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Python, opposite function urllib.urlencode

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python

urllib

How can I convert data after processing urllib.urlencode to dict? urllib.urldecode does not exist.

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Artyom Avatar asked Aug 22 '10 18:08

Artyom


2 Answers

As the docs for urlencode say,

The urlparse module provides the functions parse_qs() and parse_qsl() which are used to parse query strings into Python data structures.

(In older Python releases, they were in the cgi module). So, for example:

>>> import urllib >>> import urlparse >>> d = {'a':'b', 'c':'d'} >>> s = urllib.urlencode(d) >>> s 'a=b&c=d' >>> d1 = urlparse.parse_qs(s) >>> d1 {'a': ['b'], 'c': ['d']} 

The obvious difference between the original dictionary d and the "round-tripped" one d1 is that the latter has (single-item, in this case) lists as values -- that's because there is no uniqueness guarantee in query strings, and it may be important to your app to know about what multiple values have been given for each key (that is, the lists won't always be single-item ones;-).

As an alternative:

>>> sq = urlparse.parse_qsl(s) >>> sq   [('a', 'b'), ('c', 'd')] >>> dict(sq) {'a': 'b', 'c': 'd'} 

you can get a sequence of pairs (urlencode accepts such an argument, too -- in this case it preserves order, while in the dict case there's no order to preserve;-). If you know there are no duplicate "keys", or don't care if there are, then (as I've shown) you can call dict to get a dictionary with non-list values. In general, however, you do need to consider what you want to do if duplicates are present (Python doesn't decide that on your behalf;-).

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Alex Martelli Avatar answered Sep 19 '22 18:09

Alex Martelli


Python 3 code for Alex's solution:

>>> import urllib.parse >>> d = {'a':'b', 'c':'d'} >>> s = urllib.parse.urlencode(d) >>> s 'a=b&c=d' >>> d1 = urllib.parse.parse_qs(s) >>> d1 {'a': ['b'], 'c': ['d']} 

The alternative:

>>> sq = urllib.parse.parse_qsl(s) >>> sq [('a', 'b'), ('c', 'd')] >>> dict(sq) {'a': 'b', 'c': 'd'} 

parse_qsl is reversible:

>>> urllib.parse.urlencode(sq) 'a=b&c=d' 
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phobie Avatar answered Sep 18 '22 18:09

phobie