Match any string between two different tokens

Question

I'm trying to match a part of a string between two different tokens. They might be multiple occurrences of the tokens in a string.

Sample text (tokens are italic, text to match is bold):

This is [begin-match] a sample text [end-match] with some [begin-match] tokens and normal [end-match] text.

I have the following regex, which would work if the tokens were { and }:

/{([^}]+)}/g

I can't get this to work the [begin-match] and [end-match] tokens. It seems that the lack of negative lookbehind in Javascript is a big loss.

I can't figure out how to substitute the { and } for the tokens [begin-match] and [end-match]. How can I match on those?

You probably also need to know [What special characters must be escaped in regular expressions?](http://stackoverflow.com/questions/399078/what-special-characters-must-be-escaped-in-regular-expressions) — HamZa, May 04 '15 at 10:19

Sebastian Simon · Accepted Answer · 2015-05-04T10:09:46.697

4

/x(.*?)y/g where x is the beginning token and y the ending token.

This RegEx means: match anything (.), any number of times (*), as few times as possible (?).

A direct example from your question would be:

/\[begin-match\](.*?)\[end-match\]/g

The sample text is now in the first capturing group.

edited May 04 '15 at 10:09

answered May 04 '15 at 10:00

Sebastian Simon

score 2 · Answer 2 · answered May 04 '15 at 10:00

2

\[begin-match\]((?:(?!\[end-match\]).)*)\[end-match\]

You can try this.See demo.

answered May 04 '15 at 10:00

vks

1

That's it :) Thanks! (I have to wait to accept your answer) – Roemer May 04 '15 at 10:02
Instead of the `(?!...)` construction, you can use `(.*?)` to perform a non-greedy match: `\[begin-match\](.*?)\[end-match\]` – robertklep May 04 '15 at 10:02
@robertklep That would be nearly identical to my answer. – Sebastian Simon May 04 '15 at 10:02
Yeah, it is really over-complicated. See http://stackoverflow.com/questions/1454913/regular-expression-to-find-a-string-included-between-two-characters-while-exclu and https://regex101.com/r/kE6eB2/1 – Wiktor Stribiżew May 04 '15 at 10:03
@Xufox yes, it would :-) But showing how to escape brackets is also part of a solution. – robertklep May 04 '15 at 10:03

2 Answers2