Use:

var=`expr "$var" : "^\ *\(.*[^ ]\)\ *$"`

它去掉了开头和结尾的空格，我认为这是最基本的解决方案。不是Bash内置的，但'expr'是coreutils的一部分，所以至少不需要像sed或AWK这样的独立实用程序。

使用Bash的扩展模式匹配功能(shopt -s extglob)，你可以这样使用:

修剪{# # * ()}

删除任意数量的前导空格。

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
}'

# Strip leading and trailing white space (new line inclusive).
trim(){
    [[ "$1" =~ [^[:space:]](.*[^[:space:]])? ]]
    printf "%s" "$BASH_REMATCH"
}

OR

# Strip leading white space (new line inclusive).
ltrim(){
    [[ "$1" =~ [^[:space:]].* ]]
    printf "%s" "$BASH_REMATCH"
}

# Strip trailing white space (new line inclusive).
rtrim(){
    [[ "$1" =~ .*[^[:space:]] ]]
    printf "%s" "$BASH_REMATCH"
}

# Strip leading and trailing white space (new line inclusive).
trim(){
    printf "%s" "$(rtrim "$(ltrim "$1")")"
}

OR

# Strip leading and trailing specified characters.  ex: str=$(trim "$str" $'\n a')
trim(){
    if [ "$2" ]; then
        trim_chrs="$2"
    else
        trim_chrs="[:space:]"
    fi

    [[ "$1" =~ ^["$trim_chrs"]*(.*[^"$trim_chrs"])["$trim_chrs"]*$ ]]
    printf "%s" "${BASH_REMATCH[1]}"
}

OR

# Strip leading specified characters.  ex: str=$(ltrim "$str" $'\n a')
ltrim(){
    if [ "$2" ]; then
        trim_chrs="$2"
    else
        trim_chrs="[:space:]"
    fi

    [[ "$1" =~ ^["$trim_chrs"]*(.*[^"$trim_chrs"]) ]]
    printf "%s" "${BASH_REMATCH[1]}"
}

# Strip trailing specified characters.  ex: str=$(rtrim "$str" $'\n a')
rtrim(){
    if [ "$2" ]; then
        trim_chrs="$2"
    else
        trim_chrs="[:space:]"
    fi

    [[ "$1" =~ ^(.*[^"$trim_chrs"])["$trim_chrs"]*$ ]]
    printf "%s" "${BASH_REMATCH[1]}"
}

# Strip leading and trailing specified characters.  ex: str=$(trim "$str" $'\n a')
trim(){
    printf "%s" "$(rtrim "$(ltrim "$1" "$2")" "$2")"
}

OR

建立在moskit的expr soulution…

# Strip leading and trailing white space (new line inclusive).
trim(){
    printf "%s" "`expr "$1" : "^[[:space:]]*\(.*[^[:space:]]\)[[:space:]]*$"`"
}

OR

# Strip leading white space (new line inclusive).
ltrim(){
    printf "%s" "`expr "$1" : "^[[:space:]]*\(.*[^[:space:]]\)"`"
}

# Strip trailing white space (new line inclusive).
rtrim(){
    printf "%s" "`expr "$1" : "^\(.*[^[:space:]]\)[[:space:]]*$"`"
}

# Strip leading and trailing white space (new line inclusive).
trim(){
    printf "%s" "$(rtrim "$(ltrim "$1")")"
}

有一个解决方案只使用Bash内置的通配符:

var="    abc    "
# remove leading whitespace characters
var="${var#"${var%%[![:space:]]*}"}"
# remove trailing whitespace characters
var="${var%"${var##*[![:space:]]}"}"   
printf '%s' "===$var==="

下面是同样的包装在一个函数中:

trim() {
    local var="$*"
    # remove leading whitespace characters
    var="${var#"${var%%[![:space:]]*}"}"
    # remove trailing whitespace characters
    var="${var%"${var##*[![:space:]]}"}"
    printf '%s' "$var"
}

你传递要以引号形式修剪的字符串，例如:

trim "   abc   "

这个解决方案的一个优点是它可以与任何posix兼容的shell一起工作。

参考

从Bash变量中删除前导和尾随空格(原始源代码)

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。