FastAPI - 模拟路径函数没有效果

问题描述 投票:0回答:2

我有一个简单的 FastAPI 应用程序,我正在尝试使用

pytest
为其创建测试。

我的目标是测试应用程序在出现不同错误时的行为方式。

我的应用程序中有一个简单的健康检查路线:

from fastapi import APIRouter

router = APIRouter()


@router.get("/health")
async def health():
    return "It's working ✨"

现在在我的 pytest 模块中,我正在尝试修补上述函数,以便它引发不同的错误。 我正在使用

unittest.mock
但我的行为非常奇怪。

import pytest
from unittest import mock

from fastapi import HTTPException
from starlette.testclient import TestClient

import app.api.health
from app.main import app  # this is my application (FastAPI instance) with the `router` attached


@pytest.fixture()
def client():
    with TestClient(app) as test_client:
        yield test_client


def test_simple(client):
    def mock_health_function():
        raise HTTPException(status_code=400, detail='gibberish')

    with mock.patch('app.api.health.health', mock_health_function):
        response = client.get(HEALTHCHECK_PATH)

        with pytest.raises(HTTPException):  # this check passes successfully - my exception is raised
            app.api.health.health()

    assert response.status_code != 200  # this check does not pass. The original function was called as if nothing was patched

尽管测试内部调用了完全相同的函数,但当我到达端点时,API 测试客户端仍然调用原始函数。

为什么测试中不直接调用函数时

mock.patch
无法正常工作?

或者也许我应该以不同的方式解决我的问题?

python pytest python-unittest fastapi starlette
2个回答
11
投票

您可以使用

monkeypatch
夹具来修补您的功能。

首先拉出你要修补的代码部分:

from fastapi import FastAPI

app = FastAPI()


def response():
    return "It's working ✨"


@app.get("/health")
async def health():
    return response()

然后在你的测试中使用monkeypatch

import pytest

from fastapi import HTTPException
from starlette.testclient import TestClient

from app import main


def mocked_response():
    raise HTTPException(status_code=400, detail='gibberish')


@pytest.fixture()
def client():
    from app.main import app

    with TestClient(app) as test_client:
        yield test_client


def test_simple(client, monkeypatch):

    monkeypatch.setattr(main, "response", mocked_response)
    resp = client.get("/health")
    assert resp.status_code == 400
    assert resp.json()["detail"] == "gibberish"

另一种方法是使用 Dependencydependency_overrides。 这可能不适用于所有场景,但对于您给定的用例来说它确实有效。

from fastapi import FastAPI,  Depends

app = FastAPI()


def response():
    return "It's working ✨"


@app.get("/health")
async def health(resp=Depends(response)):
    return resp

在您的测试客户端中,您现在可以像这样覆盖依赖项:

import pytest

from fastapi import HTTPException
from starlette.testclient import TestClient

from app.main import response


def mocked_response():
    raise HTTPException(status_code=400, detail='gibberish')


@pytest.fixture()
def client():
    from app.main import app
    app.dependency_overrides[response] = mocked_response

    with TestClient(app) as test_client:
        yield test_client


def test_simple(client):

    resp = client.get("/health")

    assert resp.status_code == 400
    assert resp.json()["detail"] == "gibberish"

如果您需要向响应函数添加参数,您可以使用闭包模式

def response_closure():
    def response(arg):
        return arg
    return response


@app.get("/health")
async def health(resp=Depends(response_closure)):
    return resp("It's working ✨")

0
投票
import pytest
from unittest import mock

from fastapi import HTTPException
from starlette.testclient import TestClient

import app.api.health
from app.main import app  # this is my application (FastAPI instance) with the `router` attached


@pytest.fixture()
def client():
    with TestClient(app) as test_client:
        yield test_client


def test_simple(client):
    def mock_health_function():
        raise HTTPException(status_code=400, detail='gibberish')

    with mock.patch('app.api.health.health', mock_health_function):
        response = client.get(HEALTHCHECK_PATH)

        with pytest.raises(HTTPException):  # this check passes successfully - my exception is raised
            app.api.health.health()

        assert response.status_code != 200  # this check does not pass. The original function was called as if nothing was patched

您正在使用 with 语句将补丁用作上下文管理器,尝试将断言语句放入 with 块中,如上所示,我猜它应该会通过。

让我知道这是否有效!

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