I'd like to add the 'http' scheme name in front of a given url string if it's missing. Otherwise, leave the url alone so I thought urlparse was the right way to do this. But whenever there's no scheme and I use get url, I get /// instead of '//' between the scheme and domain.
>>> t = urlparse.urlparse('www.example.com', 'http')
>>> t.geturl()
'http:///www.example.com' # three ///
How do I convert this url so it actually looks like:
'http://www.example.com' # two //
Short answer (but it's a bit tautological):
>>> urlparse.urlparse("http://www.example.com").geturl()
'http://www.example.com'
In your example code, the hostname is parsed as a path not a network location:
>>> urlparse.urlparse("www.example.com/go")
ParseResult(scheme='', netloc='', path='www.example.com/go', params='', \
query='', fragment='')
>>> urlparse.urlparse("http://www.example.com/go")
ParseResult(scheme='http', netloc='www.example.com', path='/go', params='', \
query='', fragment='')
If you want to use urlparse as you were intending to, the closest "correct" equivalent is to use "//www.example.com" as the urlstring. Such a urlstring is unambiguously an absolute path without a scheme, so you could then supply "http" as the default scheme. I suppose you could do this by detecting whether your URL includes the string "//" and if not, prepending "//" on the front.
来源:https://stackoverflow.com/questions/7289481/urlparse-urlparse-returning-3-instead-of-2-after-scheme