Django 身份验证：员工索引和用户索引（2 个模板）

Question

我目前有 2 个功能可用，一个是 is_staff 用户的特定视图，另一个是任何用户的特定视图。现在工作人员登录后，他们可以将路径 /dashboard 添加到 URL 并可以访问该路线，而普通用户无法查看。这是意料之中的好事。

问题：我想在登录时将 user.is_staff 直接重定向到此路径 /dashboard，而不是继续首先索引。

我认为我必须以某种方式合并这些功能： (views.py)

  1 @staff_member_required
  2 def staff_dashboard(request):
  3     users = User.objects.all()
  4     customer = Customer.objects.all()
  5     accounts = Account.objects.all()
  6     context = {
  7             'users': users,
  8             'customer': customer,
  9             'accounts': accounts
 10             }
 11     return render(request, 'bank_app/dashboard.html', context)
 12
 13
 14 @login_required
 15 def index(request):
 16     customer = Customer.objects.filter(user=request.user)
 17     accounts = Account.objects.filter(user=request.user)
 18     context = {
 19             'customer': customer,
 20             'accounts': accounts
 21             }
 22     return render(request, 'bank_app/index.html', context)

(urls.py:)

urlpatterns = [
         path('dashboard', views.staff_dashboard, name='staff_dashboard'),
         path('', views.index, name='index'),
         path('create', views.create, name='create'),
         path('createaccount', views.createaccount, name='createaccount'),
         path('details/<int:pk>', views.details, name='details'),
        ]

当用户 is_staff 时，显示“dashboard/”(dashboard.html)，而不是 ""(index.html)

*我还有下面看到的扩展模板，但这并没有解决问题：

 28    <p>Welcome {{ user }}</p>
 29
 30     {% if user.is_staff  %}
 31         {% block staffcontent  %}
 32         {% endblock %}
 33     {% else %}
 34         {% block content %}
 35         {% endblock %}
 36     {% endif %}
 37
 38       {% block footer %}
 39       {% endblock %}

################### 更新 ############### #####

到目前为止这是有效的，但我仍然需要重构：

10 @login_required  
11 def index(request):  
12     if request.user.is_staff:  
##13        @staff_member_required  
##14       def staff_dashboard(request):  
15             users = User.objects.all()  
16             customer = Customer.objects.all()  
17             accounts = Account.objects.all()  
18             context = {  
19                     'users': users,  
20                'customer': customer,  
21                     'accounts': accounts  
22 }  
23             return render(request, 'bank_app/dashboard.html', context)  
##24             return HttpResponseRedirect('staff_dashboard')  
25     else:  
##26          def user_dashboard():  
27             customer = Customer.objects.filter(user=request.user)  
28             accounts = Account.objects.filter(user=request.user)  
29             context = {  
30                     'customer': customer,  
31                     'accounts': accounts  
32                     }  
33             return render(request, 'bank_app/index.html', context)  
34

我被指示在声明中使用装饰器 - 教师要求：/ - （因为只有工作人员可以查看部分内容）并使用 HttpResponse 来显示正确的 html。知道如何重构这个吗？

我正在尝试添加在注释掉的行 (##) 中看到的逻辑

(urls.py):

5 urlpatterns = [
6           path('', views.index, name='index'),
##7         path('staff_dashboard', views.index, name='staff_dashboard'),
##8         path('user_dashboard/', views.index, name='user_dashboard'),

############## 解决方案###################

按照评论中的建议，我已修复 urls.py 并将请求拆分为 views.py:

  1 @login_required
  2 def index(request):
  3     if request.user.is_staff:
  4         return HttpResponseRedirect('staff_dashboard/')
  5     else:
  6         return HttpResponseRedirect('user_dashboard/')
  7
  8
  9 @staff_member_required
 10 def staff_dashboard(request):
 11     users = User.objects.all()
 12     customer = Customer.objects.all()
 13     accounts = Account.objects.all()
 14     context = {
 15             'users': users,
 16             'customer': customer,
 17             'accounts': accounts
 18             }
 19     return render(request, 'bank_app/dashboard.html', context)
 20
 21
 22 @login_required
 23 def user_dashboard(request):
 24     customer = Customer.objects.filter(user=request.user)
 25     accounts = Account.objects.filter(user=request.user)
 26     context = {
 27             'customer': customer,
 28             'accounts': accounts
 29             }
 30     return render(request, 'bank_app/index.html', context)

(urls.py):

  1 urlpatterns = [
  2         path('', views.index, name='index'),
  3         path('staff_dashboard/', views.staff_dashboard, name='staff_dashboard'),
  4         path('user_dashboard/', views.user_dashboard, name='user_dashboard'),

Answer 1

尝试这样的事情。

@login_required
 def index(request):
      if request.user.is_superuser:
           return redirect('staff_dashboard')
      else:
          customer = Customer.objects.filter(user=request.user)
          accounts = Account.objects.filter(user=request.user)
          context = {
             'customer': customer,
            'accounts': accounts
             }

          return render(request, 'bank_app/index.html', context)

要检查用户是否 is_staff 你可以这样做 user.is_staff 或 user.is_superuser.
让我知道它是否有效。

Django 身份验证：员工索引和用户索引（2 个模板）

Django authentication: index for staff and index for user (2 templates)

python

django

django-templates

django-views

django-authentication