Question

Using Ruby, I am writing a regular expression, and I need to be a able to remove any colon that appears between parentheses. I understand that I can use

"This is a (string :)".sub!(/\([^\)]*:/, '') 

to do this, but the problem is that this function will also remove the context along with it. Is there any way to specify that I only want it to remove the colon and not the entire matching expression?

Answer 1

So some regular expression engines support what are called look-ahead and look-behind matches that will match but not consume characters. Ruby does support look-ahead, but not look-behind (which is more difficult to do in a performant way), which means you could quite easily stick with sub and remove a colon that precedes a closing parenthesis, but only without ensuring it is after an opening parenthesis:

string = 'This is a (string :)'
string.sub /:(?=\))/, ''
# => 'This is a (string )'

The alternative would be to use subpattern capturing (which happens automatically when you use grouping in your regular expression) to rebuild the string without the undesirable portion, in this case the colon:

string.sub /(\([^:]+):\)/, '\1)'

The \1 is a back-reference to what is matched in the first group, which is delimited by the parentheses that are not escaped. You can see here I didn’t bother capturing the closing parenthesis in a second group, opting instead simply to include it in the substitution. This works well in this case because it will not change, but if you don’t know that the colon will appear at the end of the parentheses-enclosed content, you would need a second group:

string.sub /(\([^:]+):([^)]+\))/, '\1\2'