Ruby regex: "grab a string if it doesn't follow ..."

Question

My regular expression captures quoted phrases:

"([^"]*)" 

I want to improve it by ignoring quotes followed by ", -" (comma, space and dash in that particular order).

How to do it?

Test: http://rubular.com/r/xls6vN1w92

Answer 1

This should do it using Negative Lookahead :

 "(?!, -)([^"]*)"(?!, -) 

A little bad, but it works. You want to make sure that none of the quotes is followed by your line, otherwise the match will start with closed quotes.

http://rubular.com/r/yFMyUKJOHL

Answer 2

Regex

"(.*?)"(?!, -)

Working example

http://rubular.com/r/9kOmZLxLfy

Answer 3

It is impossible to do in your context, its open end. The only way to make it out is to use it not the way you want, but its still unacceptable premise.

/"([^"]*?)"(?!, -)|"[^"]*?"(?=, -)/

Then check for capture group 1 in each match, something like this:

 $rx = qr/"([^"]*?)"(?!, -)|"[^"]*?"(?=, -)/; while (' "ingnore me", - "but not me" ' =~ /$rx/g) { print "'$1'\n" if defined $1 } 

Answer 4

Add (?!...) to the end of the regular expression:

 "([^"\n]*)"(?!, -)