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How to redirect CBV login_required to a specific route?

I have a HomeView Class which I added a login_required, but I don't know how to redirect it with the custom login page that I made.
Here are the urls.py and settings.py in my project:
myblog\urls.py
from django.contrib import admin
from django.urls import path,include
from .settings import DEBUG, STATIC_URL, STATIC_ROOT, MEDIA_URL, MEDIA_ROOT
from django.conf.urls.static import static
# from django.conf import settings
urlpatterns = [
path('admin/', admin.site.urls),
path('', include('myblog_app.urls')),
path('members/', include('django.contrib.auth.urls')),
path('members/', include('members.urls')),
]
# + static(settings.MEDIA_URL,document_root=settings.MEDIA_ROOT)
if DEBUG:
urlpatterns += static(STATIC_URL,document_root =STATIC_ROOT )
urlpatterns += static(MEDIA_URL,document_root = MEDIA_ROOT)
myblog_app\urls.py
from django.urls import path
from django.contrib.auth.decorators import login_required
from .views import HomeView,ArticleDetailView,AddPostView,UpdatePostView,DeletePostView,AddCategoryView,CategoryView,CategoryListView,LikeView,AddCommentView
urlpatterns = [
path('', login_required(HomeView.as_view()),name='home'),
path('article/<int:pk>', ArticleDetailView.as_view(),name='article-detail'),
path('add_post/', AddPostView.as_view(),name='add_post'),
path('article/<int:pk>/comment/', AddCommentView.as_view(),name='add_comment'),
path('article/edit/<int:pk>', UpdatePostView.as_view(),name='update_post'),
path('article/<int:pk>/remove', DeletePostView.as_view(),name='delete_post'),
path('add_category/', AddCategoryView.as_view(),name='add_category'),
path('category/<str:cats>', CategoryView,name='category'),
path('category-list', CategoryListView,name='category_list'),
path('like/<int:pk>', LikeView,name='like_post'),
]
members/urls.py
from django.urls import path
from .views import UserRegisterView,UserEditView,UserLoginView,PasswordsChangeView,ShowProfilePageView,EditProfilePageView
from django.contrib.auth import views as auth_views
from . import views
urlpatterns = [
path('login/',UserLoginView, name='login'),
path('register/',UserRegisterView.as_view(), name='register'),
path('edit_profile/',UserEditView.as_view(), name='edit_profile'),
# path('password/',auth_views.PasswordChangeView.as_view(template_name='registration/change-password.html')),
path('password/',PasswordsChangeView.as_view(template_name='registration/change-password.html')),
path('password_success/',views.password_success,name='password_success'),
path('<int:pk>/profile/',ShowProfilePageView.as_view(),name='show_profile_page'),
path('<int:pk>/edit_profile_page/',EditProfilePageView.as_view(),name='edit_profile_page'),
]
settings.py
# Default primary key field type
# https://docs.djangoproject.com/en/3.2/ref/settings/#default-auto-field
DEFAULT_AUTO_FIELD = 'django.db.models.BigAutoField'
LOGIN_REDIRECT_URL='home'
LOGOUT_REDIRECT_URL='login'
What it's doing now is that it basically just redirects to the
path('', login_required(HomeView.as_view()),name='home') and it doesn't know where to get the login url from, so instead it should redirect to path('login/',UserLoginView, name='login') but I have no idea how to do it

Okay I found the answer, so I just add in login_url='([login url path])' so that it redirects to the login url that I wanted... now it should look like this:
path('',login_required(HomeView.as_view(),login_url='/members/login/'),name='home'),

Using the URLconf defined in products.urls, Django tried these URL patterns, in this order:

*from django.urls import path, include
from django.contrib import admin
from django.conf import settings
from django.conf.urls.static import static
urlpatterns = [
path('admin/', admin.site.urls),
path('products/',include('products.urls')),
path('accounts/',include('accounts.urls')),
]*
The above code of pyshop/url.py
I could not login to admin page.
from django.urls import path
from . import views
urlpatterns = [
path('', views.index),
path('products/new',views.new),
path('products/',views.products),
path('accounts/register',views.register),
]
The code is of product/urls.py.

Old question, but I run into the same problem, so just in case I can help someone else.
I would put in the pyshop/url.py, this code (see the admin path is the only one with an actual url):
from django.urls import path, include
from django.contrib import admin
from django.conf import settings
from django.conf.urls.static import static
urlpatterns = [
path('admin/', admin.site.urls),
path('',include('products.urls')),
path('',include('accounts.urls')),
]
Then in your product/urls.py (See all of the routes end with a slash):
from django.urls import path
from . import views
urlpatterns = [
path('', views.index, name="product"),
path('products/new/',views.new, name="new_product"),
path('products/',views.products, name="products"),
path('accounts/register/',views.register, name="register"),
]

django 2.2.5 import url from one app to another app

I would like to use the url from products app (products/urls.py) inside of search app url (search/urls.py) to search for items/products using search bar. I've attempted this example on django docs but it's importing a view to url in the same app, and I've also attempted this example but it looks to be a solution for an older version of django but i am using the latest version of django at time time 2.2.5.
The error message I am receiving in terminal is coming from search/urls.py:
path('', views.ProductListView.as_view(), name='list'),
AttributeError: module 'search.views' has no attribute
'ProductListView'
I understand search.views does not have an attribute "ProductListView" but, products.views does, which is why i'm trying to import products.views in search/urls.py.
products/urls.py
from django.urls import path, re_path
from .import views
app_name = "products"
urlpatterns = [
path('', views.ProductListView.as_view(), name='list'),
re_path(r'^products/(?P<slug>[\w-]+)/$', views.ProductDetailSlugView.as_view(), name='detail'),
]
search/urls.py
from django.urls import path
from .import views
from products.views import ProductListView
urlpatterns = [
path('', views.ProductListView.as_view(), name='list'),
]
ecommerce/urls.py (main app)
from django.conf import settings
from django.conf.urls.static import static
from django.contrib import admin
from django.urls import path, include, re_path
# from products.views import ProductDetailView
from .views import home, about, contact
urlpatterns = [
path('admin/', admin.site.urls),
path('', home, name='home'),
path('about/', about, name='about'),
path('contact/', contact, name='contact'),
path('account/', include('allauth.urls'), name='login'),
path('register/', include('allauth.urls'), name='register'),
path('products/', include('products.urls', namespace='products')),
path('search/', include('search.urls', namespace='search')),
# path('', include('products.urls'), name='products-featured'),
# path('', include('products.urls'), name='featured-details'),
# path('', include('products.urls'), name='featured-slug-details'),
] + static(settings.STATIC_URL, document_root=settings.STATIC_ROOT) + static(settings.MEDIA_URL, document_root=settings.MEDIA_ROOT)

You have:
from products.views import ProductListView
Therefore you should use ProductListView, not views.ProductListView
urlpatterns = [
path('', ProductListView.as_view(), name='list'),
...
]
Note you can remove the from .import views import unless you are using views somewhere else in search/urls.py
An alternative is to use import as, so that you can import multiple views.py from different apps in the same module:
from products import views as product_views
urlpatterns = [
path('', product_views.ProductListView.as_view(), name='list'),
]

How to modifying url.py to run in latest django version?

urls.py
from django.conf.urls import patterns, include, url from django.conf.urls.i18n import i18n_patterns from django.utils.translation import ugettext_lazy as _ from django.contrib import admin admin.autodiscover()
urlpatterns = patterns('',
url(r'^admin/', include(admin.site.urls)),
url(r'^i18n/', include('django.conf.urls.i18n')), )
urlpatterns += i18n_patterns('',
(_(r'^dual-lang/'), include('duallang.urls')),
(r'^', include('home.urls')), )
from django.conf import settings
if 'rosetta' in settings.INSTALLED_APPS:
urlpatterns += patterns('',
url(r'^rosetta/', include('rosetta.urls')),
)
Getting ImportError: cannot import name 'patterns' from
'django.conf.urls'

you must convert pattern to list, url can be conveted to path or re_path like this:
from django.urls import path, include
urlpatterns = [
path('admin/', include(admin.site.urls)),
path('i18n/', include('django.conf.urls.i18n'))
]
i18n_pattern can be used like this:
from django.conf.urls.i18n import i18n_patterns
from django.urls import path, include
urlpatterns += i18n_patterns(
path('', include('home.urls'))
)

How to change the default Django page to my own one?

I'm learning Django,and I followed a guidebook.I want to change the default Django page into my own one.It has took me 2 hours to solve this,and nothing worked.
The project is called learning_log,and the app is called learning_logs.
Here's what I'm trying to do:
1.add module learning_logs.urls:
"""
Definition of urls for learning_log.
"""
from datetime import datetime
from django.urls import path
from django.contrib import admin
from django.contrib.auth.views import LoginView, LogoutView
from app import forms, views
#added
from django.conf.urls import include, url
import learning_logs.views
from django.urls import path,re_path
app_name='learning_logs'
urlpatterns =[
path('', views.home, name='home'),
path('contact/', views.contact, name='contact'),
path('about/', views.about, name='about'),
path('login/',
LoginView.as_view
(
template_name='app/login.html',
authentication_form=forms.BootstrapAuthenticationForm,
extra_context=
{
'title': 'Log in',
'year' : datetime.now().year,
}
),
name='login'),
path('logout/', LogoutView.as_view(next_page='/'), name='logout'),
path('admin/', admin.site.urls),
#added
re_path(r'',include('learning_logs.urls',namespace='learning_logs'))
]
To include the module,I've tried:
url(r'',include('learning_logs.urls',namespacec='learning_logs'))
path('',include('learning_logs.urls',namespacec='learning_logs'))
path('',include('learning_logs.urls'))
path('',learning_logs.urls)
path('',learning_logs.views)
But none of them worked.
2.Create urls.py in learning_logs.here's the code:
"""define learning_logs's url mode"""
from django.urls import path,re_path
from . import views
from django.conf.urls import include,url
urlpatterns=[
#homepage：
re_path(r'(?P^$)',views.index,name='index')]
#path('',include('learning_logs.views.index'))]
As you can see,I also tried many times.
3.Write views.py in learning_logs
from django.shortcuts import render
# Create your views here.
def index(request):
"""homepage of learning logs"""
return render(request,'learning_logs/index.html')
4.write HTML document in learning_logs/templates/learning_logs/index.html
code skipped,for it never came up.
Here is my document tree:
tree
What I want is that show my own homepage at localhost:xxxxx/
How to solve this?

it's simple. just change this
urlpatterns =[
path('', views.home, name='home'),
to this
urlpatterns =[
path('', learning_logs.views.index, name='home'), #correct path to your index view

I don't understand what you're trying to do, why don't you just use:
urlpatterns =[
path('', views.home, name='home'),
And change the name so it directs you to that home page instead?
E.g.
urlpatterns =[
path('', views.learning_logs, name='learning logs'),
Also, since learning logs IS in the views file you shouldn't use
path('',learning_logs.views)
Does that make sense?