Lua：将字符串拆分为单词，除非引用

Question

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]]

我该怎么做？

编辑： 只是为了补充问题，这个想法是将参数从一个程序传递到另一个程序。这是我正在处理的拉取请求，目前正在审核中：https://github.com/mpv-player/mpv/pull/1619

Answer 1

可能有一些方法可以通过巧妙的解析来做到这一点，但另一种方法可能是跟踪一个简单的状态并根据对引用片段的检测来合并片段。这样的事情可能会奏效：

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

已更新以处理混合引号和转义引号。更新以删除引号。已更新以处理引用的单词。

Answer 2

试试这个：

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

Answer 3

Lua 模式不足以正确处理此任务。这是一个LPeg solution adapted from the Lua Lexer。它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

Lua：将字符串拆分为单词，除非引用

Lua: split string into words unless quoted

string

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lua-patterns