如果里面的模式匹配，则抓取完整的正则表达式字

Question

How do I retrieve an entire word that has a specific portion of it that matches a regex?

For example, I have the below text. Using ^.[\\.\\?\\!:;,]{2,} , I match the first 3, but not the last. The last should be matched as well, but $ doesn't seem to produce anything.

a!!!!!!
n.......
c..,;,;,,

huhuhu..

I want to get all strings that have an occurrence of certain characters equal to or more than twice. I produced the aforementioned regex, but on Rubular it only matches the characters themselves, not the entire string. Using ^ and $

I've read a few stackoverflow posts similar, but not quite what I'm looking for.

Answer 1

Change your regex to:

/^.*[.?!:;,]{2,}/gm

ie match 0 more character before 2 of those special characters.

RegEx Demo

Answer 2

If I understand well you are trying to match an entire string that contains at least the same punctuation character two times:

^.*?([.?!:;,])\1.*

Note: if your string has newline characters, change .* to [\\s\\S]*

The trick is here:

([.?!:;,])   # captures the punct character in group 1
\1           # refers to the character captured in group 1