Question

i am trying to write a regex that produces the content in a string that is NOT in parentheses or brackets. The parentheses is always a year, and the brackets could contain any normal characters, upper and lower case. i was going about it by finding the brackets and parentheses and then [^\regex] to escape it (is this right?)

here’s the string:

s = 'Some words (1999) [THINGS]

and the regex:

/[^(\(\d{4}\))|\[.*\]]/

but this includes the characters inside the brackets see (http://rubular.com/r/bbpcnnGgCI)

everything works up until adding the [^\regex]

for example, this works to get (1999):

>> puts s.match(/\(\d{4}\)/)
(1999)  

and for whats in brackets:

>> puts s.match(/\[.*\]/)
[THINGS]

but put them together using | for “or”:

>> puts s.match(/\(\d{4}\)|\[.*\]/)
(1999)

…it just matches the parentheses and its contents.

what’s going on here?

what am i doing wrong here?

Answer 1

Try this /\(.+/ which will match everything from the opening ( onwards. If you strip that out, you’re left with 'Some words' which should be what you need?

Two points

1. I may be misunderstanding the question
2. You need something more complicated if there’s any possibility of an ( appearing earlier in the string.

By the way, I find this rather valuable when trying to come up with Regex patterns.

Edit This pattern should only match stuff in brackets even if there is a stray bracket earlier in the string.

string.gsub(/(\(|\[).+(\)|\])/, '')