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Comments (4)

gruns avatar gruns commented on June 2, 2024

username and password should be None when no username or password
were provided. That's a bug. Great find.

host (and scheme and netloc) should be None for URLs comprised
only of a path, query, or fragment. That's another bug. Excellent find.

To get a URL query string of furl.query.params, use furl.querystr or
str(furl.query). furl.query.params is an ordered multivalue
dictionary for direct manipulation of query parameters.

I'll incorporate a fix with None defaults for username, password,
host, scheme, and netloc into furl 0.3.4.

Thanks for bringing these issues to my attention xhujerr.

from furl.

xhujerr avatar xhujerr commented on June 2, 2024

Ok so with None and empty string it's the other way around.

With the params i was using urlparse before. urlparse, as a part of the result, returns params:

>>> urlparse.urlparse("http://www.google.com.mx/search;blabla?client=opera&q=furl&sourceid=opera&ie=utf-8&oe=utf-8&channel=suggest")
ParseResult(scheme='http', netloc='www.google.com.mx', path='/search', params='blabla', query='client=opera&q=furl&sourceid=opera&ie=utf-8&oe=utf-8&channel=suggest', fragment='')

I'm not sure how often you can practically encounter this part of url, but these params don't seem to be part of query. You may decide to ignore this part, because even in rfc3986 is only mentioned:
" ... URI producing applications
often use the reserved characters allowed in a segment to delimit
scheme-specific or dereference-handler-specific subcomponents. For
example, the semicolon (";") and equals ("=") reserved characters are
often used to delimit parameters and parameter values applicable to
that segment..."

There is also stackoverflow question What are the URL parameters? (element at position #3 in urlparse result) about these params.

from furl.

gruns avatar gruns commented on June 2, 2024

furl doesn't support URL parameters for now. They're not commonly used.

If their use grows, I'll happily add them (or accept a pull request).

If you need access to URL parameters, you can use urlparse.urlparse(), as you mentioned previously.

from furl.

gruns avatar gruns commented on June 2, 2024

furl v0.3.4 fixes these bugs.

>>> from furl import furl
>>> f = furl()
>>> f.scheme, f.username, f.password, f.host, f.netloc
(None, None, None, None, None)
>>> f.load('http://www.pumps.com/')
furl('http://www.pumps.com/')
>>> f.scheme, f.username, f.password, f.host, f.netloc
('http', None, None, 'www.pumps.com', 'www.pumps.com')

Update with

pip install furl --upgrade

Thank you for bringing these issues to my attention xhujerr.

from furl.

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