TypeError:login（）获得意外的关键字参数“template\u name”

Question

我将我的项目从Django 1.11升级到2.2，做了所有的更改，但带来了新的错误，说login（）得到了一个意想不到的关键字参数template_name。它与Django 1.11的前一个版本运行良好（所有其他网址都在工作，只有着陆页给出了错误）。我找不到任何关于这个问题的参考资料。以下是该问题的错误、网址和视图。

Internal Server Error: /
Traceback (most recent call last):
  File "C:\Users\User\venv\lib\site-packages\django\core\handlers\exception.py", line 34, in inner
    response = get_response(request)
  File "C:\Users\User\venv\lib\site-packages\django\core\handlers\base.py", line 115, in _get_response
    response = self.process_exception_by_middleware(e, request)
  File "C:\Users\User\venv\lib\site-packages\django\core\handlers\base.py", line 113, in _get_response
    response = wrapped_callback(request, *callback_args, **callback_kwargs)
  File "C:\Users\User\PycharmProjects\MyWebsite\landing\views.py", line 43, in landing_validation
    login_response = login(request, template_name='landing.html')
TypeError: login() got an unexpected keyword argument 'template_name'
[19/Apr/2019 16:20:00] "GET / HTTP/1.1" 500 71948

着陆\urls.py

from django.conf.urls import url
from landing.views import landing_validation

app_name='landing'
urlpatterns = [
    url(r'^$', landing_validation, name='landing')
]

着陆\views.py

from django.contrib.auth import login 

def landing_validation(request):
  login_response = login(request, template_name='landing.html')

  return login_response

C:\Users\User\venv\Lib\site packages\django\contrib\auth\uuuuu init\uuuuuu。派克

def login(request, user, backend=None):

    """
    Persist a user id and a backend in the request. This way a user doesn't
    have to reauthenticate on every request. Note that data set during
    the anonymous session is retained when the user logs in.
    """
    session_auth_hash = ''
    if user is None:
        user = request.user
    if hasattr(user, 'get_session_auth_hash'):
        session_auth_hash = user.get_session_auth_hash()

    if SESSION_KEY in request.session:
        if _get_user_session_key(request) != user.pk or (
                session_auth_hash and
                not constant_time_compare(request.session.get(HASH_SESSION_KEY, ''), session_auth_hash)):
            # To avoid reusing another user's session, create a new, empty
            # session if the existing session corresponds to a different
            # authenticated user.
            request.session.flush()
    else:
        request.session.cycle_key()

    try:
        backend = backend or user.backend
    except AttributeError:
        backends = _get_backends(return_tuples=True)
        if len(backends) == 1:
            _, backend = backends[0]
        else:
            raise ValueError(
                'You have multiple authentication backends configured and '
                'therefore must provide the `backend` argument or set the '
                '`backend` attribute on the user.'
            )
    else:
        if not isinstance(backend, str):
            raise TypeError('backend must be a dotted import path string (got %r).' % backend)

    request.session[SESSION_KEY] = user._meta.pk.value_to_string(user)
    request.session[BACKEND_SESSION_KEY] = backend
    request.session[HASH_SESSION_KEY] = session_auth_hash
    if hasattr(request, 'user'):
        request.user = user
    rotate_token(request)
    user_logged_in.send(sender=user.__class__, request=request, user=user)

Answer 1

您正在使用的login函数实际上不是一个视图。它只是一个常规函数，您可以使用它来验证用户身份并设置会话cookie等。

请阅读文档了解其用法：https://docs.djangoproject.com/en/2.2/topics/auth/default/#django.contrib.auth.login

根据您的代码判断，您似乎想要使用LoginView。

from django.contrib.auth.views import LoginView

def landing_validation(request):
    login_response = LoginView.as_view()(request, template_name='landing.html')

    return login_response

Answer 2

在Django 2.2应用程序中，我通过使用LoginView而不是login修复了类似错误。

文件用户/URL。py:

from django.conf.urls import url
# from django.contrib.auth.views import login
from django.contrib.auth.views import LoginView

from . import views

app_name = 'users'
urlpatterns = [
    # url(r'^login/$', login, {'template_name': 'users/login.html'}, name='login'),
    url(r'^login/$', LoginView.as_view(template_name='users/login.html'), name='login'),
]

Answer 3

没有必要做landing_validation查看这条线解决了这个问题。

url(r'^$', LoginView.as_view(template_name='landing.html'), name='landing')