如何将mixin应用于Django应用程序中的所有cbv？

Question

假设我想使用自己创建的LoginRequiredMixin和UserPermissionMixin，并将它们应用于应用程序中的所有视图。这只是一个例子，我可能也有一些混音添加一些上下文或做其他事情

我可以手动执行，例如以下视图：

class MyCreateView(LoginRequiredMixin, UserPermissionMixin, CreateView)

但是，由于我有很多视图，并且可能有其他特定的视图混合，所以这会变得混乱并且难以管理

想到的一个解决方案是为泛型视图创建新类：

class DecoratedCreateView(LoginRequiredMixin, UserPermissionMixin, CreateView):
    pass


class DecoratedDetailView(LoginRequiredMixin, UserPermissionMixin, DetailView):
    pass


class DecoratedUpdateView(LoginRequiredMixin, UserPermissionMixin, UpdateView):
    pass


class DecoratedDeleteView(LoginRequiredMixin, UserPermissionMixin, DeleteView):
    pass

然后，使用这些作为我的通用视图：

class MyCreateView(DecoratedCreateView)

这是个好办法吗？我是否必须在上面的类中添加任何方法，还是将它们留空就可以按预期工作？ 有没有其他方法可以实现这一点，也许在url.py中

Answer 1

你的方法很好。我在一些项目中这样做有点不同：

myapp/views/generic.py

from django.views.generic import (
    CreateView as BaseCreateView,
    DetailView as BaseDetailView,
    UpdateView as BaseUpdateView,
    DeleteView as BaseDeleteView,
)

__all__ = ['MyappMixin', 'CreateView', 'DetailView', 'UpdateView', 'DeleteView']


class MyappMixin(LoginRequiredMixin, UserpermissionMixin):
    pass


class CreateView(MyappMixin, BaseCreateView):
    pass


class DetailView(MyappMixin, BaseDetailView):
    pass


class UpdateView(MyappMixin, BaseUpdateView):
    pass


class DeleteView(MyappMixin, BaseDeleteView):
    pass

myapp/views/base.py

from .generic import CreateView

class MyCreateView(CreateView):
    pass

它工作良好，没有太多的麻烦，并允许您轻松地跳过混合异常，如果需要的话

根据用例，另一种解决方案可能是使用middlewares或context processors

class MyMiddleware:

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

    def __call__(self, request):
        is_in_myapp = request.resolver_match.app_name == 'myapp'
        if is_in_myapp and not request.user.is_authenticated:
            response = HttpResponse("Permission denied", status=403)
        else:
            response = self.get_response(request)
        return response