"trim"函数删除所有水平空白:

ltrim () {
    if [[ $# -eq 0 ]]; then cat; else printf -- '%s\n' "$@"; fi | perl -pe 's/^\h+//g'
    return $?
}

rtrim () {
    if [[ $# -eq 0 ]]; then cat; else printf -- '%s\n' "$@"; fi | perl -pe 's/\h+$//g'
    return $?
}

trim () {
    ltrim "$@" | rtrim
    return $?
}

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<

我见过脚本只是使用变量赋值来完成工作:

$ xyz=`echo -e 'foo \n bar'`
$ echo $xyz
foo bar

空格会自动合并和修剪。必须小心shell元字符(潜在的注入风险)。

我还建议在shell条件句中使用双引号变量替换:

if [ -n "$var" ]; then

因为变量中的-o或其他内容可能会修改测试参数。

这是我见过的最简单的方法。它只使用Bash，只有几行，regexp很简单，它匹配所有形式的空白:

if [[ "$test" =~ ^[[:space:]]*([^[:space:]].*[^[:space:]])[[:space:]]*$ ]]
then 
    test=${BASH_REMATCH[1]}
fi

下面是一个用于测试的示例脚本:

test=$(echo -e "\n \t Spaces and tabs and newlines be gone! \t  \n ")

echo "Let's see if this works:"
echo
echo "----------"
echo -e "Testing:${test} :Tested"  # Ugh!
echo "----------"
echo
echo "Ugh!  Let's fix that..."

if [[ "$test" =~ ^[[:space:]]*([^[:space:]].*[^[:space:]])[[:space:]]*$ ]]
then 
    test=${BASH_REMATCH[1]}
fi

echo
echo "----------"
echo -e "Testing:${test}:Tested"  # "Testing:Spaces and tabs and newlines be gone!"
echo "----------"
echo
echo "Ah, much better."

我创建了以下函数。我不确定printf的可移植性如何，但这个解决方案的美妙之处在于，您可以通过添加更多字符代码来指定什么是“空白”。

    iswhitespace()
    {
        n=`printf "%d\n" "'$1'"`
        if (( $n != "13" )) && (( $n != "10" )) && (( $n != "32" )) && (( $n != "92" )) && (( $n != "110" )) && (( $n != "114" )); then
            return 0
        fi
        return 1
    }

    trim()
    {
        i=0
        str="$1"
        while (( i < ${#1} ))
        do
            char=${1:$i:1}
            iswhitespace "$char"
            if [ "$?" -eq "0" ]; then
                str="${str:$i}"
                i=${#1}
            fi
            (( i += 1 ))
        done
        i=${#str}
        while (( i > "0" ))
        do
            (( i -= 1 ))
            char=${str:$i:1}
            iswhitespace "$char"
            if [ "$?" -eq "0" ]; then
                (( i += 1 ))
                str="${str:0:$i}"
                i=0
            fi
        done
        echo "$str"
    }

#Call it like so
mystring=`trim "$mystring"`

这里有一个trim()函数，用于修整和规范化空白

#!/bin/bash
function trim {
    echo $*
}

echo "'$(trim "  one   two    three  ")'"
# 'one two three'

还有一种使用正则表达式的变体。

#!/bin/bash
function trim {
    local trimmed="$@"
    if [[ "$trimmed" =~ " *([^ ].*[^ ]) *" ]]
    then 
        trimmed=${BASH_REMATCH[1]}
    fi
    echo "$trimmed"
}

echo "'$(trim "  one   two    three  ")'"
# 'one   two    three'