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

#!/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]
#[]

其他回答

我创建了以下函数。我不确定printf的可移植性如何,但这个解决方案的美妙之处在于,您可以通过添加更多字符代码来指定什么是“空白”。

    iswhitespace()
    {
        n=`printf "%d\n" "'$1'"`
        if (( $n != "13" )) && (( $n != "10" )) && (( $n != "32" )) && (( $n != "92" )) && (( $n != "110" )) && (( $n != "114" )); then
            return 0
        fi
        return 1
    }

    trim()
    {
        i=0
        str="$1"
        while (( i < ${#1} ))
        do
            char=${1:$i:1}
            iswhitespace "$char"
            if [ "$?" -eq "0" ]; then
                str="${str:$i}"
                i=${#1}
            fi
            (( i += 1 ))
        done
        i=${#str}
        while (( i > "0" ))
        do
            (( i -= 1 ))
            char=${str:$i:1}
            iswhitespace "$char"
            if [ "$?" -eq "0" ]; then
                (( i += 1 ))
                str="${str:0:$i}"
                i=0
            fi
        done
        echo "$str"
    }

#Call it like so
mystring=`trim "$mystring"`

我发现我需要从混乱的sdiff输出中添加一些代码来清理它:

sdiff -s column1.txt column2.txt | grep -F '<' | cut -f1 -d"<" > c12diff.txt 
sed -n 1'p' c12diff.txt | sed 's/ *$//g' | tr -d '\n' | tr -d '\t'

这将删除尾随空格和其他不可见字符。

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})"

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 ${1+"$@"}
}

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