ParseUrl

1681 단어 parse
#!/usr/bin/python
# coding:utf-8

import re
import urlparse

#   url
def ParseUrl(url):
    if not re.search(r"^http[s]?://",url):
        if ":443" in url:
            url = "https://" + url
        else:
            url = "http://" + url
        return url

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