python 正则获取url地址

Question

sample：<a rpos="" cpos="title" href="http://xxx.com/a.html" style="font-family:Arial,SimSun,sans-serif;font-size:16px;color:#0000cc; text-decoration:none;" target="_blank">

我想获取http://xxx.com/a.html 请问如何写正则表达式

Answer 1

/<a[^>]*href="([^"]+)"[^>]*>/

Answer 2

from bs4 import BeautifulSoup

soup = BeautifulSoup(sample)

url = soup.find('a').get('href')

Answer 3

这还不简单

text = '<a rpos="" cpos="title" href="http://xxx.com/a.html" style="font-family:Arial,SimSun,sans-serif;font-size:16px;color:#0000cc; text-decoration:none;" target="_blank">'

urlPattern = r'(href="http://[\s\S]+.html")'

pattern = re.compile(urlPattern)

match = re.findall(pattern,text,0)