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

创建一个数组而不是变量,这将修剪所有的空格,制表符和换行符:

arr=( $(hg st -R "$path") )
if [[ -n "${arr[@]}" ]]; then
    printf -- '%s\n' "${arr[@]}"
fi

其他回答

你可以使用tr删除换行符:

var=`hg st -R "$path" | tr -d '\n'`
if [ -n $var ]; then
    echo $var
done

这就是我所做的,结果完美而简单:

the_string="        test"
the_string=`echo $the_string`
echo "$the_string"

输出:

test

使用AWK:

echo $var | awk '{gsub(/^ +| +$/,"")}1'

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

答案有很多,但我仍然认为我刚刚写的剧本值得一提,因为:

it was successfully tested in the shells bash/dash/busybox shell it is extremely small it doesn't depend on external commands and doesn't need to fork (->fast and low resource usage) it works as expected: it strips all spaces and tabs from beginning and end, but not more important: it doesn't remove anything from the middle of the string (many other answers do), even newlines will remain special: the "$*" joins multiple arguments using one space. if you want to trim & output only the first argument, use "$1" instead if doesn't have any problems with matching file name patterns etc

脚本:

trim() {
  local s2 s="$*"
  until s2="${s#[[:space:]]}"; [ "$s2" = "$s" ]; do s="$s2"; done
  until s2="${s%[[:space:]]}"; [ "$s2" = "$s" ]; do s="$s2"; done
  echo "$s"
}

用法:

mystring="   here     is
    something    "
mystring=$(trim "$mystring")
echo ">$mystring<"

输出:

>here     is
    something<