Pytest - 从另一个装置调用一个装置

Question

我有一个夹具返回某种类型的对象，我在另一个文件中定义了另一个夹具，它基本上使用该对象来做其他事情。但是我无法从我的第一个装置中返回对象。

file-1

def fixture_1(s, **kwargs):
    def hook(s, **kwargs):
        p_b = s.get()
        p = p_b.build()
        yield p
    return hook

file-2 conftest.py

@pytest.fixture(scope='module')
def fixture_1(s, **kwargs):
    def hook(s, **kwargs):
        #Default implementation is no-op.
        pass
    return hook

@pytest.fixture(scope='module')
def fixture_2(s,b_p):
    some_p = fixture_1(s)
    current_status = s.start(some_p)

    print(current_status)
    yield current_status

我想基本上检索p返回的对象file-1 fixture_1并在file-2 fixture_2夹具中使用它。

Answer 1

包含简单示例

Py.test 支持直接开箱即用的夹具调用其他夹具

将固定装置放在 conftest.py 或测试文件中：

conftest.py：

@pytest.fixture(scope="session")
def fixture_A():
    some_obj = create_obj()
    return some_obj  # or use yield some_obj in case you want to destruct

@pytest.fixture(scope="session")
def fixture_B(fixture_A):
   this_is_some_obj = fixture_A
   # do something
   another = {}
   return this_is_some_obj, another

测试示例.py：

@pytest.fixture(scope="session")
def fixture_C(fixture_B):
   return fixtureB

def test_finally_the_test(fixture_C):
    some_obj, another = fixture_C
 

值得一提的是，上面的固定装置将被调用一次（即使多个测试使用这些固定装置） - 这是由于每个固定装置的“会话”范围，就像这些固定装置是单例一样（如果与 OOP 相比）

另一个注意事项 pytest 知道运行装置的顺序（它检查依赖关系 - 所以这里没有什么特别要做的）

Answer 2

您似乎使用了错误的 pytest 装置（查看您的参数名称）。

我建议您浏览https://docs.pytest.org/en/latest/fixture.html

您的问题有两种解决方案：

###
# file_1
def not_really_a_fixture(s, **kwargs): # just some hook generator
    def hook(s, **kwargs):
        p_b = s.get()
        p = p_b.build()
        yield p
    return hook


###
# conftest.py
from file_1 import not_really_a_fixture

@pytest.fixture(scope='module')
def fixture_2(s,b_p): # s and b_p should be names of fixtures that need to run before this
    some_p = not_really_a_fixture(s)
    current_status = s.start(some_p)

    print(current_status)
    yield current_status

和：

###
# file_1
@pytest.fixture(scope='module')
def fixture_1(s): # s is name of another fixture
    # there is no point in having **kwargs as arg in pytest fixture
    def hook(s, **kwargs):
        #Default implementation is no-op.
        pass
    return hook

###
# conftest.py
from file_1 import fixture_1

@pytest.fixture(scope='module')
def fixture_2(s,b_p,fixture_1): # s and b_p should be names of fixtures that need to run before this
    # in fixture_1 is value returned by fixture_1, that means your hook func
    current_status = s.start(fixture_1)

    print(current_status)
    yield current_status