Regular Expression to capture entire list as one item- javascript

Question

I am working on js filter for string text and I want to use regex to capture entire list of list items, so I can count them and iterate throught them.

Please see what I am trying to achieve: https://regex101.com/r/Jd3JYW/2/

I know how to capture lines separately using this regex (^\*[^\*](.*)$)/gm but I need to capture entire list so I can differentiate between two lists, so I can make each of them start from n.1.

I expected the following code to work ((^\*[^\*](.*)$)+)/gm I thought that this wil capture first item (as it works for separate lines) and also any following untill patern is broken. However it is not working, I am still getting separate lines captured. Do you have any idea guys, how to capture whole list? Thank you.

Adam, I posted a variation of the regex that also supports the last line as a part of the list. — Wiktor Stribiżew
– Wiktor Stribiżew, Commented Feb 26, 2018 at 20:27
Thank you Wiktor. I really appreciate your help as well as explanation :) — Adam Ilčišák
– Adam Ilčišák, Commented Feb 26, 2018 at 20:31

Wiktor Stribiżew · Accepted Answer · 2018-02-26 20:22:18Z

2

You use $ to match line endings, but it is a zero width assertion. You should use some consuming pattern to match line endings, like [\r\n]+:

/^(?:\*[^*\n].*(?:[\r\n]+|$))+/gm

See the regex demo

Details

^ - start of a line
(?:^\*[^*\n].*(?:[\r\n]+|$))+ - 1 or more sequences of:
- \* - a * char
- [^*\n] - a char other than LF and *
- .* - any 0+ chars other than line break chars, as many as possible
- (?:[\r\n]+|$) - either 1+ LF or CR symbols or end of a line (to match the last line).

answered Feb 26, 2018 at 20:22

Wiktor Stribiżew

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