# Trim whitespace from both ends of specified parameter

trim () {
    read -rd '' $1 <<<"${!1}"
}

# Unit test for trim()

test_trim () {
    local foo="$1"
    trim foo
    test "$foo" = "$2"
}

test_trim hey hey &&
test_trim '  hey' hey &&
test_trim 'ho  ' ho &&
test_trim 'hey ho' 'hey ho' &&
test_trim '  hey  ho  ' 'hey  ho' &&
test_trim $'\n\n\t hey\n\t ho \t\n' $'hey\n\t ho' &&
test_trim $'\n' '' &&
test_trim '\n' '\n' &&
echo passed

#!/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]
#[]

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'

这没有不必要的通配符问题，而且，内部空白是未修改的(假设$IFS被设置为默认值，即' \t\n')。

它一直读取到第一个换行符(但不包括换行符)或字符串的结尾，以先到者为准，并删除任何前导和尾随空格以及\t字符的混合。如果你想保留多行(同时去掉开头和结尾换行符)，请使用read -r -d " var << eof;但是请注意，如果您的输入恰好包含\neof，它将在之前被切断。(其他形式的空白，即\r、\f和\v，即使您将它们添加到$IFS，也不会被剥离。)

read -r var << eof
$var
eof

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。