How to check if a user is logged in (how to properly use user.is_authenticated)?

PythonDjangoAuthentication

Python Problem Overview


I am looking over this website but just can't seem to figure out how to do this as it's not working. I need to check if the current site user is logged in (authenticated), and am trying:

request.user.is_authenticated

despite being sure that the user is logged in, it returns just:

>

I'm able to do other requests (from the first section in the url above), such as:

request.user.is_active

which returns a successful response.

Python Solutions


Solution 1 - Python

Update for Django 1.10+:

is_authenticated is now an attribute in Django 1.10.

The method was removed in Django 2.0.

For Django 1.9 and older:

is_authenticated is a function. You should call it like

if request.user.is_authenticated():
    # do something if the user is authenticated

As Peter Rowell pointed out, what may be tripping you up is that in the default Django template language, you don't tack on parenthesis to call functions. So you may have seen something like this in template code:

{% if user.is_authenticated %}

However, in Python code, it is indeed a method in the User class.

Solution 2 - Python

Django 1.10+

Use an attribute, not a method:

if request.user.is_authenticated: # <-  no parentheses any more!
    # do something if the user is authenticated

The use of the method of the same name is deprecated in Django 2.0, and is no longer mentioned in the Django documentation.


Note that for Django 1.10 and 1.11, the value of the property is a CallableBool and not a boolean, which can cause some strange bugs. For example, I had a view that returned JSON

return HttpResponse(json.dumps({
    "is_authenticated": request.user.is_authenticated()
}), content_type='application/json') 

that after updated to the property request.user.is_authenticated was throwing the exception TypeError: Object of type 'CallableBool' is not JSON serializable. The solution was to use JsonResponse, which could handle the CallableBool object properly when serializing:

return JsonResponse({
    "is_authenticated": request.user.is_authenticated
})

Solution 3 - Python

Following block should work:

    {% if user.is_authenticated %}
        <p>Welcome {{ user.username }} !!!</p>       
    {% endif %}

Solution 4 - Python

In your view:

{% if user.is_authenticated %}
<p>{{ user }}</p>
{% endif %}

In you controller functions add decorator:

from django.contrib.auth.decorators import login_required
@login_required
def privateFunction(request):

Solution 5 - Python

If you want to check for authenticated users in your template then:

{% if user.is_authenticated %}
    <p>Authenticated user</p>
{% else %}
    <!-- Do something which you want to do with unauthenticated user -->
{% endif %}

Solution 6 - Python

to check if user is logged-in (authenticated user) in views.py file, use "is_authenticated" method, as the following example:

def login(request):
    if request.user.is_authenticated:
        print('yes the user is logged-in')
    else:
        print('no the user is not logged-in')

to check if user is logged-in (authenticated user) in your html templates file you can use it also as the following example :

 {% if user.is_authenticated %}
    Welcome,{{request.user.first_name}}           

 {% endif %}

this is just example , and change it based on your requirements.

i hope this helpful for you .

Solution 7 - Python

For Django 2.0+ versions use:

    if request.auth:
       # Only for authenticated users.

For more info visit https://www.django-rest-framework.org/api-guide/requests/#auth

request.user.is_authenticated() has been removed in Django 2.0+ versions.

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionRickView Question on Stackoverflow
Solution 1 - PythonBrian NealView Answer on Stackoverflow
Solution 2 - PythonMark ChackerianView Answer on Stackoverflow
Solution 3 - PythonSopanView Answer on Stackoverflow
Solution 4 - PythonCubiczxView Answer on Stackoverflow
Solution 5 - PythonSuyash KumarView Answer on Stackoverflow
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Solution 7 - PythonJatin GoyalView Answer on Stackoverflow