如何使用Ruby从字符串中删除括号内的所有符号？

Question

I have a string "Some words, some other words (words in brackets)"

How can I completely remove words in brackets with brackets too, to get "Some words, some other words " string as result?

I'm newbie for regexp but I promise to learn to they works )

Thank for help!

Answer 1

Try this:

# irb
irb(main):001:0> x = "Some words, some other words (words in brackets)"
=> "Some words, some other words (words in brackets)"
irb(main):002:0> x.gsub(/\(.*?\)/, '')
=> "Some words, some other words "

Answer 2

Because of the greedyness of the "*" if there is more than on pair of brackets everything within will be deleted:

s = "Some words, some other words (words in brackets) some text and more ( text in brackets)"
=> "Some words, some other words (words in brackets) some text and more ( text in brackets)" 

ruby-1.9.2-p290 :007 > s.gsub(/\(.*\)/, '')
=> "Some words, some other words " 

A more stable solution would be:

/\(.*?\)/
ruby-1.9.2-p290 :008 > s.gsub(/\(.*?\)/, '')
=> "Some words, some other words  some text and more "

Leaving the text between groups of brackets intact.

Answer 3

String#[] :

>>  "Some words, some other words (words in brackets)"[/(.*)\(/, 1] 
    #=> "Some words, some other words "

The regexp means: group everything (.*) before the first open bracket \\( , and the argument 1 means: take the first group.

If you need to match also the closed bracket you can use /(.*)\\(.*\\)/ , but this will return nil if the string does not contain one of the brackets.

/(.*)(\\(.*\\))?/ matches also strings which not contain brackets.