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Question

我需要开发Django中间件并创建类似这样的类：

import re
from django.conf import settings
from django.shortcuts import redirect
from django.contrib.auth import logout


class LoginRequiredMiddleware:

    def __init__(self, get_response):
        self.get_response = get_response

    def __call__(self, request):
        response = self.get_response(request)
        return response

    def process_view(self, request, view_func, view_args, view_kwargs):
        assert hasattr(request, 'user')

        path = request.path_info
        print(path)

        p = path.lstrip('/')
        compiled_login_url = re.compile(settings.LOGIN_URL)

        print(compiled_login_url.match(p))


        if not request.user.is_authenticated:
            if path == '/':
                print("==============> INDEX PAGE")

现在，在这里，我想比较网址的路径。我必须确定路径是否与 settings.LOGIN_URL 匹配。我曾尝试使用正则表达式，但它们没有帮助。

这是我的网址

urlpatterns = [
              path('admin/', admin.site.urls),
              path('', views.index, name='index'),
              path('<slug:slug>/', include('app.urls')),
              path('<slug:slug>/accounts/login/', views.login, name='login'),
              path('<slug:slug>/accounts/logout/', views.logout, name='logout'),
              path('<slug:slug>/dashboard/', 'views.dashboard', name='dashboard'),
          ] + static(settings.MEDIA_URL, document_root=settings.MEDIA_ROOT)

我想免除登录和注销并在中间件中进行检查，我也需要子弹。

views.py

def user_login(request, slug):
    """
    Login user using email and password

    :param request: WSGI Request
    :param slug: to verify the facility
    :return: login page or redirect
    """

    context = {}
    try:
        facility = models.Facility.objects.get(slug=slug)
        context['facility'] = facility

    except models.Facility.DoesNotExist:
        return redirect('index')

    if request.method == "POST":
        email = request.POST['email']
        password = request.POST['password']

        user = authenticate(request, username=email, password=password)
        if user is not None:
            if user.profile.facility.slug == slug:
                login(request, user)
                return redirect('dashboard', slug=slug)
            else:
                context['error'] = "You don't belong to this facility"

        else:
            context['error'] = "Invalid email or password"

    return render(request, 'app/user/login.html', context)


@login_required
def dashboard(request, slug):
    """
    Display the dashboard of facility and list down bird
    view for the daily, weekly, monthly and yearly progress

    :param request:
    :return:
    """
    context = {}

    try:
        facility = models.Facility.objects.get(slug=slug)
        context['facility'] = facility
    except models.Facility.DoesNotExist:
        return render(request, 'app/404.html', status=404)

index.html

<!DOCTYPE html>
<html lang="en">
<head>
    <meta charset="UTF-8">
    <title>Title</title>
</head>
<body>
<ul>
    {% for f in facilities %}
        <li><a href="{% url 'login' f.slug %}">{{ f.name }}</a></li>
    {% empty %}
        <li>No Facility</li>
    {% endfor %}
</ul>
</body>
</html>

Answer 1

您可以将Django的身份验证视图一个一个地添加到您的urlpatterns中，并用login_exempt装饰每个。

OR

如果您希望将检查移入中间件，则可以尝试以下操作：

urls.py

urlpatterns = [
    path('admin/', admin.site.urls),
    path('', include("myapp.urls")),
    path('accounts/', include('django.contrib.auth.urls', namespace="accounts"))
]

middleware.py

class LoginRequiredMiddleware:            # <-- Custom Middleware
    def __init__(self, get_response):
        self.get_response = get_response

    def __call__(self, request):
        return self.get_response(request)

    def process_view(self, request, view_func, view_args, view_kwargs):

        if getattr(view_func, 'login_exempt', False):
            return

        if 'accounts' in request.resolver_match.namespaces:
            return

其他可能性可能是检查中间件中的url或视图名称，但也许最干净的方法是使用装饰器并逐个添加必要的视图。

OR

您可以为此使用django-decorator-include。

$ pip install django-decorator-include

在你project/url.py

from decorator_include import decorator_include
from project/custom_middleware import login_exempt


urlpatterns = [
    path('admin/', admin.site.urls),
    path('', include("myapp.urls")),
    # do this 
    path('accounts/', decorator_include(login_exempt, 'django.contrib.auth.urls'))


]

不要忘记将软件包添加到requirements.txt

比较Django 2中间件中的url路径

1 个答案:

OR

OR