How do I make part of a regular expression optional in Ruby?

Question

To match the following:

On Mar 3, 2011 11:05 AM, "mr person" 
wrote:

I have the following regular expression:

/(On.* (?:Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sep|Oct|Nov|Dec) \d{1,2}, [12]\d{3}.* at \d{1,2}:\d{1,2} (?:AM|PM),.*wrote:)/m

Is there a way to make the at optional? so if it's there great, if not, it still matches?

+1 negaging the -1, I just don't think its right is all, sorry... — , Mar 09 '11 at 01:59
@sln: Did you mean "negating"? Also, why do you feel the downvote is unjustified? — Andrew Grimm, Mar 09 '11 at 02:12

kindall · Accepted Answer · 2011-03-09T00:21:18.810

24

Sure. Put it in parentheses, put a question mark after it. Include one of the spaces (since otherwise you'll be trying to match two spaces if the "at" is missing.) (at )? (or as someone else suggested, (?:at )? to avoid it being captured).

edited Mar 09 '11 at 00:21

answered Mar 09 '11 at 00:06

kindall

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score 18 · Answer 2 · answered Mar 09 '11 at 00:07

18

Don't forget (?:) to make sure the bracketed expression doesn't get captured

(?:at)?

answered Mar 09 '11 at 00:07

Andrew Grimm

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This works for complex expressions too e.g. /(\w+(\[\w+\](?:\[\d+\])?)+)(?:\[\w+\])/; – Donato Nov 11 '14 at 20:58
I never knew how to make it not get captured when I used parens for optional items, thanks! – Sam Lehman Nov 18 '16 at 16:39

DigitalRoss · Answer 3 · 2011-03-09T00:27:58.603

4

Sure, you just need to group the optional part...

   (at )*

And, ok, I guess that will match at at at at, so you might want to just do:

   (at )?

edited Mar 09 '11 at 00:27

answered Mar 09 '11 at 00:08

DigitalRoss

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That should be a `?`, not an `*`. – Andrew Grimm Mar 09 '11 at 00:11

score 2 · Answer 4 · answered Mar 09 '11 at 00:32

Others got your answer. This is just an aside re: Regular Expressions.

When you say "conditions" in regular expressions, it refers to the regex language. Like any language, its a branch in code execution, but the code is a different regular expression path, the "code" of regular expressions.

So in psudo code: if (evaluation is true) do this regular sub-expression, else do this other sub-expression.

This conditional exists in advanced regular expression engines ... Perl.
Perl uses the most advanced regular expression engine that exists. In version 6 and beyond it will be an integral part of the language, where code and expression intermingle seamlessly.

Perl 5.10 has this construct:
(?(condition)yes-pattern|no-pattern).

Edit Just a warning that where Perl goes, every other language follows as far as regular expression.

+1. I went ahead and edited the question to use "optional" instead. — Alan Moore, Mar 09 '11 at 06:01

How do I make part of a regular expression optional in Ruby?

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