doctest (python) 中的模拟 Y (from X import Y)

发布于 2024-08-20 12:38:28 字数 612 浏览 7 评论 0原文

我正在尝试创建一个带有模拟功能的文档测试，该功能位于单独的模块中并按以下方式导入

from foomodule import foo

def bar():
    """
    >>> from minimock import mock
    >>> mock('foo', nsdicts=(bar.func_globals,), returns=5)
    >>> bar()
    Called foo()
    10
    """
    return foo() * 2


import doctest
doctest.testmod()

foomodule.py:

def foo():
    raise ValueError, "Don't call me during testing!"

这失败了。

如果我将 import 更改为 import foomodule 并在任何地方使用 foomodule.foo 然后就可以了。

但是对于通过上述方式导入的模拟函数有什么解决方案吗？

原文

I'm trying to create a doctest with mock of function that resides in a separate module
and that is imported as bellow

from foomodule import foo

def bar():
    """
    >>> from minimock import mock
    >>> mock('foo', nsdicts=(bar.func_globals,), returns=5)
    >>> bar()
    Called foo()
    10
    """
    return foo() * 2


import doctest
doctest.testmod()

foomodule.py:

def foo():
    raise ValueError, "Don't call me during testing!"

This fails.

If I change import to import foomodule
and use foomodule.foo everywhere
Then it works.

But is there any solution for mocking function imported the way above?

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秋叶绚丽 2024-08-27 12:38:28

您刚刚遇到了最好不要从“内部”模块导入对象的众多原因之一——仅导入模块本身（可能来自包内）。我们已将此规则纳入 Google 风格指南（发布于此处) 我衷心向每一位 Python 程序员推荐它。

话虽这么说，您需要做的就是获取刚刚用模拟替换的 foomodule.foo 并将其粘贴到当前模块中。我不记得足够多的 doctest 内部来确认是否

   >>> import foomodule
   >>> foo = foomodule.foo

足够了——尝试一下，如果不起作用，那就做是

   >>> import foomodule
   >>> import sys
   >>> sys.modules[__name__].foo = foomodule.foo

的，这是一团糟，但造成这种混乱的原因是看起来无辜的from foomodule import foo -- 避免这种情况，你的生活会更简单、更有效率;-)。

You've just met one of the many reasons that make it best to never import object from "within" modules -- only modules themselves (possibly from within packages). We've made this rule part of our style guidelines at Google (published here) and I heartily recommend it to every Python programmer.

That being said, what you need to do is to take the foomodule.foo that you've just replaced with a mock and stick it in the current module. I don't recall enough of doctest's internal to confirm whether

   >>> import foomodule
   >>> foo = foomodule.foo

will suffice for that -- give it a try, and if it doesn't work, do instead

   >>> import foomodule
   >>> import sys
   >>> sys.modules[__name__].foo = foomodule.foo

yeah, it's a mess, but the cause of that mess is that innocent-looking from foomodule import foo -- eschew that, and your life will be simpler and more productive;-).

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