Detect opening and closing brackets

Question

From a string:

"(book(1:3000))"

I need to exclude opening and closing brackets and match:

"book(1:3000)"

using regular expression.

I tried this regular expression:

/([^',()]+)|'([^']*)'/

which detects all characters and integers excluding brackets. The string detected by this regex is:

"book 1:3000"

Is there any regex that disregards the opening and closing brackets, and gives the entire string?

@Amadan in a hope Rails provides some magic, like a dedicated method doing exactly that? — Aleksei Matiushkin, Jun 15 '18 at 10:12
It would probably help if you could provide some _context_. For example, what does that string represent, where does it come from, what is `book(...)` (looks like a function call), what is `1:3000`, what's the pattern that you want to match (explain it or show some examples), is there always an additional pair of parentheses surrounding the substring you're interested in? — Stefan, Jun 15 '18 at 10:32
instead of regex black magic, you could just do `new_str = str[1..-2]` or maybe `new_str = str.start_with?('(') ? str[1..-2] : str` if that is what your data consistently looks like. — Kimmo Lehto, Jun 15 '18 at 13:14

score 2 · Answer 1 · answered Jun 15 '18 at 09:51

2

Build the regexp that explicitly states exactly what you want to extract: alphanumerics, followed by the opening parenthesis, followed by digits, followed by a colon, followed by digits, followed by closing parenthesis:

'(book(1:3000))'[/\w+\(\d+:\d+\)/]
#⇒ "book(1:3000)"

answered Jun 15 '18 at 09:51

Aleksei Matiushkin

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No Sir, It is not working. Here is the link `http://rubular.com/r/RipuLVfSXy` for the tested `regex`. – Thananjaya S Jun 15 '18 at 09:57
2

Sure it works. The link contains some crap that has nothing to do with a regexp I provided. – Aleksei Matiushkin Jun 15 '18 at 10:00
1

@ThananjayaChakravarthy the square brackets belong to the [method](http://ruby-doc.org/core-2.5.1/String.html#method-i-5B-5D), not the regexp. It's `str[regexp]` with `str` being `'(book(1:3000))'` and `regexp` being `/\w+\(\d+:\d+\)/`. Here's a working example: http://rubular.com/r/7UpYX5ijQS – Stefan Jun 15 '18 at 10:19

Kimmo Lehto · Answer 2 · 2018-06-15T13:10:37.937

"(book(1:3000))"[/^\(?(.+?\))\)?/, 1]
=> "book(1:3000)"
"book(1:3000)"[/^\(?(.+?\))\)?/, 1]
=> "book(1:3000)"

The regex split on multiple lines for easier reading:

/
 ^      # start of string
  \(?   # character (, possibly (?)
  (     # start capturing
    .+? # any characters forward until..
    \)  # ..a closing bracket
  )     # stop capturing
/x      # end regex with /x modifier (allows splitting to lines)

1. Look for a possible ( in the beginning of string and ignore it. 2. Start capturing 3. Capture until and including the first )

But this is where it fails:

"book(1:(10*30))"[/^\(?(.+?\))\)?/, 1]
=> "book(1:(10*30)"

If you need something like that, you probably need to use a recursive regex as described in another stackoverflow answer.

Detect opening and closing brackets

2 Answers2