import reu = "http[s]?://(?:[a-zA-Z]|[0-9]|[$-_@.&+]|[!*\(\),]|(?:%[0-9a-fA-F][0-9a-fA-F]))+"def match_by_re(str): ???pattern = re.compile(u, re.S) ???res = pattern.findall(string=str) ???if res: return res[0]
正则匹配URL地址
原文地址：https://www.cnblogs.com/li1992/p/10438208.html