在pytest中如何编写返回其他函数AsyncGenerator的fixture？

正确示例 #

以下是可正常运行的异步生成器Fixture测试用例：

from collections.abc import AsyncGenerator

import pytest

@pytest.fixture()
async def fixture() -> AsyncGenerator[str, None]:
    yield "a"

@pytest.mark.asyncio
async def test(fixture: str):
    assert fixture[0] == "a"

错误示例及报错 #

当尝试让Fixture返回另一个函数生成的异步生成器时，会触发类型错误：

from collections.abc import AsyncGenerator

import pytest

async def _fixture() -> AsyncGenerator[str, None]:
    yield "a"

@pytest.fixture()
async def fixture() -> AsyncGenerator[str, None]:
    return _fixture()

@pytest.mark.asyncio
async def test(fixture: str):
    assert fixture[0] == "a"

报错信息：

>     assert fixture[0] == "a"
E     TypeError: 'async_generator' object is not subscriptable

问题原因与修正 #

你忽略的核心点是：异步Fixture的生成器需要通过yield（或yield from）传递产出值，而非直接返回生成器对象。

错误代码中，return _fixture()会把异步生成器本身返回给测试函数，导致测试里的fixture变量是async_generator对象，而非生成器产出的字符串"a"，自然无法执行下标操作。

正确写法是用yield from迭代内部异步生成器，传递它的产出值：

from collections.abc import AsyncGenerator

import pytest

async def _fixture() -> AsyncGenerator[str, None]:
    yield "a"

@pytest.fixture()
async def fixture() -> AsyncGenerator[str, None]:
    yield from _fixture()  # 用yield from替代return

@pytest.mark.asyncio
async def test(fixture: str):
    assert fixture[0] == "a"

内容的提问来源于stack exchange，提问作者sttawm