Formulation of regex to match js libraries and version

Question

I'm trying to formulate how to write one regex to match this: I would like to write a regex to match on words like: backbone,jquery,bootstrap,yui and match on javascript library versions like 0.5.0 or 1.8 (as shown). How can I reasonably do this with python 3? JavaScript questions and answers, JavaScript questions pdf, JavaScript question bank, JavaScript questions and answers pdf, mcq on JavaScript pdf, JavaScript questions and solutions, JavaScript mcq Test , Interview JavaScript questions, JavaScript Questions for Interview, JavaScript MCQ (Multiple Choice Questions)

Answer 1

Maybe a starting point: use two expressions, one for the version, one for the lib: import re urls = """https://ajax.cdnjs.com/ajax/libs/backbone.js/0.5.0/backbone-min.js https://ajax.cdnjs.com/ajax/libs/backbone-0.5.0.js https://ajax.cdnjs.com/ajax/libs/jquery-min.js https://ajax.cdnjs.com/ajax/libs/jquery-1.6.3.js https://ajax.cdnjs.com/ajax/libs/backbone.js/0.5.0/backbone-min.js https://stackpath.bootstrapcdn.com/bootstrap/4.0.2/js/jquery.validate.js http://code.jquery.com/mobile/1.0rc2/jquery.mobile-1.0rc2.min.js"> http://ajax.googleapis.com/ajax/libs/jqueryui/1.9.2/jquery-ui.min.js http://ajax.googleapis.com/ajax/libs/jqueryui/1.8/jquery-ui.min.js http://yui.yahooapis.com/3.5.0/build/yui/yui-min.js""" rx_version = re.compile(r'[-/](\d+[.\w]*)[/.]') rx_lib = re.compile(r'/([a-z]+)[^/]+$') for url in urls.split("\n"): version, lib = None, None # version m = rx_version.search(url) if m: version = m.group(1) m = rx_lib.search(url) if m: lib = m.group(1) print(lib, version) This yields backbone 0.5.0 backbone 0.5.0 jquery None jquery 1.6.3 backbone 0.5.0 jquery 4.0.2 jquery 1.0rc2 jquery 1.9.2 jquery 1.8 yui 3.5.0