super() in Python 3

[Rhodri James]

[Rearranged and snipped so this makes any kind of sense]

On 16/07/2019 16:43, אורי wrote:
> On Tue, Jul 16, 2019 at 3:13 PM Rhodri James <rhodri at kynesim.co.uk> wrote:
>> On 16/07/2019 11:08, אורי wrote:
>>> 2. I want to override a function called build_suite in an inherited
>> class.
>>> The function receives an argument "test_labels" which I want to change (I
>>> define it if it's not defined), but I don't do anything with the argument
>>> "extra_tests". Is it possible to include "extra_tests" in *args, **kwargs
>>
>> Yes.
>>
>>> and how?
>>
>> Don't list it in your parameters :-)
>>
>> def build_suite(self, test_labels=None, *args, **kwargs):
>>       ...
>>       return super().build_suite(test_labels=test_labels, *args, **kwargs)
 >
 > Thanks for your explanation. But I tried your code and it doesn't work
 > (with Django==1.11.22):
 >
 >    File "<...>\site-packages\django\test\runner.py", line 600, in 
run_tests
 >      suite = self.build_suite(test_labels, extra_tests)
 >    File "<...>\speedy\core\base\test\models.py", line 35, in build_suite
 >      return super().build_suite(test_labels=test_labels, *args, **kwargs)
 > TypeError: build_suite() got multiple values for argument 'test_labels'

Ah, sorry, I was under the impression that you were calling the function 
with "test_labels" and "extra_tests" as keyword arguments.  You can 
solve the immediate problem by omitting the "=test_labels" bit:

   return super().build_suite(test_labels, *args, **kwargs)

assuming that "test_labels" is the first positional parameter.  If it 
isn't, you've got trouble anyway :-) and you'll need to do some more 
involved parameter handling.

-- 
Rhodri James *-* Kynesim Ltd