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

[Guido van Rossum]

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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