Lua:将字符串拆分为单词,除非被引用。

我有以下代码来分割一个字符串中的空格:

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

结果:

I
am
'the
text'

但我需要做到这一步:

I
am
the text --[[是的,没有引号]]

我该如何做到这一点?

编辑: 补充问题,我的想法是将一个程序的参数传递给另一个程序。这是我正在处理的拉请求,目前正在审核中:https://github.com/mpv-player/mpv/pull/1619

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用户1442917
用户1442917

可能有一些巧妙的解析方式,但另一种方法可能是通过跟踪简单状态并基于检测到引用片段合并片段。可以尝试这种方式:

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, buf, 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。更新以删除引号。更新以处理带引号的单词。

2015-02-22 23:29:45
用户107090
用户107090
text = [[I am 'the text' and '' here is "another text in quotes" and this is the end]]

local e = 0
while true do
    local b = e+1
    b = text:find("%S",b)
    if b==nil then break end
    if text:sub(b,b)=="'" then
        e = text:find("'",b+1)
        b = b+1
    elseif text:sub(b,b)=='"' then
        e = text:find('"',b+1)
        b = b+1
    else
        e = text:find("%s",b+1)
    end
    if e==nil then e=#text+1 end
    print("["..text:sub(b,e-1).."]")
end

尝试一下:

text = [[我是文本'这里是""引号中的另一个文本"这就是结尾"]]

local e = 0
while true do
    local b = e+1
    b = text:find("%S",b)
    if b==nil then break end
    if text:sub(b,b)=="'" then
        e = text:find("'",b+1)
        b = b+1
    elseif text:sub(b,b)=='"' then
        e = text:find('"',b+1)
        b = b+1
    else
        e = text:find("%s",b+1)
    end
    if e==nil then e=#text+1 end
    print("["..text:sub(b,e-1).."]")
end
2015-02-23 01:27:29
用户2726734
用户2726734

Lua模式不足以处理这个任务,这里提供一个改编自Lua LexerLPEG解决方案,可以处理单引号和双引号。

local lpeg = require 'lpeg'

local P, S, C, Cc, Ct = lpeg.P, lpeg.S, lpeg.C, lpeg.Cc, lpeg.Ct

local function token(id, patt) return Ct(Cc(id) * C(patt)) end

local singleq = P "'" * ((1 - S "'\r\n\f\\") + (P '\\' * 1)) ^ 0 * "'"
local doubleq = P '"' * ((1 - S '"\r\n\f\\') + (P '\\' * 1)) ^ 0 * '"'

local white = token('whitespace', S('\r\n\f\t ')^1)
local word = token('word', (1 - S("' \r\n\f\t\""))^1)

local string = token('string', singleq + doubleq)

local tokens = Ct((string + white + word) ^ 0)

input = [["This is a string" 'another string' these are words]]
for _, tok in ipairs(lpeg.match(tokens, input)) do
  if tok[1] ~= "whitespace" then
     if tok[1] == "string" then
        print(tok[2]:sub(2,-2)) -- cut off quotes
     else
       print(tok[2])
     end
  end
end

输出:

This is a string
another string
these
are
words
2015-02-23 13:14:09