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django login_required decorator class view

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Ashvin answered on February 17, 2022 Popularity 5/10 Helpfulness 7/10

answer django login_required decorator class view

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django login_required decorator class view

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#urls.py
from django.contrib.auth.decorators import login_required
from django.urls import path

path('/path',login_required(views.PostListView.as_view(),
                            login_url='login-page'),name='homepage')

Popularity 5/10 Helpfulness 7/10 Language python

Source: Grepper

Tags: class decorator python view

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Contributed on Dec 12 2023

Ashvin

0 Answers Avg Quality 2/10

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Django login_required decorator

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from django.contrib.auth.decorators import login_required

@login_required
def my_view(request):
    ...
    


# login_required() does the following:

# If the user isn’t logged in, redirect to settings.LOGIN_URL,
# passing the current absolute path in the query string.
# Example: /accounts/login/?next=/polls/3/. If the user is logged in,
# execute the view normally. The view code is free to assume the user is logged in.
# By default, the path that the user should be redirected to upon successful authentication is stored in a query string parameter called "next".
# If you would prefer to use a different name for this parameter, 
# login_required() takes an optional redirect_field_name parameter:




from django.contrib.auth.decorators import login_required

@login_required(redirect_field_name='my_redirect_field')
def my_view(request):
    ...




# Note that if you provide a value to redirect_field_name,
# you will most likely need to customize your login template as well,
# since the template context variable which stores the redirect path will use the value of redirect_field_name as its key rather than "next" (the default).




login_required() also takes an optional login_url parameter. Example:

from django.contrib.auth.decorators import login_required

@login_required(login_url='/accounts/login/')
def my_view(request):
    ...




# Note that if you don’t specify the login_url parameter,
# you’ll need to ensure that the settings.LOGIN_URL and your login view are properly associated.
# For example, using the defaults,
# add the following lines to your URLconf:




from django.contrib.auth import views as auth_views

path('accounts/login/', auth_views.LoginView.as_view()),




# The settings.LOGIN_URL also accepts view function names and named URL patterns.
# This allows you to freely remap your login view within your URLconf without having to update the setting.

Popularity 9/10 Helpfulness 5/10 Language python

Source: docs.djangoproject.com

Tags: decorator deco

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Contributed on Jun 19 2022

OHIOLee

0 Answers Avg Quality 2/10

login_required on class django

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You can add the decorator in the urls.py

from django.contrib.auth.decorators import login_required
 url(r'^workers/$', login_required(views.RootWorkerView.as_view()))

Popularity 9/10 Helpfulness 5/10 Language python

Source: stackoverflow.com

Tags: class clas

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Contributed on Feb 17 2022

BlueMoon

0 Answers Avg Quality 2/10

django login required class based views

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class MyView(...):
    # this is a generic view
    ...

my_view = login_required(MyView.as_view())

#and using my_view in your URLconf?
#Like so Myview = login_required(MyView.as_view(), login_url='login_url')

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Source: code.djangoproject.com

Tags: class clas

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Contributed on Apr 17 2022

Miguel Mehgoss

0 Answers Avg Quality 2/10

login_required on class django

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now you can use Django builtin LoginRequiredMixin

from django.contrib.auth.mixins import LoginRequiredMixin

class MyView(LoginRequiredMixin, View):
    login_url = '/login/'
    redirect_field_name = 'redirect_to'

Popularity 9/10 Helpfulness 1/10 Language python

Source: stackoverflow.com

Tags: class clas

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Contributed on Feb 17 2022

BlueMoon

0 Answers Avg Quality 2/10

django login_required decorator class view

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