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'

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'

# 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

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)

Use:

trim() {
    local orig="$1"
    local trmd=""
    while true;
    do
        trmd="${orig#[[:space:]]}"
        trmd="${trmd%[[:space:]]}"
        test "$trmd" = "$orig" && break
        orig="$trmd"
    done
    printf -- '%s\n' "$trmd"
}

它适用于各种空格，包括换行符， 不需要修改shop。 它保留内部空白，包括换行符。

单元测试(用于手动检查):

#!/bin/bash

. trim.sh

enum() {
    echo "   a b c"
    echo "a b c   "
    echo "  a b c "
    echo " a b c  "
    echo " a  b c  "
    echo " a  b  c  "
    echo " a      b  c  "
    echo "     a      b  c  "
    echo "     a  b  c  "
    echo " a  b  c      "
    echo " a  b  c      "
    echo " a N b  c  "
    echo "N a N b  c  "
    echo " Na  b  c  "
    echo " a  b  c N "
    echo " a  b  c  N"
}

xcheck() {
    local testln result
    while IFS='' read testln;
    do
        testln=$(tr N '\n' <<<"$testln")
        echo ": ~~~~~~~~~~~~~~~~~~~~~~~~~ :" >&2
        result="$(trim "$testln")"
        echo "testln='$testln'" >&2
        echo "result='$result'" >&2
    done
}

enum | xcheck

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