM在评论中注意到的那样

string='    wordAA
    >four spaces before<
 >one space before<    '
echo "$string" | sed -e 's/^[ \t]*//' | sed -e 's/[ \t]*$//'

输出:

wordAA
>four spaces before<
>one space before<

所以如果你介意保留这些空格,请使用我回答开头的函数!

d)解释sed语法“find and replace”在函数trim中使用的多行字符串:

sed -n '
# If the first line, copy the pattern to the hold buffer
1h
# If not the first line, then append the pattern to the hold buffer
1!H
# If the last line then ...
$ {
    # Copy from the hold to the pattern buffer
    g
    # Do the search and replace
    s/^[ \t]*//g
    s/[ \t]*$//g
    # print
    p
}'

让我们定义一个包含开头、结尾和中间空格的变量:

FOO=' test test test '
echo -e "FOO='${FOO}'"
# > FOO=' test test test '
echo -e "length(FOO)==${#FOO}"
# > length(FOO)==16

如何删除tr中的所有空格(由[:space:]表示):

FOO=' test test test '
FOO_NO_WHITESPACE="$(echo -e "${FOO}" | tr -d '[:space:]')"
echo -e "FOO_NO_WHITESPACE='${FOO_NO_WHITESPACE}'"
# > FOO_NO_WHITESPACE='testtesttest'
echo -e "length(FOO_NO_WHITESPACE)==${#FOO_NO_WHITESPACE}"
# > length(FOO_NO_WHITESPACE)==12

如何仅删除前导空格:

FOO=' test test test '
FOO_NO_LEAD_SPACE="$(echo -e "${FOO}" | sed -e 's/^[[:space:]]*//')"
echo -e "FOO_NO_LEAD_SPACE='${FOO_NO_LEAD_SPACE}'"
# > FOO_NO_LEAD_SPACE='test test test '
echo -e "length(FOO_NO_LEAD_SPACE)==${#FOO_NO_LEAD_SPACE}"
# > length(FOO_NO_LEAD_SPACE)==15

如何删除尾随空格:

FOO=' test test test '
FOO_NO_TRAIL_SPACE="$(echo -e "${FOO}" | sed -e 's/[[:space:]]*$//')"
echo -e "FOO_NO_TRAIL_SPACE='${FOO_NO_TRAIL_SPACE}'"
# > FOO_NO_TRAIL_SPACE=' test test test'
echo -e "length(FOO_NO_TRAIL_SPACE)==${#FOO_NO_TRAIL_SPACE}"
# > length(FOO_NO_TRAIL_SPACE)==15

如何删除前导和尾随空格-链种子:

FOO=' test test test '
FOO_NO_EXTERNAL_SPACE="$(echo -e "${FOO}" | sed -e 's/^[[:space:]]*//' -e 's/[[:space:]]*$//')"
echo -e "FOO_NO_EXTERNAL_SPACE='${FOO_NO_EXTERNAL_SPACE}'"
# > FOO_NO_EXTERNAL_SPACE='test test test'
echo -e "length(FOO_NO_EXTERNAL_SPACE)==${#FOO_NO_EXTERNAL_SPACE}"
# > length(FOO_NO_EXTERNAL_SPACE)==14

或者,如果你的bash支持它,你可以替换echo -e "${FOO}" | sed…有了sed…<<<${FOO},像这样(用于尾随空格):

FOO_NO_TRAIL_SPACE="$(sed -e 's/[[:space:]]*$//' <<<${FOO})"

Trim()删除空白(以及制表符、不可打印字符;为了简单起见,我正在考虑使用空格)。我的解决方案是:

var="$(hg st -R "$path")" # I often like to enclose shell output in double quotes
var="$(echo "${var}" | sed "s/\(^ *\| *\$\)//g")" # This is my suggestion
if [ -n "$var" ]; then
 echo "[${var}]"
fi

sed命令只修剪开头和结尾的空白,但它也可以被输送到第一个命令,导致:

var="$(hg st -R "$path" | sed "s/\(^ *\| *\$\)//g")"
if [ -n "$var" ]; then
 echo "[${var}]"
fi