CODEWITHSUNDEEP
pythonurlparse
Dan Holman
Dan Holman

Reputation: 835

urlparse.urlparse returning 3 '/' instead of 2 after scheme

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 //

Upvotes: 9

Views: 1869

Answers (2)

mgiuca
mgiuca

Reputation: 21357

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.

Upvotes: 2

miku
miku

Reputation: 188114

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='')

Upvotes: 6

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