Split a string where the separators can be escaped

Question

This is quite a simple regex, but I can't get my head around how I'd expand this regex so that it would allow me to use my delimiter character as long as it is escaped in the string. Here's what I have:

// Contents of str is exactly '|1|2|\|Three and Four\||5'
str.match(/[^|]/);

// Looking for: ['1', '2', '|Three and Four|', '5']

So currently my regex selects everything that isn't a | character and I get an array of each item. But what I want to do is to ignore the | character as a separator if it has been escaped first with \, but of course I don't want the \ to come through.

I know this'll be marked as a duplicate of the billion other regex questions, but I've tried to apply other solutions on here to my own, and played around with regex101.com. Alas, my Regex Fu is not strong.

P.s. Anyone know of any good resources to learn JS flavoured regex?

Answer 1

This should do it:

var str =  '|1|2|\\|Three and Four\\||5';
str.match(/((\\\|)|[^|])+/gi)

my output is this:

 ["1", "2", "\|Three and Four\|", "5"]

What I did was to create a pattern matching the \| string in first sub-pattern then match anything that is not a |. I escaped the the \ too because otherwise writing that string in javascript would just parse them to the escaped character.

Answer 2

If using a JavaScript with a regex engine that supports negative look-behinds (eg. Chrome), and in a case of only a single/simple escape shown, and no method to escape-the-escape, it's possible to use a relatively simple negative look-behind:

'|1|2|\\|Three and Four\\||5'.split(/(?<!\\)\|/)

# -> ["", "1", "2", "\|Three and Four\|", "5"]

This says to - in Chrome which supports negative look-behinds - split on a "|" that is not preceded by a "\".

Here is a method to convert a look-behind to a look-ahead for engine compatibility. Variations are also dicussed in RegEx needed to split javascript string on "|" but not "\|".

However, as pointed out, the above doesn't touch the \| sequence and thus leaves in the escape sequence.

---

Alternatively, a multistep approach can also solve this, which can also takes care of the escape character as part of the process.

1. Replace the escaped separators with an "alternate" character/string
2. Split on the remaining (non-escaped) separators
3. Convert the "alternate" character/string back in the individual components

In code,

str = '|1|2|\\|Three and Four\\||5'

# replace \| -> "alternative"
# this assumes that \\| (escape-the-escape) is not allowed
rep = str.replace(/\\[|]/g, '~~~~')

# replace back, without any of the escapes
res = rep.split('|').map(function (f) { return f.replace(/~~~~/g, "|") })

# res -> ["", "1", "2", "|Three and Four|", "5"]

Answer 3

Paul G Mihai's answer works fine but does not capture empty strings: a||b|c would return [ "a", "b", "c" ], instead of [ "a", "", "b", "c" ] as one might want.

Elaborating from his solution, here is a way to get also the empty strings, mimicking the same behaviour of split():

str.match(
  /((\\\|)|[^\|])*/gi
).filter(
  (e, i, a) => !(i > 0 && e == "" && a[i-1] != "")
)

What I do here is using match() with the same pattern, but allowing zero-length matches (* instead of +).

This gives me an array of matches with an empty string element for each separator found and at the end of the string, e.g.: a|b|c would return [ "a", "", "b", "", "c", "" ].

Then I filter() it, discarding any empty string element which comes after a non empty string element, so I get rid of the unwanted items.

This seems to handle edge cases correctly as well:

a||b|c         → ["a", "", "b", "c"]
a|b|||c        → ["a", "b", "", "", "c"]
a|b\|b|c|      → ["a", "b\|b", "c", ""]
|a|\|b\||c|    → ["", "a", "\|b\|", "c", ""]
(empty string) → [""]