Mocking a Celery task call in Python with patch

Question

Patching a Celery task call with a mocked return value returns `` instead of the expected `return_value` defined by `mock_task.get.return_value = "value"`. However, the mocked task functions correctly within my unit test. Here is the unit test where I am patching my Celery task: ``` def test_foo(self): mock_task = Mock() mock_task.get = Mock(return_value={'success': True}) print mock_task.get() # outputs {'success': True} with patch('app.tasks.my_task.delay', new=mock_task) as mocked_task: foo() # this calls the mocked task with an argument, 'input from foo' mock_tasked.assert_called_with('input from foo') # works ``` And here is the function being tested: ``` def foo(): print tasks.my_task.delay # shows a Mock object, as expected # now let's call get() on the mocked task: task_result = tasks.my_task.delay('input from foo').get() print task_result # => # unexpectedly, this does not return {'success': True} if task…

Accepted Answer

Unfortunately I know almost nothing about Celery, but it looks like the problem is with mocking.

You have:

```
tasks.my_task.delay('input from foo').get()
```

After `patch('app.tasks.my_task.delay', new=mock_task)` it becomes:

```
mock_task('input from foo').get()
```

Which is not the same as:

```
mock_task.get()
```

You should change your mock creation to:

```
mock_task().get = Mock(return_value={'success': True})
```

New Mock instance is created by default when you access existing Mock attribute or call it. So we can simplify it a little:

```
mock_task().get.return_value = {'success': True}
```

Mocking a Celery task call in Python with patch

Solution

Problem