[Python-Dev] Should urlencode() sort the query parameters (if they come from a dict)?

Sat Aug 18 22:23:47 CEST 2012

On Saturday, August 18, 2012, MRAB wrote:
> On 18/08/2012 18:34, Guido van Rossum wrote:
>>> On Sat, Aug 18, 2012 at 6:28 AM, Christian Heimes <lists at cheimes.de>
>> wrote:
>>>>> Am 17.08.2012 21:27, schrieb Guido van Rossum:
>>>>>>> I wonder if it wouldn't make sense to change urlencode() to generate
>>>> URLs that don't depend on the hash order, for all versions of Python
>>>> that support PYTHONHASHSEED? It seems a one-line fix:
>>>>>>>> query = query.items()
>>>>>>>> with this:
>>>>>>>> query = sorted(query.items())
>>>>>>>> This would not prevent breakage of unit tests, but it would make a
>>>> much simpler fix possible: simply sort the parameters in the URL.
>>>>>>>>>> I vote -0. The issue can also be addressed with a small and simple
>>> helper function that wraps urlparse and compares the query parameter. Or
>>> you cann urlencode() with `sorted(qs.items)` instead of `qs` in the
>>> application.
>>>>>>> Hm. That's actually a good point.
>>>> The order of query string parameter is actually important for some
>>> applications, for example Zope, colander+deform and other form
>>> frameworks use the parameter order to group parameters.
>>>>>> Therefore I propose that the query string is only sorted when the query
>>> is exactly a dict and not some subclass or class that has an items()
>>> method.
>>>>>> if type(query) is dict:
>>> query = sorted(query.items())
>>> else:
>>> query = query.items()
>>>>>>> That's already in the bug I filed. :-) I also added that the sort may
>> fail if the keys mix e.g. bytes and str (or int and str, for that
>> matter).
>>>> One possible way around that is to add the class names, perhaps only if
> sorting raises an exception:
>> def make_key(pair):
> return type(pair[0]).__name__, type(pair[1]).__name__, pair
>> if type(query) is dict:
> try:
> query = sorted(query.items())
> except TypeError:
> query = sorted(query.items(), key=make_key)
> else:
> query = query.items()

Doesn't strike me as necessary.
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