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

Python有一个函数strip(),它的工作原理与PHP的trim()相同,因此我们可以使用一点内联Python来创建一个易于理解的实用程序:

alias trim='python -c "import sys; sys.stdout.write(sys.stdin.read().strip())"'

这将删除开头和结尾的空白(包括换行符)。

$ x=`echo -e "\n\t   \n" | trim`
$ if [ -z "$x" ]; then echo hi; fi
hi

其他回答

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

我必须测试一个命令的结果(数字),但似乎变量的结果包含空格和一些不可打印的字符。因此,即使经过“修整”,这种比较也是错误的。 我通过从变量中提取数值部分来解决它:

numerical_var=$(echo ${var_with_result_from_command} | grep -o "[0-9]*")

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'
# 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")")"
}

为了删除字符串开头和结尾的所有空格(包括行末字符):

echo $variable | xargs echo -n

这也将删除重复的空格:

echo "  this string has a lot       of spaces " | xargs echo -n

Produces: 'this string has a lot of spaces'