我有一个shell脚本与这段代码:

var=`hg st -R "$path"`
if [ -n "$var" ]; then
    echo $var
fi

但是条件代码总是执行,因为hg st总是打印至少一个换行符。

是否有一个简单的方法从$var中剥离空白(如PHP中的trim())?

or

有没有处理这个问题的标准方法?

我可以使用sed或AWK,但我认为有一个更优雅的解决方案来解决这个问题。


当前回答

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。

其他回答

# 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")")"
}

让我们定义一个包含开头、结尾和中间空格的变量:

FOO=' test test test '
echo -e "FOO='${FOO}'"
# > FOO=' test test test '
echo -e "length(FOO)==${#FOO}"
# > length(FOO)==16

如何删除tr中的所有空格(由[:space:]表示):

FOO=' test test test '
FOO_NO_WHITESPACE="$(echo -e "${FOO}" | tr -d '[:space:]')"
echo -e "FOO_NO_WHITESPACE='${FOO_NO_WHITESPACE}'"
# > FOO_NO_WHITESPACE='testtesttest'
echo -e "length(FOO_NO_WHITESPACE)==${#FOO_NO_WHITESPACE}"
# > length(FOO_NO_WHITESPACE)==12

如何仅删除前导空格:

FOO=' test test test '
FOO_NO_LEAD_SPACE="$(echo -e "${FOO}" | sed -e 's/^[[:space:]]*//')"
echo -e "FOO_NO_LEAD_SPACE='${FOO_NO_LEAD_SPACE}'"
# > FOO_NO_LEAD_SPACE='test test test '
echo -e "length(FOO_NO_LEAD_SPACE)==${#FOO_NO_LEAD_SPACE}"
# > length(FOO_NO_LEAD_SPACE)==15

如何删除尾随空格:

FOO=' test test test '
FOO_NO_TRAIL_SPACE="$(echo -e "${FOO}" | sed -e 's/[[:space:]]*$//')"
echo -e "FOO_NO_TRAIL_SPACE='${FOO_NO_TRAIL_SPACE}'"
# > FOO_NO_TRAIL_SPACE=' test test test'
echo -e "length(FOO_NO_TRAIL_SPACE)==${#FOO_NO_TRAIL_SPACE}"
# > length(FOO_NO_TRAIL_SPACE)==15

如何删除前导和尾随空格-链种子:

FOO=' test test test '
FOO_NO_EXTERNAL_SPACE="$(echo -e "${FOO}" | sed -e 's/^[[:space:]]*//' -e 's/[[:space:]]*$//')"
echo -e "FOO_NO_EXTERNAL_SPACE='${FOO_NO_EXTERNAL_SPACE}'"
# > FOO_NO_EXTERNAL_SPACE='test test test'
echo -e "length(FOO_NO_EXTERNAL_SPACE)==${#FOO_NO_EXTERNAL_SPACE}"
# > length(FOO_NO_EXTERNAL_SPACE)==14

或者,如果你的bash支持它,你可以替换echo -e "${FOO}" | sed…有了sed…<<<${FOO},像这样(用于尾随空格):

FOO_NO_TRAIL_SPACE="$(sed -e 's/[[:space:]]*$//' <<<${FOO})"

Read已经修整了空白,所以在bash中你可以这样做:

$ read foo <<< "   foo  bar   two spaces follow   "
$ echo ".$foo."
.foo  bar   two spaces follow.

POSIX兼容的版本稍长一些

$ read foo << END
   foo  bar   two spaces follow   
END
$ echo ".$foo."
.foo  bar   two spaces follow.

这修整了前端和末端的多个空间

whatever=${whatever%% *}

无论= ${无论# *}

使用这个简单的Bash参数展开:

$ x=" a z     e r ty "
$ echo "START[${x// /}]END"
START[azerty]END