简体   繁体   中英

Error getting route from urls.py

I'm testing the application. It is necessary to test the method of processing requests coming to the address ' http://127.0.0.1:8000/api/v1/test/api_address/ '. Tell me, please, as through reverse () the full address to the client

class MyTestCase(APITestCase):

    def setUp(self):
        self.message = {
            'username': 'user_name',
            'password': 'user_password',
        }

    def test_get_token(self):
        response = self.client.post(reverse('api_address'), self.message)
        self.assertEqual(response.status_code, status.HTTP_201_CREATED)

Code for urls.py :

users_router = DefaultRouter()
users_router.register(r'test', TestViewSet, 'test')
users_router.register(r'test/api_address', APIAddressRequestSet, 'api_address')

with the current implementation, reverse ('map address') does not work, falling with an error:

django.urls.exceptions.NoReverseMatch: Reverse for 'api_address' not found. 'api_address' is not a valid view function or pattern name.

The url names for DefaultRouter are automatically generated, check the docs .

Set a base_name first:

# urls.py    
users_router = DefaultRouter()
users_router.register(r'test', TestViewSet, base_name='test')
users_router.register(r'test/api_address', APIAddressRequestSet,
                      base_name='api_address')

Now your urls are reverse-accessible via reverse('test-list') reverse('test-detail') , etc. Check the table in the docs for the other names.

Your updated test:

class MyTestCase(APITestCase):

    def setUp(self):
        self.message = {
            'username': 'user_name',
            'password': 'user_password',
        }

    def test_get_token(self):
        # note the appended `-list` to the url name
        response = self.client.post(reverse('api_address-list'), self.message)
        self.assertEqual(response.status_code, status.HTTP_201_CREATED)

The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM