如何将 request.user 导入 Django 中的 TemplateView

How to import request.user into TemplateVew in djnago

我需要知道如何使用渲染页面的 TemplateView 方法检查用户是否经过身份验证。

我已将其添加到我的上下文处理器中:

    TEMPLATE_CONTEXT_PROCESSORS = [
    'django.contrib.auth.context_processors.auth',
]

我的 views.py 目前看起来像:

    from django.shortcuts import render
from CreateVuln.forms import *
from django.views.generic import TemplateView
from django.template import RequestContext

from pages.decorators import *

vulnform = vulnform


class Dashboard(TemplateView):
    template_name = 'vuln_pages/Create_Vuln.html'

    def get(self, request):
        Outform = {
                     'vulnform': vulnform,

                     }
        return render(request, self.template_name, Outform)

    def post(self, request):
            forminput = vulnform(request.POST or None)


            if forminput.is_valid():
                forminput.save()
                forminput = vulnform()



                inoutform = {
                             'vulnform': forminput,

                             }
                return render(request, self.template_name, inoutform, )

            else:

                inoutform = {
                             'vulnform': vulnform,

                             }

                return render(request, self.template_name,inoutform )

# Create your views here.

class ViewVulns(TemplateView):

    template_name = 'vuln_pages/Create_Vuln.html'

    def get(self, request):

        return render(request, self.template_name)

我想使页面无法通过 GET 请求查看,也无法通过 POST 请求更新。 我试过使用 RequestContext。但是文档看起来很复杂,我似乎无法理解如何使用它。

你可以试试这个:if request.user.is_authenticated:

使用LoginRequiredMixin混合

If a view is using this mixin, all requests by non-authenticated users will be redirected to the login page or shown an HTTP 403 Forbidden error, depending on the raise_exception parameter.

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


class Dashboard(<b>LoginRequiredMixin,</b> TemplateView):
    # your code