Lua：除非引用,否则将字符串拆分为单词

所以我有以下代码在空格之间拆分字符串：

text = "I am 'the text'"
for string in text:gmatch("%s+") do
    print(string)
end

结果：

I
am
'the
text'

但我需要这样做：

I
am
the text --[[yep,without the quotes]]

我怎样才能做到这一点？

编辑：只是为了补充问题,我们的想法是将参数从程序传递到另一个程序.这是我正在工作的拉取请求,目前正在审核中：https://github.com/mpv-player/mpv/pull/1619

解决方法

有可能通过巧妙的解析来实现这一点,但另一种方法可能是跟踪简单状态并根据引用片段的检测合并片段.这样的事情可能有用：

local text = [[I "am" 'the text' and "some more text with '" and "escaped \" text"]]
local spat,epat,buf,quoted = [=[^(['"])]=],[=[(['"])$]=]
for str in text:gmatch("%s+") do
  local squoted = str:match(spat)
  local equoted = str:match(epat)
  local escaped = str:match([=[(\*)['"]$]=])
  if squoted and not quoted and not equoted then
    buf,quoted = str,squoted
  elseif buf and equoted == quoted and #escaped % 2 == 0 then
    str,quoted = buf .. ' ' .. str,nil,nil
  elseif buf then
    buf = buf .. ' ' .. str
  end
  if not buf then print((str:gsub(spat,""):gsub(epat,""))) end
end
if buf then print("Missing matching quote for "..buf) end

这将打印：

I
am
the text
and
some more text with '
and
escaped \" text

更新以处理混合和转义引号.已更新以删除引号.更新以处理引用的单词.