您可以使用老式的tr。例如，它返回git存储库中已修改文件的数量，删除空白。

MYVAR=`git ls-files -m|wc -l|tr -d ' '`

这修整了前端和末端的多个空间

whatever=${whatever%% *}

无论= ${无论# *}

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'

从Bash指南的通配符部分

在参数展开中使用extglob

 #Turn on extended globbing  
shopt -s extglob  
 #Trim leading and trailing whitespace from a variable  
x=${x##+([[:space:]])}; x=${x%%+([[:space:]])}  
 #Turn off extended globbing  
shopt -u extglob  

下面是相同的函数封装在函数中(注意:需要引用传递给函数的输入字符串):

trim() {
    # Determine if 'extglob' is currently on.
    local extglobWasOff=1
    shopt extglob >/dev/null && extglobWasOff=0 
    (( extglobWasOff )) && shopt -s extglob # Turn 'extglob' on, if currently turned off.
    # Trim leading and trailing whitespace
    local var=$1
    var=${var##+([[:space:]])}
    var=${var%%+([[:space:]])}
    (( extglobWasOff )) && shopt -u extglob # If 'extglob' was off before, turn it back off.
    echo -n "$var"  # Output trimmed string.
}

用法:

string="   abc def ghi  ";
#need to quote input-string to preserve internal white-space if any
trimmed=$(trim "$string");  
echo "$trimmed";

如果我们将函数更改为在subshell中执行，我们不必担心检查extglob的当前shell选项，我们可以只设置它而不影响当前shell。这极大地简化了函数。我还更新了位置参数“就地”，所以我甚至不需要一个局部变量

trim() {
    shopt -s extglob
    set -- "${1##+([[:space:]])}"
    printf "%s" "${1%%+([[:space:]])}" 
}

so:

$ s=$'\t\n \r\tfoo  '
$ shopt -u extglob
$ shopt extglob
extglob         off
$ printf ">%q<\n" "$s" "$(trim "$s")"
>$'\t\n \r\tfoo  '<
>foo<
$ shopt extglob
extglob         off

还有一个单元测试的解决方案，它从stdin中删除$IFS，并适用于任何输入分隔符(甚至$'\0'):

ltrim()
{
    # Left-trim $IFS from stdin as a single line
    # $1: Line separator (default NUL)
    local trimmed
    while IFS= read -r -d "${1-}" -u 9
    do
        if [ -n "${trimmed+defined}" ]
        then
            printf %s "$REPLY"
        else
            printf %s "${REPLY#"${REPLY%%[!$IFS]*}"}"
        fi
        printf "${1-\x00}"
        trimmed=true
    done 9<&0

    if [[ $REPLY ]]
    then
        # No delimiter at last line
        if [ -n "${trimmed+defined}" ]
        then
            printf %s "$REPLY"
        else
            printf %s "${REPLY#"${REPLY%%[!$IFS]*}"}"
        fi
    fi
}

rtrim()
{
    # Right-trim $IFS from stdin as a single line
    # $1: Line separator (default NUL)
    local previous last
    while IFS= read -r -d "${1-}" -u 9
    do
        if [ -n "${previous+defined}" ]
        then
            printf %s "$previous"
            printf "${1-\x00}"
        fi
        previous="$REPLY"
    done 9<&0

    if [[ $REPLY ]]
    then
        # No delimiter at last line
        last="$REPLY"
        printf %s "$previous"
        if [ -n "${previous+defined}" ]
        then
            printf "${1-\x00}"
        fi
    else
        last="$previous"
    fi

    right_whitespace="${last##*[!$IFS]}"
    printf %s "${last%$right_whitespace}"
}

trim()
{
    # Trim $IFS from individual lines
    # $1: Line separator (default NUL)
    ltrim ${1+"$@"} | rtrim ${1+"$@"}
}

"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 $?
}