如何模拟在 python 中为特定路径打开的文件？

Question

So I know that in my unit test I can mock a context manager open(), ie:

with open('file_path', 'r') as stats:

mocked with

with mock.patch('builtins.open', mock.mock_open(read_data=mock_json)):

but is there a way for me to only mock it for a specific file path? Or maybe some other way to ensure that the context manager gets called with the correct path in a unit test?

Answer 1

To mock open only for a specific path, you have to provide your own mock object that handles open differently, depending on the path. Assuming we have some function:

def do_open(path):
    with open(path, "r") as f:
        return f.read()

where open shall be mocked to return a file with the content "bar" if path is "foo", but otherwise just work as usual, you could do something like this:

from unittest import mock
from my_module.do_open import do_open

builtin_open = open  # save the unpatched version

def mock_open(*args, **kwargs):
    if args[0] == "foo":
        # mocked open for path "foo"
        return mock.mock_open(read_data="bar")(*args, **kwargs)
    # unpatched version for every other path
    return builtin_open(*args, **kwargs)

@mock.patch("builtins.open", mock_open)
def test_open():
    assert do_open("foo") == "bar"
    assert do_open(__file__) != "bar"

If you don't want to save the original open in a global variable, you could also wrap that into a class:

class MockOpen:
    builtin_open = open

    def open(self, *args, **kwargs):
        if args[0] == "foo":
            return mock.mock_open(read_data="bar")(*args, **kwargs)
        return self.builtin_open(*args, **kwargs)

@mock.patch("builtins.open", MockOpen().open)
def test_open():
    ...