修复损坏的urls

Question

有没有人知道修复“坏掉的”urls的库。当我尝试打开一个url时，比如

http://www.domain.com/../page.html
http://www.domain.com//page.html
http://www.domain.com/page.html#stuff

urllib2.urlopen会阻塞并给我一个HTTPError回溯。有没有人知道可以修复这类问题的库？

Answer 1

比如……怎么样？

import re
import urlparse

urls = '''
http://www.domain.com/../page.html
http://www.domain.com//page.html
http://www.domain.com/page.html#stuff
'''.split()

def main():
  for u in urls:
    pieces = list(urlparse.urlparse(u))
    pieces[2] = re.sub(r'^[./]*', '/', pieces[2])
    pieces[-1] = ''
    print urlparse.urlunparse(pieces)

main()

正如您所希望的，它确实会发出：

http://www.domain.com/page.html
http://www.domain.com/page.html
http://www.domain.com/page.html

如果我理解正确的话，它看起来与你的需求大致相符。