Optimal Regular Expression: match sets of lines starting with ...

Question

Alright, this one's interesting. I have a solution, but I don't like it.

The goal is to be able to find a set of lines that start with 3 periods - not an individual line, mind you, but a collection of all the lines in a row that match. For example, here's some matches (each match is separated by a blank line):

...

...hello

...
...hello
...world
...
...wazzup?
...

My solution is as follows:

^\.\.\..*(\n\.\.\..*)*$

It matches all those, so it's what I'm using for now - however, it looks kinda silly to repeat the \.\.\..* pattern. Is there a simpler way?

Please test your regex before submitting it, rather than submit what "should work." For example, I tried the following first:

(^\.\.\..*$)+

which only returned individual lines, even though in my mind it looks like it would do the trick - I guess I just don't understand regex internals. (And no, I didn't need to set any flags to get ^ and $ to match line boundaries, since I'm implementing this in Ruby.)

So I'm not totally sure there's a good answer, but one would be much appreciated - thanks in advance!

Answer 1

In most regex implementations you can shorten \.\.\. using \.{3} so your solution would turn into \.{3}.*(\n\.{3}.*)*.

Answer 2

What you already have is already simple and understandable. Keep in mind that more "clever" RegExps may very well be slower and undoubtedly less readable.

Assuming lines are terminated by a \n:

((^|\n)\.{3}[^\n]*)+

I am not familiar with Ruby, so depending on how it returns matches you might need to "nonmatch" groups:

((?:(?:^|\n)\.{3}[^\n]*)+)

Answer 3

You are pretty close to a solution with (^\.\.\..*$)+, but because the + modifier is on the outside of the group, it is getting overwritten each time and you are only left with the last line. Try wrapping it in an outer group: ((^\.\.\..*$)+) and looking at the first submatch and ignoring the inner one.

Combined with the other suggestion: ((^\.{3}.*$)+)

Answer 4

^([.]{3}.*$\n?)+

This doesn't really need $ in there.