Parse custom URIs with urlparse (Python)

Question

My application creates custom URIs (or URLs?) to identify objects and resolve them. The problem is that Python's urlparse module refuses to parse unknown URL schemes like it parses http.

If I do not adjust urlparse's uses_* lists I get this:

>>> urlparse.urlparse("qqqq://base/id#hint")
('qqqq', '', '//base/id#hint', '', '', '')
>>> urlparse.urlparse("http://base/id#hint")
('http', 'base', '/id', '', '', 'hint')

Here is what I do, and I wonder if there is a better way to do it:

import urlparse

SCHEME = "qqqq"

# One would hope that there was a better way to do this
urlparse.uses_netloc.append(SCHEME)
urlparse.uses_fragment.append(SCHEME)

Why is there no better way to do this?

Solution 1

I think the problem is that URI's don't all have a common format after the scheme. For example, mailto: urls aren't structured the same as http: urls.

I would use the results of the first parse, then synthesize an http url and parse it again:

parts = urlparse.urlparse("qqqq://base/id#hint")
fake_url = "http:" + parts[2]
parts2 = urlparse.urlparse(fake_url)

Solution 2

You can also register a custom handler with urlparse:

import urlparse

def register_scheme(scheme):
    for method in filter(lambda s: s.startswith('uses_'), dir(urlparse)):
        getattr(urlparse, method).append(scheme)

register_scheme('moose')

This will append your url scheme to the lists:

uses_fragment
uses_netloc
uses_params
uses_query
uses_relative

The uri will then be treated as http-like and will correctly return the path, fragment, username/password etc.

urlparse.urlparse('moose://username:password@hostname:port/path?query=value#fragment')._asdict()
=> {'fragment': 'fragment', 'netloc': 'username:password@hostname:port', 'params': '', 'query': 'query=value', 'path': '/path', 'scheme': 'moose'}

Solution 3

There is also library called furl which gives you result you want:

>>>import furl
>>>f=furl.furl("qqqq://base/id#hint");
>>>f.scheme
'qqqq' 

>>> f.host
'base'  
>>> f.path
Path('/id')
>>>  f.path.segments
['id']
>>> f.fragment                                                                                                                                                                                                                                                                 
Fragment('hint')   
>>> f.fragmentstr                                                                                                                                                                                                                                                              
'hint'

Solution 4

The question appears to be out of date. Since at least Python 2.7 there are no issues.

Python 2.7.10 (default, May 23 2015, 09:40:32) [MSC v.1500 32 bit (Intel)] on win32
>>> import urlparse
>>> urlparse.urlparse("qqqq://base/id#hint")
ParseResult(scheme='qqqq', netloc='base', path='/id', params='', query='', fragment='hint')

Solution 5

Try removing the scheme entirely, and start with //netloc, i.e.:

>>> SCHEME="qqqq"
>>> url="qqqq://base/id#hint"[len(SCHEME)+1:]
>>> url
'//base/id#hint'
>>> urlparse.urlparse(url)
('', 'base', '/id', '', '', 'hint')

You won't have the scheme in the urlparse result, but you know the scheme anyway.

Also note that Python 2.6 seems to handle this url just fine (aside from the fragment):

$ python2.6 -c 'import urlparse; print urlparse.urlparse("qqqq://base/id#hint")'
ParseResult(scheme='qqqq', netloc='base', path='/id#hint', params='', query='', fragment='')

View more solutions

Comments

[u0b34a0f6ae]

I perfer my own workaround to this one; I would have to do this roundtrip all the time in my custom URL module.

[yantrab]

Fair enough: I didn't like relying on internals of the module, but I reasonable engineers can differ!

[Vladimir Mihailenco]

But query still is not parsed properly... Thanks anyway.

[knickum]

Wanted to back this up further, as of 04/26/2016; also parses beyond the basics shown above: weird_scheme = 'qqq://username:[email protected]/some/path?params=key#fr‌​ag_ment'. Then parse and show username: urlparse(weird_scheme).username #'username' or show query: urlparse(weird_scheme).query) #'params=key'