将函数装饰器转换为 Class Mixin django

Question

我尝试将 cutoms 装饰器从函数转换为 Class Mixin，因为我使用 CVB 并且我想继承此 Mixin 以检查某些页面的用户状态。我有这个装饰器来检查用户是否经过身份验证，如果用户经过身份验证并尝试访问具有此装饰器的页面，它将被重定向到仪表板，如果没有，那么 il 将有权访问该页面。我将此 class 编写为装饰器的 class 版本，如果用户已登录，则它可以正常工作，但如果未登录并尝试访问该页面，则会出现此错误：

在 /auth/login/ 处配置不当 UserLoginView 缺少 permission_required 属性。定义 UserLoginView.permission_required，或覆盖 UserLoginView.get_permission_required().

这是装饰器：

def is_authenticated(view_func):
    def wrapper_func(request, *args, **kwargs):
        if request.user.is_authenticated:
            return redirect('dashboard')
        else:
            return view_func(request, *args, **kwargs)
    return wrapper_func

class版本：

class IsAuthenticatedMixin(PermissionRequiredMixin):
    def dispatch(self, request, *args, **kwargs):
        if request.user.is_authenticated:
            return redirect('dashboard')
        return super().dispatch(request, *args, **kwargs)

继承这个mixin的视图

class IndexFormView(IsAuthenticatedMixin, CreateView):
    permission_required = 'view'
    template_name = 'home/index.html'
    form_class = NewsletterForm

    def post(self, request, *args, **kwargs):
        email = request.POST['email']
        if Newsletter.objects.filter(email=email).exists():
            messages.warning(
                request, 'This email is already subscribed in our system')
        else:
            Newsletter.objects.create(
                email=email)
            messages.success(request,
                             'Your email was subscribed in our system, you\'ll hear from us as soon as possible !')
        return super().post(request, *args, **kwargs)

    def get_context_data(self, **kwargs):
        context = super().get_context_data(**kwargs)
        context['testimonials'] = Testimonial.objects.order_by('-created_date')[:9]
        return context


class AboutTemplateView(IsAuthenticatedMixin, TemplateView):
    template_name = 'home/about.html'

    def get_context_data(self, **kwargs):
        context = super().get_context_data(**kwargs)
        context['banner_page_title'] = 'About Us'
        context['page_location'] = 'home / about'
        return context

class UserResgistrationCreateView(IsAuthenticatedMixin, CreateView):
    template_name = 'home/auth/register.html'
    form_class = UserRegistrationForm
    success_url = reverse_lazy('login')


class UserLoginView(IsAuthenticatedMixin, LoginView):
    template_name = 'home/auth/login.html'

Answer 1

发生这种情况的原因是因为您从 PermissionRequiredMixin [Django-doc] subclass，这意味着在您使用 IsAuthenticatedMixin 的所有视图中，您需要提供一个 permission_required class 属性。这里的错误是关于 UserLoginView:

class UserLoginView(IsAuthenticatedMixin, LoginView):
    template_name = 'home/auth/login.html'
    <b>permission_required = …</b>

因此您需要为 … 输入一个值。但同样的问题会出现在其他基于 class 的视图中，您从 IsAuthenticatedMixin 继承，并且您没有为 permission_required.

提供值

然而，这就是为什么要在这里编写自定义 mixin。你所做的是由 LoginRequiredMixin [Django-doc] 实现的。唯一不同的是重定向 url，但您可以指定：

from django.contrib.auth.mixins import <b>LoginRequiredMixin</b>
from django.urls import reverse_lazy

class IsAuthenticatedMixin(<b>LoginRequiredMixin</b>):
    login_url = reverse_lazy('dashboard')
    redirect_field_name = None

在需要使用权限的视图中，也可以subclass the PermissionRequiredMixin:

from django.contrib.auth.mixins import PermissionRequiredMixin

class IndexFormView(IsAuthenticatedMixin, <b>PermissionRequiredMixin</b>, CreateView):
    <b>permission_required = 'view'</b>
    template_name = 'home/index.html'
    form_class = NewsletterForm

将函数装饰器转换为 Class Mixin django

Converting a function decorator to a Class Mixin django

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