我使用Connexion Flask框架来构建微服务。我想使用py.test
为我的应用程序编写测试。
在pytest-flask
文档中,它说要在conftest.py
中创建一个像这样创建应用程序的工具:
conftest.py
import pytest
from api.main import create_app
@pytest.fixture
def app():
app = create_app()
return app
在我的测试中,我使用client
这样的夹具:
test_api.py
def test_api_ping(client):
res = client.get('/status')
assert res.status == 200
但是当我运行py.test
时,我收到以下错误消息:
==================================== ERRORS ====================================
_______________________ ERROR at setup of test_api_ping ________________________
request = <SubRequest '_monkeypatch_response_class' for <Function 'test_api_ping'>>
monkeypatch = <_pytest.monkeypatch.MonkeyPatch instance at 0x7f9f76b76518>
@pytest.fixture(autouse=True)
def _monkeypatch_response_class(request, monkeypatch):
"""Set custom response class before test suite and restore the original
after. Custom response has `json` property to easily test JSON responses::
@app.route('/ping')
def ping():
return jsonify(ping='pong')
def test_json(client):
res = client.get(url_for('ping'))
assert res.json == {'ping': 'pong'}
"""
if 'app' not in request.fixturenames:
return
app = request.getfuncargvalue('app')
monkeypatch.setattr(app, 'response_class',
> _make_test_response_class(app.response_class))
E AttributeError: 'App' object has no attribute 'response_class'
如何让py.test
工作?这是我的create_app
函数:
main.py
import connexion
def create_app():
app = connexion.App(__name__, port=8002,)
app.add_api('swagger.yaml')
return app
if __name__ == "__main__":
create_app().run()
答案 0 :(得分:10)
<强> test_api.py 强>
import pytest
import connexion
flask_app = connexion.FlaskApp(__name__)
flask_app.add_api('swagger.yml')
@pytest.fixture(scope='module')
def client():
with flask_app.app.test_client() as c:
yield c
def test_health(client):
response = client.get('/health')
assert response.status_code == 200
<强> swagger.yml 强>
swagger: '2.0'
info:
title: My API
version: '1.0'
consumes:
- application/json
produces:
- application/json
schemes:
- https
paths:
/health:
get:
tags: [Health]
operationId: api.health
summary: Health Check
responses:
'200':
description: Status message from server describing current health
<强> api.py 强>
def health():
return {'msg': 'ok'}, 200
使用swagger-tester的另一种解决方案:
<强> test_api.py 强>
from swagger_tester import swagger_test
authorize_error = {
'get': {
'/health': [200],
}
}
def test_swagger():
swagger_test('swagger.yml', authorize_error=authorize_error)
关于这个库的一件好事是你可以使用你的规范中提供的例子。但我不认为它与connexion.RestyResolver
开箱即用:您必须在每个端点指定OperationId。
答案 1 :(得分:0)
import pytest
from json import JSONEncoder
import pytest
from connexion import App
SWAGGER_PATH = "path_to_directory_that_containes_swagger_file"
@pytest.fixture
def app():
app = App(__name__, specification_dir=SWAGGER_PATH)
app.app.json_encoder = JSONEncoder
app.add_api("swagger.yaml")
app_client = app.app.test_client()
return app_client
def test_health(app) -> None:
"""
:except: success
"""
response = app.get("/health", content_type='application/json')
assert response.status_code == 200
有关更多信息,请检查this。