Question

我有一个返回此JSON响应的API

{
    "message": "Staff name and password pair not match",
    "errors": {
        "resource": "Login",
        "field": "staff_authentication",
        "code": "invalid",
        "stack_trace": null
    }
}

使用pytest，我想构建一个JSON对象的副本，并确保它完全相同

import pytest
import json
from collections import namedtuple
from flask import url_for
from myapp import create_app

@pytest.mark.usefixtures('client_class')
class TestAuth:

    def test_login(self, client):
        assert client.get(url_for('stafflogin')).status_code == 405
        res = self._login(client, 'no_such_user', '123456')
        assert res.status_code == 422
        response_object = self._json2obj(res.data)
        assert response_object.message == 'Staff name and password pair not match'
        invalid_password_json = dict(message="Staff name and password pair not match",
                                    errors=dict(
                                        resource="Login",
                                        code="invalid",
                                        field="staff_authentication",
                                        stack_trace=None,)
                                    )
        assert self._ordered(response_object) == self._ordered(invalid_password_json)

    def _login(self, client, staff_name, staff_password):
        return client.post('/login',
            data=json.dumps(dict(staff_name=staff_name, staff_password=staff_password)),
            content_type='application/json',
            follow_redirects=True)

    def _json_object_hook(self, d): return namedtuple('X', d.keys())(*d.values())
    def _json2obj(self, data): return json.loads(data, object_hook=self._json_object_hook)

    def _ordered(self, obj):
        if isinstance(obj, dict):
            return sorted((k, self._ordered(v)) for k, v in obj.items())
        if isinstance(obj, list):
            return sorted(self._ordered(x) for x in obj)
        else:
            return obj

pytest表示2个对象不相等。

>       assert self._ordered(response_object) == self._ordered(invalid_password_json)
E       AssertionError: assert X(message='St...k_trace=None)) == [('errors', [(...r not match')]
E         At index 0 diff: 'Staff name and password pair not match' != ('errors', [('code', 'invalid'), ('field', 'staff_authentication'), ('resource', 'Login'), ('stack_trace', None)])
E         Full diff:
E         - X(message='Staff name and password pair not match', errors=X(resource='Login', field='staff_authentication', code='invalid', stack_trace=None))
E         + [('errors',
E         +   [('code', 'invalid'),
E         +    ('field', 'staff_authentication'),
E         +    ('resource', 'Login'),
E         +    ('stack_trace', None)]),
E         +  ('message', 'Staff name and password pair not match')]

tests/test_app.py:31: AssertionError
=========================== 1 failed in 0.22 seconds ===========================

如何将新创建的JSON对象与响应相同？

Answer 1

我没有将JSON响应转换为 Object ，而是使用json.loads()将其转换为 Dictionary ，并进行比较。

def test_login(self, client):
        res = return client.post('/login',
            data=json.dumps(dict(staff_name='no_such_user', staff_password='password')),
            content_type='application/json',
            follow_redirects=True)
        assert res.status_code == 422
        invalid_password_json = dict(message="Staff name and password pair not match",
                                    errors=dict(
                                        resource="Login",
                                        code="invalid",
                                        field="staff_authentication",
                                        stack_trace=None,),
                                    )
        assert json.loads(res.data) == invalid_password_json

这样，我不必担心JSON响应中的空格差异，以及JSON结构的排序。只需让Python的Dictionary比较函数检查是否相等。

Answer 2

如果确实需要两个doctionaries之间的字面值，值相等，那么比较它们的json序列化结果会更简单，否则你需要对dicts及其值进行一些递归比较

注意：由于python中的dicts是未排序的集合，因此您需要将sort_keys=True传递给json.dumps，有关详细信息，请参阅this question

Python：比较pytest中的两个JSON对象

2 个答案: