CODEWITHSUNDEEP
rubyregex
Thananjaya S
Thananjaya S

Reputation: 1689

Detect opening and closing brackets

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?

Upvotes: 0

Views: 1769

Answers (2)

Kimmo Lehto
Kimmo Lehto

Reputation: 6041

"(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.

Upvotes: 2

Aleksei Matiushkin
Aleksei Matiushkin

Reputation: 121010

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)"

Upvotes: 2

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