如何修复 Django 中的“名称‘csrf’未定义”错误

Question

I took this code in the WEB but it was for Django 1.9. I'm using 2.1 in my project.

I import this:

from django.shortcuts import render_to_response, reverse
from django.views import View
from django.core.mail import send_mail
from .forms import ContactForm
from blog import settings

class EContactsView(View):
    template_name = 'home/contacts.html'

def get(self, request, *args, **kwargs):
    context = {}
    context.update(csrf(request))
    context['contact_form'] = ContactForm()

    return render_to_response(template_name=self.template_name, context=context)

def post(self, request, *args, **kwargs):
    context = {}

    form = ContactForm(request.POST)

    if form.is_valid():
        email_subject = 'EVILEG :: Сообщение через контактную форму '
        email_body = "С сайта отправлено новое сообщение\n\n" \
                     "Имя отправителя: %s \n" \
                     "E-mail отправителя: %s \n\n" \
                     "Сообщение: \n" \
                     "%s " % \
                     (form.cleaned_data['name'], form.cleaned_data['email'], form.cleaned_data['message'])

        send_mail(email_subject, email_body, settings.EMAIL_HOST_USER, ['target_email@example.com'], fail_silently=False)

    return render_to_response(template_name=self.template_name, context=context)

name 'csrf' is not defined / traceback: img

Answer 1

Don't use render_to_response . It was obsolete, even in Django 1.9.

Use render instead. Then you don't have to do anything in the view to handle csrf protection.

Change the import to

from django.shortcuts import render

Change the get method to:

def get(self, request, *args, **kwargs):
    context = {}
    context['contact_form'] = ContactForm()
    return render(request, template_name=self.template_name, context=context)

and the final line of the post method to:

return render(request, template_name=self.template_name, context=context)

Finally, you might prefer to use FormView here. The example in the docs is for a contact form.