I have a function to check if the user is a staff:
class VerificateUser():
def allowed_user(request):
if request.user.is_staff:
return True
In my template, I’m going to show this section only for staff users, but this is not working.
{% url if not allowed_user %}
<h1>
Welcome to the show
</h1>
If do something like it, works:
```html
{% if not request.user.is_staff %}
<h1>
Welcome to the show
</h1>
But I need to use a view function to clean my code because I’m probably going to add more conditionals.
CodePudding user response:
Since you are using a class-based view
then I would suggest updating the context
dictionary which you could use in the html
template to determine whether or not the user
is allowed or not. For example., Within the views.py
.
class VerificateUser():
# Update the context
def get_context_data(self, **kwargs):
context = super().get_context_data(**kwargs)
# Call the allowed_user() and return whatever value, passing that value it to the context variable
context['allowed_user'] = self.allowed_user()
return context
# Checking if the user is allowed here
def allowed_user():
if self.request.user.is_staff:
return True
return False
Now within the html
file, you can reference that allowed_user
from the context variable.
{% if allowed_user %}
<h1>Hi, you are allowed to see here...</h1>
{% endif %}
That should do the trick.
CodePudding user response:
You can do this sort of thing in many ways, but simply you can do by the following way-
{% if request.user.is_authenticated %}
<p>Welcome,{{request.user.first_name}} </p>
{% endif %}
request
object is by default available in all of your Django templates context.