Python 如何为templateView创建URL？

如何使用TemplateView为表单编写url。我编写了一个方法来验证并通过表单传递公司的详细信息。使用我试图访问HTML字段的表单对象

Form.py

View.py

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我想使用TemplateView运行我的company_details.html文件。我无法为相同的内容编写url。请建议模板视图只有get方法

def get(self, request, *args, **kwargs):
    return render(request,self.template_name, {'form': self.form})

---

如果您有get和post方法，请使用FormView

在URL.py中尝试了什么？为什么要使用模板视图而不是create视图来为您处理表单？from django.urls从django.urls导入路径从django.views导入包含从django.views.generic导入模板视图从。导入视图app_name='astra'urlpatterns=[path，TemplateView.as_viewtemplate_name='astra/index.html'，name='home'，path'contact'，views.ContactView.as_view，path'company'，views.GetCompanyView.as_view，]

class GetCompanyView(TemplateView):
    template_name = "astra/company_details.html"
    form = CompanyDetailsForm()

    def get_context_data(self,**kwargs):
        context = super().get_context_data(**kwargs)
        context['form']=self.form
        return context

    def company_details(request):
        if request.method =="POST":
            form = CompanyDetailsForm(request.POST)
            if form.is_valid():
                company_name = form.cleaned_data['company_name']
                contact_person = form.cleaned_data['contact_person']
                email = form.cleaned_data['email']
                employee_count = form.cleaned_data['employee_count']
                mobile_number = form.cleaned_data['mobile_number']
                try:
                    form.save()
                    send_mail(company_name,contact_person,email,employee_count,mobile_number,['salesastra500@gmail.com'])
                except BadHeaderError:
                        return BadHeaderError
            return render(request,'astra/company_details.html',{'form':form})
        else:
            return render(request,'astra/company_details.html')

def get(self, request, *args, **kwargs):
    return render(request,self.template_name, {'form': self.form})