Django模板中的Django CreateUpdateView实现

问题描述

我对django Class based views非常陌生，试图将其与现有项目集成。 我的目标是使用与Class based view和Create相同的Update学生申请表。 我正在尝试从solution的@scubabuddha的答案中整合CreateUpdateView。

views.py

from createupdateview import CreateUpdateView

class StudentView(CreateUpdateView):
   template_name="admission.html"
   Model = Student
   form_class = StudentForm
  
  def get(self,request,*args,**kwargs):
    return self.post(request,**kwargs)

  def post(self,**kwargs):
    forms = {"userform": UserForm(request.POST or None),guardian...,contact...,# other 5-6 forms}
    
    if request.POST:
      invalid_forms = [form.is_valid() for form in self.forms.values()]
      if not all(invalid_forms):
        messages.error(request,'You must fill up all the valid mandatory form fields!')
        return render(request,'admission.html',forms)
      #-- Logic to validate and save each form
      ...
      return render(request,forms)  
    return render(request,forms)

这对于CreateView来说是完美的工作，但是无法理解如何将其用于UpdateView或editview。如果用户点击editview，则{{form|crispy}}也应显示详细信息，以允许用户编辑表单。

urls.py（我也想将2个以下的URL合并为1，我们可以这样做吗？）

from django.urls import path
from students import views 
from students.views import StudentList,StudentView

urlpatterns = [
    
    path('',StudentList.as_view(),name='students'),path('add/',StudentView.as_view(),name='addview'),path('<int:pk>/edit/',name='editview'),...
]

我想以UpdateView形式显示所有学生详细信息-

admission.html

<form class="form" name="admissionForm" id="admissionForm" method="post" 
            enctype="multipart/form-data" action="{% url 'addview' %}"> 
 {% csrf_token %}
  <div class="pages">
     <h4 class="mt-2 mb-3">Personal information</h4>
      {% include "student_errors.html" %}                
      {{userform|crispy}}      #-- It should display student details 
      {{guardian_form|crispy}}    
      {{contact_form|crispy}}    
      ....
      <div class="btn_container">
          <button type="submit" class="btn btn-info float-right btn-next">Submit</button>  
      </div>
 </div>

P.S。在这里，我保留了最少的代码，在实际生产中，它的代码很大。 （将Django应用1.9迁移到3.0）

解决方法

您从this solution使用的CreateUpdateView继承自ModelFormMixin，并希望仅处理一种形式（初始化，form_class，保存等）。并且在您的代码中，您正在重写get()和post()方法，因此从CreateUpdateView继承是没有意义的。

您可以使用简单的View（未经测试）来完成此操作：

from django.http import Http404
from django.views.generic import View
from django.shortcuts import render

class StudentView(View):
    template_name = "admission.html"

    def get_object(self):
        if self.kwargs.get('pk'):
            try:
                obj = Student.objects.get(pk=pk)
            except Student.DoesNotExist:
                raise Http404("No student found matching the query")
            else:
                return obj
        return None # create view

    def get_view(self,request,*args,**kwargs):
        forms = {"userform": UserForm(request.POST or None,instance=self.object),guardian...,contact...} # other 5-6 forms}
        if request.POST:
            invalid_forms = [form.is_valid() for form in self.forms.values()]
            if not all(invalid_forms):
                messages.error(request,'You must fill up all the valid mandatory form fields!')
            else:
                #-- Logic to validate and save each form
        return render(request,self.template_name,forms)  

    def get(self,**kwargs):
        self.object = self.get_object()
        return self.get_view(request,**kwargs)

    def post(self,**kwargs)