Python mock function with only specific argument

Question

I'm new to Python and I'm trying to mock a function only when a specific argument is passed. If other than the desired argument is passed, I'd like to call the original function instead.

In Python 2.7 I tried something like this:

from foo import config def test_something(self): original_config = config # config is a Module. def side_effect(key): if key == 'expected_argument': return mocked_result else: return original_config.get(key) config.get = Mock(side_effect=side_effect) # actualy_test_something...

It won't work 'cause original_config is not a copy of config. It references the same module ending up in an infinite loop. I could try cloning the original config module instead but that seems to be overkill.

Is there something similar to RSpec's mocks I could use? e.g:

obj.stub(:message).with('an_expected_argument').and_return('a_mocked_result')

Any help would be appreciated. Thanks.

Then I do not understand how it is executed. Both branches of if above it end with return so this line does not seem to be ever reached. — Petr Gladkikh
– Petr Gladkikh, Commented Nov 12, 2017 at 10:33

Martijn Pieters · Accepted Answer · 2014-11-30 03:35:45Z

You'd need to store a reference to the unpatched function first:

def test_something(self): original_config_get = config.get def side_effect(key): if key == 'expected_argument': return mocked_result else: return original_config_get(key) config.get = Mock(side_effect=side_effect)

Here original_config_get references the original function before you replaced it with a Mock() object.

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