Lua: split string into words unless quoted

问题

So I have the following code to split a string between whitespaces:

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

The result:

I
am
'the
text'

But I need to do this:

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

How can I do this?

Edit: just to complement the question, the idea is to pass parameters from a program to another program. Here is the pull request that I am working, currently in review: https://github.com/mpv-player/mpv/pull/1619

回答1:

There may be ways to do this with clever parsing, but an alternative way may be to keep track of a simple state and merge fragments based on detection of quoted fragments. Something like this may work:

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

This will print:

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

Updated to handle mixed and escaped quotes. Updated to remove quotes. Updated to handle quoted words.

回答2:

Try this:

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

回答3:

Lua Patterns aren't powerful to handle this task properly. Here is an LPeg solution adapted from the Lua Lexer. It handles both single and double quotes.

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

Output:

This is a string
another string
these
are
words

来源：https://stackoverflow.com/questions/28664139/lua-split-string-into-words-unless-quoted

标签

string

lua

lua-patterns