Regular Expression for extracting strings in quotes (under certain requirements)

Question

maybe the problems seems bathetic, but at the moment I have a real problem in extracting strings from quotes, but fulfilling certain criteria. These are

• Just the text in quotes should be extracted, but not the quotes itself
  e.g. I live in "Germany" -> Germany
• If there are no quotes the text should be returned completely
  e.g. I live in Germany -> I live in Germany
• If the quotes are protected by \ they should be treated as normal characters and be extracted, but without the backslash
  e.g. I live in \"Germany\" -> I live in "Germany"
• If the last quotation is missing, the rest of the text should be returned
  e.g. I "live in Germany -> live in Germany

All my experiments failed. The best guess of mine was:

    (?<=(?<!\\)").*?(?=(?<!\\)")

The problem of it is that if there are no quotes at all, the extracted string is empty.

Thanks a lot for the help
Best regards
cerebro

Answer 1

For the test cases written in your question following regex should work for you:

^([^"\n\\]*)(?:\\(")|"|)([^"\n\\]*)(?:\\(")|"|)(.*)$

You need to concatenate group # 1, 2, 3, 4, 5 to get your captured string.

Live Demo: http://www.rubular.com/r/u6m08cRo1v