按空格分割字符串，但忽略双引号或单引号之间的部分

Question

我需要将连续空格处的输入字符串拆分为字符串列表。输入可能包含单引号或双引号字符串，必须忽略它们。

如何在空格处分割字符串，但忽略带引号的字符串，所以分割的结果

me   you   "us and them"   'everyone else' them'

返回这个？

me
you
us and them
everyone else
them

这个问题的重复，但也需要忽略单引号字符串。

Answer 1

这个优秀的解决方案已被修改为忽略单引号字符串，并删除每个参数中的所有前导和尾随引号。

$people  = 'me   you   "us and them"   ''everyone else'' them'

$pattern = '(?x)
  [ ]+              # Split on one or more spaces (greedy)
  (?=               # if followed by one of the following:
    (?:[^"'']|      #   any character other a double or single quote, or 
    (?:"[^"]*")|    #   a double-quoted string, or
    (?:''[^'']*'')) #   a single-quoted string.
  *$)               # zero or more times to the end of the line.
'  
   
[regex]::Split($people, $pattern) -replace '^["'']|["'']$', ''

结果：

me
you
us and them
everyone else
them

简而言之，只要后面的所有内容都是非引号或带引号的字符串，此正则表达式就会匹配空格字符串 - 有效地将带引号的字符串视为单个字符。