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.