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I am having issue with a reg ex expression and can't find the answer to my question.

I am trying to build a reg ex pattern that will pull in any matches that have # around them. for example #match# or #mt# would both come back.

This works fine for that. #.*?# However I don't want matches on ## to show up. Basically if there is nothing between the pound signs don't match.

Hope this makes sense. Thanks.

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  • Is it assumed that there will be no spaces inside of the hashes? Commented Mar 4, 2015 at 23:08
  • What would be##matched##in##this##and##that###or#the#other#thing# ? So, it looks like you left out %90 of it. Commented Mar 4, 2015 at 23:13
  • Please see my edit below Commented Mar 4, 2015 at 23:20
  • @sln 89.79591836734694% to be precise :) Commented Mar 4, 2015 at 23:48

3 Answers 3

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Please use + to match 1 or more symbols:

#+.+#+

UPDATE: If you want to only match substrings that are enclosed with single hash symbols, use:

(?<!#)#(?!#)[^#]+#(?!#)

See regex demo

Explanation:

  • (?<!#)#(?!#) - a # symbol that is not preceded with a # (due to the negative lookbehind (?<!#)) and not followed by a # (due to the negative lookahead (?!#))
  • [^#]+ - one or more symbols other than # (due to the negated character class [^#])
  • #(?!#) - a # symbol not followed with another # symbol.
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3 Comments

I tried that and get matches in between. This entire line then matched ##test## matches. #test#
Please edit your question to include some concrete examples of the input strings and the desired output. It is not easy to make guesses with regexes.
I have updated the regex, it works a bit differently from the one accepted as it wouldn't match #mixed# in #mixed##.
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Instead of using * to match between zero and unlimited characters, replace it with +, which will only match if there is at least one character between the #'s. The edited regex should look like this: #.+?#. Hope this helps!

Edit

Sorry for the incorrect regex, I had not expected multiple hash signs. This should work for your sentence: #+.+?#+

Edit 2

I am pretty sure I got it. Try this: (?<!#)#[^#].*?#. It might not work as expected with triple hashes though.

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4 Comments

When I ran that on this sentence from a reg ex site. Edit the Expression & Text to ##see## matches. #Roll# over matches or the expression for details. I see this as matching ##see## matches. #
Do you want the text between the double hash signs to show, or not come up at all?
I need everything to show. So for example #content# would return "#content" but ## would be ignored
Close, but still getting matches when ##match## is used fore example
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Try:

[^#]?#.+#[^#]?

The [^ character_group] construction matches any single character not included in the character group. Using the ? after it will let you match at the beginning/end of a string (since it matches the preceeding character zero or more times. Check out the documentation here

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Perhaps I didn't read your question correctly, are looking for matches on ##string## or do those need kicked out?

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