这修整了前端和末端的多个空间

whatever=${whatever%% *}

无论= ${无论# *}

当IFS变量被设置为其他变量时，我需要从脚本中删除空白。依赖Perl实现了这个目的:

# trim() { echo $1; } # This doesn't seem to work, as it's affected by IFS

trim() { echo "$1" | perl -p -e 's/^\s+|\s+$//g'; }

strings="after --> , <-- before,  <-- both -->  "

OLD_IFS=$IFS
IFS=","
for str in ${strings}; do
  str=$(trim "${str}")
  echo "str= '${str}'"
done
IFS=$OLD_IFS

var="  a b  "
echo "$(set -f; echo $var)"

>a b

创建一个数组而不是变量，这将修剪所有的空格，制表符和换行符:

arr=( $(hg st -R "$path") )
if [[ -n "${arr[@]}" ]]; then
    printf -- '%s\n' "${arr[@]}"
fi

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.

你可以简单地使用echo进行修剪:

foo=" qsdqsd qsdqs q qs   "

# Not trimmed
echo \'$foo\'

# Trim
foo=`echo $foo`

# Trimmed
echo \'$foo\'