Question

I’m trying to parse (in Ruby) what’s effectively the UNIX passwd file-format: comma delimiters, with an escape character \ such that anything escaped should be considered literally. I’m trying to use a regular expression for this, but I’m coming up short — even when using Oniguruma for lookahead/lookbehind assertions.

Essentially, all of the following should work:

a,b,c    # => ["a", "b", "c"]
\a,b\,c  # => ["a", "b,c"]
a,b,c\
d        # => ["a", "b", "c\nd"]
a,b\\\,c # => ["a", "b\,c"]

Any ideas?

The first response looks pretty good. With a file containing

\a,,b\\\,c\,d,e\\f,\\,\
g

it gives:

[["\\a,"], [","], ["b\\\\\\,c\\,d,"], ["e\\\\f,"], ["\\\\,"], ["\\\ng\n"], [""]]

which is pretty close. I don’t need the unescaping done on this first pass, as long as everything splits correctly on the commas. I tried Oniguruma and ended up with (the much longer):

Oniguruma::ORegexp.new(%{
  (?:       # - begins with (but doesn't capture)
    (?<=\A) #   - start of line
    |       #   - (or) 
    (?<=,)  #   - a comma
  )

  (?:           # - contains (but doesn't capture)
    .*?         #   - any set of characters
    [^\\\\]?    #   - not ending in a slash
    (\\\\\\\\)* #   - followed by an even number of slashes
  )*?

  (?:      # - ends with (but doesn't capture)
    (?=\Z) #   - end of line
    |      #   - (or)
    (?=,)) #   - a comma
  },

  'mx'
).scan(s)

Answer 1

Try this:

s.scan(/((?:\\.|[^,])*,?)/m)

It doesn’t translate the characters following a \, but that can be done afterwards as a separate step.