通用正则, 抓取a标签href属性

import re # url通用匹配 url_regex = re.compile('<a[^>]+href=["\'](.*?)["\']', re.IGNORECASE) # test url_str = "<a type='akfdf' href='http://www.itmeng.top' xxxx>hahah</a>" result = url_regex.findall(url_s

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