对于我的用例，最好的选择是:

grep -oP '\w+' file

基本上，这是一个匹配连续非空格字符的正则表达式。这意味着任何类型和任何数量的空白都不匹配。-o参数在不同的行上输出每个匹配的单词。

(A)要将一个句子分成它的单词(空格分隔)，你可以简单地使用默认的IFS

array=( $string )

运行以下代码片段的示例

#!/bin/bash

sentence="this is the \"sentence\"   'you' want to split"
words=( $sentence )

len="${#words[@]}"
echo "words counted: $len"

printf "%s\n" "${words[@]}" ## print array

将输出

words counted: 8
this
is
the
"sentence"
'you'
want
to
split

As you can see you can use single or double quotes too without any problem Notes: -- this is basically the same of mob's answer, but in this way you store the array for any further needing. If you only need a single loop, you can use his answer, which is one line shorter :) -- please refer to this question for alternate methods to split a string based on delimiter. (B) To check for a character in a string you can also use a regular expression match. Example to check for the presence of a space character you can use:

regex='\s{1,}'
if [[ "$sentence" =~ $regex ]]
    then
        echo "Space here!";
fi

只需使用shell内置的“set”即可。例如,

set $text

在此之后，$text中的单个单词将以$1、$2、$3等形式存在。对于健壮性，通常是这样做的

set -- junk $text
shift

处理$text为空或以破折号开头的情况。例如:

text="This is          a              test"
set -- junk $text
shift
for word; do
  echo "[$word]"
done

这个打印

[This]
[is]
[a]
[test]

另一个例子(使用Perl):

$ echo foo bar baz | perl -nE 'say for split /\s/'
foo
bar
baz

$ echo foo bar baz | sed 's/ /\n/g'

foo
bar
baz

对于我的用例，最好的选择是:

grep -oP '\w+' file

基本上，这是一个匹配连续非空格字符的正则表达式。这意味着任何类型和任何数量的空白都不匹配。-o参数在不同的行上输出每个匹配的单词。