match regex between varying number of characters

Question

The following regex matches "EXAMPLETEXT" in my example below but I'd like to be able to use \d* to match one or more digits instead of \d\d. Is that possible or is there a better method?

String:

09.04.EXAMPLETEXT.14

Regex:

(?<=\.\d\d\.)(.*)(?=\.)

I'm using Python but trying to get the regex running in regex101 first — Arno, Jul 18 '15 at 15:31
Can you give some more examples of strings you'd like your regex to match (including what substring you'd like to capture) and strings you don't want it to match? — das-g, Jul 18 '15 at 15:40
If you always have two digits followed by a dot (and don't know how many), then there is a better method to do this: `((?:\d{2}\.)*)` — maraca, Jul 18 '15 at 16:12
Or this: `(?:\d+\.)+(.*)(?:\.\d+)+` https://regex101.com/r/pP6lX7/1 — Jan, Jul 18 '15 at 16:54
didn't realize that regex behaved slightly differently in different languages and that regex101 lets you choose your language. — Arno, Jul 18 '15 at 20:48
ArnoVR, it's good form to mark the correct Answer as such. Looks like Jan called it, is that right? — CLaFarge, Jul 18 '15 at 22:50
sure I tried to mark the correct answer, maybe I need more rep for that? — Arno, Jul 20 '15 at 22:08
ah found Jan's answer below, was looking for a way to mark in the comments, oops — Arno, Jul 20 '15 at 22:12

Avinash Raj · Answer 1 · 2015-07-18T16:48:36.173

4

You can't use * or + or ? quantifiers inside lookbehind assertions in perl,java,python regexes (they won't support variable length lookbehind). But you can use those symbols inside lookbehind in c# family.

If your're on php or perl, you may use \K

\.\d*\.\K(.*)(?=\.)

Another hack in all languages is, just print all the captured chars.

\.\d*\.(.*)\.

Example for greedy:

>>> s = "09.043443.EXAMPLETEXT.14"
>>> re.search(r'\.\d*\.(.*)\.', s).group(1)
'EXAMPLETEXT'

Example for non-greedy match:

>>> re.search(r'\.\d*\.(.*?)\.', s).group(1)
'EXAMPLETEXT'

Use negated char class.

>>> re.search(r'\.\d*\.([^.]*)\.', s).group(1)
'EXAMPLETEXT'

edited Jul 18 '15 at 16:48

answered Jul 18 '15 at 15:29

Avinash Raj

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@maraca I just modified his regex `\d\d` with `\d*` without touching the functionality of greedy match. – Avinash Raj Jul 18 '15 at 15:52
1

@AvinashRaj ok, but wouldn't that be `\d{2}` (or maybe `\d+`) not completely sure what the question is, just alternatives. – maraca Jul 18 '15 at 15:55
`\K` used because to achieve the functionality of variable length lookbehind. – Avinash Raj Jul 18 '15 at 16:51

score 2 · Accepted Answer · answered Jul 18 '15 at 22:15

You don't need to use lookbehind at all for this, a regex such as this would work:

(?:\d+\.)+(.*)(?:\.\d+)+

https://regex101.com/r/pP6lX7/1

Or if you want to be able to match the string inside some other text

(?:\d+\.)+([^\s]*)(?:\.\d+)+

https://regex101.com/r/sB5mB2/1

match regex between varying number of characters

2 Answers2