这修整了前端和末端的多个空间

whatever=${whatever%% *}

无论= ${无论# *}

剥离一个前导和一个后导空间

trim()
{
    local trimmed="$1"

    # Strip leading space.
    trimmed="${trimmed## }"
    # Strip trailing space.
    trimmed="${trimmed%% }"

    echo "$trimmed"
}

例如:

test1="$(trim " one leading")"
test2="$(trim "one trailing ")"
test3="$(trim " one leading and one trailing ")"
echo "'$test1', '$test2', '$test3'"

输出:

'one leading', 'one trailing', 'one leading and one trailing'

去掉所有前导和尾随空格

trim()
{
    local trimmed="$1"

    # Strip leading spaces.
    while [[ $trimmed == ' '* ]]; do
       trimmed="${trimmed## }"
    done
    # Strip trailing spaces.
    while [[ $trimmed == *' ' ]]; do
        trimmed="${trimmed%% }"
    done

    echo "$trimmed"
}

例如:

test4="$(trim "  two leading")"
test5="$(trim "two trailing  ")"
test6="$(trim "  two leading and two trailing  ")"
echo "'$test4', '$test5', '$test6'"

输出:

'two leading', 'two trailing', 'two leading and two trailing'

Bash有一个称为参数展开的特性，它允许基于所谓的模式替换字符串(模式类似于正则表达式，但有基本的区别和限制)。 [flussence的原文:Bash有正则表达式，但它们隐藏得很好:]

下面演示如何从变量值中删除所有空白(甚至来自内部)。

$ var='abc def'
$ echo "$var"
abc def
# Note: flussence's original expression was "${var/ /}", which only replaced the *first* space char., wherever it appeared.
$ echo -n "${var//[[:space:]]/}"
abcdef

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

要从左到第一个单词删除空格和制表符，输入:

echo "     This is a test" | sed "s/^[ \t]*//"

cyberciti.biz /技巧/ delete-leading-spaces-from-front-of-each-word.html

使用这个简单的Bash参数展开:

$ x=" a z     e r ty "
$ echo "START[${x// /}]END"
START[azerty]END