Javascript多行正则表达式先行或其余字符串/字符串结尾

Question

I'm having trouble capturing all groups in the following markdown text:

### AAA
text for group AAA

### BBB
text for group BBB

### CCC
text for group CCC

The regex I'm currently using is:

/(^###\s[\s\S]*?(?=^###\s))/gm

Which uses a positive lookahead to know when to cut off each group. As a result though, it always fails to capture the last group (there's no lookahead match).

The newline characters won't be consistent, so I can't rely on those, and javascript regex doesn't have \\Z (which I believe is "very end of string" in other languages, as opposed to $ which is end of line).

How can I capture all groups here?

Answer 1

You can use negative look-ahead instead of positive : the content of a group will be every character that isn't at the beginning of a ### string.

^###\\s((?!###)[\\s\\S])*

Try it on regex101 !

Answer 2

See if JS has the \\z construct.
If not, you could use a lookahead for not a character (EOS)
(?![\\S\\s]) inside your existing lookahead.

(^###[^\\S\\r\\n][\\S\\s]*?(?=^###[^\\S\\r\\n]|(?![\\S\\s])))

Formatted:

 (                             # (1 start)
      ^ \#\#\# [^\S\r\n] 
      [\S\s]*? 
      (?=
           ^ \#\#\# [^\S\r\n]  
        |  (?! [\S\s] )
      )
 )                             # (1 end)