How can I use regex to select multiple lines and end the string after a character is found?

Asked Oct 26 '20 at 05:36

Active Oct 26 '20 at 06:23

Viewed 38 times

I'm trying to use a regex that recognizes multiple lines of text and stops when a certain character is found, in this case the # character. I used


##[a-zA-Z][\s\S]*?(?=\n.*?)

but I don't quite understand what to add to the regex to make it find multiple lines that follow a specific list pattern (* or ##)

Any help would be greatly appreciated.

Here is the test string in regex101.com

Sample Text

## SECTIONS

  * [Congress]
  * [White House]
  * [Magazine]
  * [The Agenda]
  * [Video]
  * [Podcasts]

## 2020 Elections

The expected output is everything from the first ## to the last ##. I want the returned result to include this. Sorry forgot to update the regex.

edited Oct 26 '20 at 06:23

asked Oct 26 '20 at 05:36

Jon

What is wrong with your current regex? It might be correct already, at least for certain texts. – Tim Biegeleisen Oct 26 '20 at 05:38
I'll show you from regex101.com – Jon Oct 26 '20 at 05:39
1

@Jon, you can also avoid posting images and just give us the links. This is teh case for regex101, where you can save and share the regex. It is probably available if you click on the three line on the top right of your screen. – Enlico Oct 26 '20 at 05:47
https://regex101.com/r/0WDw7f/1 – Jon Oct 26 '20 at 06:02
Your sample text doesn't even have `foo` and it is not clear what should be matched in sample text – anubhava Oct 26 '20 at 06:08
You provided the sample text, but where's the expected output? – Maya Oct 26 '20 at 06:18
1

Try this regex `##[\s\S]*?(?=##)` – anubhava Oct 26 '20 at 06:26
Thanks, is there a way to include that last line? – Jon Oct 26 '20 at 06:53
1

use: `##[\s\S]*?(?=##|\z)` to get both matches – anubhava Oct 26 '20 at 07:14