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."

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'

#!/bin/bash

function trim
{
    typeset trimVar
    eval trimVar="\${$1}"
    read trimVar << EOTtrim
    $trimVar
EOTtrim
    eval $1=\$trimVar
}

# Note that the parameter to the function is the NAME of the variable to trim, 
# not the variable contents.  However, the contents are trimmed.


# Example of use:
while read aLine
do
    trim aline
    echo "[${aline}]"
done < info.txt



# File info.txt contents:
# ------------------------------
# ok  hello there    $
#    another  line   here     $
#and yet another   $
#  only at the front$
#$



# Output:
#[ok  hello there]
#[another  line   here]
#[and yet another]
#[only at the front]
#[]

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
}'

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<

这将删除字符串中的所有空格，

 VAR2="${VAR2//[[:space:]]/}"

/替换字符串中第一次出现的空格和//所有出现的空格。也就是说，所有的空格都被- nothing取代