Is there a better way to write this URL Manipulation in Python?

Question

I'm curious if there's a simpler way to remove a particular parameter from a url. What I came up with is the following. This seems a bit verbose. Libraries to use or a more pythonic version appreciated.

parsed = urlparse(url)
if parsed.query != "":
    params = dict([s.split("=") for s in parsed.query.split("&")])
    if params.get("page"):
        del params["page"]
    url = urlunparse((parsed.scheme,
                      None,
                      parsed.path,
                      None,
                      urlencode(params.items()),
                      parsed.fragment,))
    parsed = urlparse(url)

Jacob Marble · Accepted Answer · 2013-03-14 04:19:34Z

11

Use urlparse.parse_qsl() to crack the query string. You can filter this in one go:

params = [(k,v) for (k,v) in parse_qsl(parsed.query) if k != 'page']

edited Mar 14, 2013 at 4:19

Jacob Marble

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answered May 20, 2010 at 12:09

Marcelo Cantos

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7 Comments

dnolen Over a year ago

The url manipulation here seems tortured even with your minor change.

Marcelo Cantos Over a year ago

@dnolen: I agree. Python's baked-in libraries aren't particularly good for simple URI manipulation. (Did you downvote me? If so, it hardly seems reasonable to downvote someone because of limitations in the language or its libraries.)

dnolen Over a year ago

I downvoted because your answer didn't attempt to address the actual problem, and some people seemed to think otherwise. In any case at 14.3k karma it hardly affects you.

Marcelo Cantos Over a year ago

The rep isn't the issue here; what irks me is that you think I wasn't trying to help solve your problem. The title question was: "Is there a better way to write this URL Manipulation in Python?" My answer offers precisely that.

Marcelo Cantos Over a year ago

@bvukelic: No, but then you'd have to write if x[0] …, which isn't as clear.

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Ned Batchelder · Accepted Answer · 2010-05-20 21:36:45Z

9

I've created a small helper class to represent a url in a structured way:

import cgi, urllib, urlparse

class Url(object):
    def __init__(self, url):
        """Construct from a string."""
        self.scheme, self.netloc, self.path, self.params, self.query, self.fragment = urlparse.urlparse(url)
        self.args = dict(cgi.parse_qsl(self.query))

    def __str__(self):
        """Turn back into a URL."""
        self.query = urllib.urlencode(self.args)
        return urlparse.urlunparse((self.scheme, self.netloc, self.path, self.params, self.query, self.fragment))

Then you can do:

u = Url(url)
del u.args['page']
url = str(u)

More about this: Web development peeve.

answered May 20, 2010 at 21:36

Ned Batchelder

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2 Comments

dnolen Over a year ago

A reasonable compromise. Far more useful than urlparse I would say ;)

twig Over a year ago

I've made some modifications to this class, making it a bit easier to use.

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