[英]mock function arguments in python
Lets say I have this function 可以说我有这个功能
from datetime import date
def get_next_friday(base_date=date.today()):
next_friday = ...
return next_friday
Then I have a celery task to call this function without passing in the base_date
然后我有一个芹菜任务来调用此函数,而无需传入
base_date
@celery_app.task
def refresh_settlement_date():
Record.objects.update(process_date=get_next_friday())
In the unittest I am running the refresh_settlement_date()
task, but it's not providing the base_date
when it's calling the get_next_friday()
, my question is how to mock that parameter to test the days in the future? 在单元测试中,我正在运行
refresh_settlement_date()
任务,但在调用get_next_friday()
时未提供base_date
,我的问题是如何模拟该参数以测试将来的日子?
I am trying to avoid adding parameter to become refresh_settlement_date(base_date)
as it doesn't serve real purpose but only for unittest. 我试图避免将参数添加为
refresh_settlement_date(base_date)
因为它不是真正的目的,而仅用于单元测试。
An alternative approach would to be to patch the current date. 一种替代方法是修补当前日期。
There is a relevant thread with multiple options: 有一个具有多个选项的相关线程:
My favorite option is to use a third-party module called freezegun
. 我最喜欢的选择是使用名为
freezegun
的第三方模块。
You would need only one line to add, very clean and readable: 您只需要添加一行即可,非常清晰易读:
@freeze_time("2014-10-14")
def test_refresh_settlement_date_in_the_future(self):
...
You need to @patch get_next_friday()
function and substitute it's return value with the one you need: 您需要@patch
get_next_friday()
函数,并将其返回值替换为所需的返回值:
date_in_the_future = date.today() + timedelta(50)
next_friday_in_the_future = get_next_friday(base_date=date_in_the_future)
with patch('module_under_test.get_next_friday') as mocked_function:
mocked_function.return_value = next_friday_in_the_future
# call refresh_settlement_date
I just tried this out, it seems to work: 我只是尝试了一下,它似乎有效:
first I need to copy the function: 首先,我需要复制函数:
old_get_next_friday = get_next_friday
then patch it: 然后打补丁:
with patch.object(get_next_friday) as mocked_func:
for i in range(8):
mocked_func.return_value = old_get_next_friday(date.today() + timedelta(days=i))
refresh_settlement_date()
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