In my template, I have the following:
<ul class="tabbed" id="network-tabs">
{% if user.is_authenticated %}
<li>My Account</li>
<li>Log Out</li>
{% else %}
<li>Log in</li>
<li>Register</li>
{% endif %}
</ul>
It seems to work fine, unless the page been created has a #login_required decorator, in which case the page works fine but the navigation appears as if the user is not logged in, even when they are.
You should check your view function to see where the user variable is coming from. Unless you're specifically passing user into the context from the view, that's your problem.
You do have access to request.user, though, and that will always return true in a template rendered from a view that has the #login_required decorator.
The reason I can tell you for certain that there's nothing wrong with the decorator, though, is that in the code for User and AnonymousUser (located in django.contrib.auth.models) the is_authenticated method strictly returns true for User and false for AnonymousUser. The decorator does not and cannot change that. And what that means is that your template isn't actually getting a User object where you're checking user.
To follow on from Gabriel's answer, is the user variable coming from the auth context processor? If it is, and you are using the render_to_response shortcut, you need to use a RequestContext instance.
from django.template import RequestContext
...
#login_required
def some_view(request):
# ...
return render_to_response('my_template.html',
my_data_dictionary,
context_instance=RequestContext(request))
Related
In a Laravel Blade template, I can do this to show code only if a user is logged in:
#auth
<div>Show code only if logged in.</div>
#endauth
Does Django have a template tag or something similar that does the equivalent? E.g.:
{% auth %}
<div>Show code only if logged in.</div>
{% endauth %}
In Django templates you should check for {% if user.is_authenticated %}. See this answer for more.
I'll answer my own question since I found more of what I was looking. In Django, they're called "decorators," e.g., #login_required.
from django.contrib.auth.decorators import login_required
#login_required
def my_view(request):
...
Source: https://docs.djangoproject.com/en/3.2/topics/auth/default/#the-login-required-decorator
Added: 11/02/2021.
Hi I'm writing a django project, and I write template code like this:
<ul id="nav">
<li>Home</li>
<li>Users</li>
{% if user %}
<li>Settings</li>
<li>Log Out</li>
{% else %}
<li>Log In</li>
<li>Sign Up</li>
{% endif %}
</ul>
Now in login view I write like this:
def login(request):
if user_logged_in(request):
return redirect('/')
if request.method == 'GET':
form = LogInForm()
return render_to_response(LOGIN_PATH, {'form':form}, context_instance=RequestContext(request))
But when I run the server, no user is logged in, and visit login page, it shows Settings and Log Out(there is a user object in context), but it shouldn't!
If I remove RequestContext, say return render_to_response(LOGIN_PATH, {'form':form}), it will be OK. And
return render_to_response(LOGIN_PATH, {'form':form, 'user':None}, context_instance=RequestContext(request))
is OK too. But I don't want to do it.
I know it's dirty design, well... I'm looking for suggestions and solutions. Many thanks~!
{% if user.is_authenticated %}
your tag just checks for a user object, not for an authenticated one.
check here for more informations on what you can do with an auth user :)
The default setting for TEMPLATE_CONTEXT_PROCESSORS includes "django.contrib.auth.context_processors.auth". This context processor adds a user to cotnext, which will be anonymous if the user is not provided in request so.
If you want to be able to know whether or not the user is authenticated in template the #Samuele Mattiuzzo answer is what you should use, but if you don't want, for any reason, to include the user in context, then you need to modify the default TEMPLATE_CONTEXT_PROCESSORS setting without the auth context processor.
For more information read the docs or the code.
My Django app currently has URLs which are protected by 'permission_required()' functions.
This function is called in three different ways.
As a decorator in views.py, with hardcoded parameters.
As a plain function, with autogenerated parameter, in custom Class Based Generic Views.
As a function invoking views in urls.py, with hardcoded parameters.
I'm now adding a menu system to the app, and I need to make menu entries reflect whether the user has permission to request the URL of each menu entry. (Either by greying-out or hiding said entries.)
Is there a way of query the permissions required to a URL without requesting the URL?
The only solution I've thought of so far is to replace the decorator with a parameterless 'menu_permssion_required()' decorator and hardcode all of the permissions into a Python structure. This seems like a step backwards, as my custom Class Based Generic Views already autogenerate their required permissions.
Any suggestions on how to make a menu system which reflects URL permissions for the current user?
Here is an example of how to solve your problem:
First, Create a decorator wrapper to use instead of permission_required:
from django.contrib.auth.decorators import login_required, permission_required, user_passes_test
from django.core.exceptions import PermissionDenied
from functools import wraps
from django.utils.decorators import available_attrs
def require_perms(*perms):
def decorator(view_func):
view_func.permissions = perms
#wraps(view_func, assigned=available_attrs(view_func))
def _wrapped_view(request, *args, **kwargs):
for perm in perms:
return view_func(request, *args, **kwargs)
raise PermissionDenied()
return _wrapped_view
return decorator
Then, use it to decorate your views:
#require_perms('my_perm',)
def home(request):
.....
Then, add a tag to use for your menu items:
from django.core.urlresolvers import resolve
def check_menu_permissions(menu_path, user):
view = resolve(menu_path)
if hasattr(view.func, "permissions"):
permissions = view.func.permissions
for perm in permissions:
if user.has_perm(perm):
return True # Yep, the user can access this url
else:
return False # Nope, the user cannot access this url
return True # or False - depending on what is the default behavior
And finally, in your templates, when building the menu tree:
<button href="{{ some_path }} {% if not check_menu_permissions some_path request.user %}disabled="disabled"{% endif %} />
N.B. I've not tested the last part with the tag, but I hope you got the idea. The magic thing here is to add the permissions to the view_func in the decorator, and then you can access this using resolve(path). I'm not sure how this will behave in terms of performance, but after all that's just an idea.
EDIT: Just fixed a bug in the example..
Is there a way of query the permissions required to a URL without requesting the URL?
User.has_perm() and User.has_module_perms()
Any suggestions on how to make a menu system which reflects URL permissions for the current user?
I really like this question, because it concerns anyone that makes a website with django, so I find it really relevant. I've been through that myself and even coded a menu "system" in my first django project back in 2008. But since then I tried Pinax, and one of the (so many) things I learnt from their example projects is that it is completely unnecessary bloat.
So, I have no suggestion which I would support on how to make a menu "system" which respects the request user permissions.
I do have a suggestion on how to make a simple menu which respects the request user permissions, so that might not be completely unrelated.
Just make your menu in plain HTML, it's not like it's going to change so often that it has to be generated. That will also keep your Python code simpler.
Add to settings.TEMPLATE_CONTEXT_PROCESSORS: 'django.core.context_processors.PermWrapper'
Use the {{ perms }} proxy to User.has_perms.
Example:
{% if perms.auth %}
<li class="divider"></li>
{% if perms.auth.change_user %}
<li>
{% trans 'Users' %}
</li>
{% endif %}
{% if perms.auth.change_group %}
<li>
{% trans 'User groups' %}
</li>
{% endif %}
{% endif %}
{# etc, etc #}
That's how I keep navigation simple, stupid, and out of the way. But also I always include an autocomplete not far from the menus to allows the user to navigate to any detail page easily. So, that's all I know about navigation in django projects, I'm eager to read other answers !
I had a similar issue, but it went a little deeper. Instead of just permissions, I also wanted other tests based on the lidded in user (ie, is_staff, or user.units.count() > 1). Duplicating these in the view and the template seems prone to errors.
You can introspect a view object, and see all of the decorators wrapping it, and work out if they are checks (in my case: the first argument I'd u or user). If they all pass, then allow rendering the link.
Get all decorators wrapping a function describes the technique in a little more detail. You can find the app that wraps this up into a handy replacement for {% url %} at Django-menus.
I am having problems with user authentication for my django site. I have a log-in screen that seems to work. When the user clicks log-in, I call the django.contrib.auth.login and it seems to work fine. However on subsequent pages have no knowledge that there is a user logged in. Example {% user.is_authenticated %} is false. There are also some menu functions that I want to be available for logged in users such as my-account and logout. Those functions are not available, except on the log-in page. Which is really strange.
This seems to be a user context problem. But I'm not sure how I am supposed to be passing a context around to ensure that my login is stable. Does anyone know at could be going on here? Any advice?
---------part of base.html------------
<!--- The following doesn't register even though I know I'm authenticated -->
{% if user.is_authenticated %}
<div id="menu">
<ul>
<li>My Customers</li>
<li>Customer Actions</li>
<li>My Account</li>
</ul>
</div>
{% endif %}
---------my views.py -----------------
# Should I be doing something to pass the user context here
def customer_list(request):
customer_list = Customer.objects.all().order_by('lastName')[:5]
c = Context({
'customer_list': customer_list,
})
t = loader.get_template(template)
return HttpResponse(t.render(cxt))
If you're using Django 1.3, you can use the render() shortcut, which automatically includes RequestContext for you.
from django.shortcuts import render
def customer_list(request):
customer_list = Customer.objects.all().order_by('lastName')[:5]
return render(request, "path_to/template.html",
{'customer_list': customer_list,})
In this case, you could go one step further, and use the generic ListView:
from django.views.generic import ListView
class CustomerList(Listview):
template_name = 'path_to/template.html'
queryset = Customer.objects.all().order_by('lastName')[:5]
Use a RequestContext.
As Daniel Suggested, use the RequestContext... or better, just use the render_to_response shortcut:
from django.template import RequestContext
from django.shortcuts import render_to_response
def customer_list(request):
customer_list = Customer.objects.all().order_by('lastName')[:5]
return render_to_response(
"path_to/template.html",
{'customer_list':customer_list,},
context_instance=RequestContext(request))
I have project in Django 1.3. In order to show username in all pages I use such tags in base.html
{% if user.is_authenticated %}
<p>Welcome, {{ user.username }}.
loggout</p>
{% else %}
loggin
{% endif %}
But if I dont return context_instance=RequestContext(request) from view value of user in template is empty. The 'django.contrib.auth.context_processors.auth' is included to TEMPLATE_CONTEXT_PROCESSORS.
Is it possible automaticaly include user to all templates?
since django 1.3. use shortcuts.render function and dont warry about requestcontext including to your views
You've given the answer yourself. As long as you use a RequestContext, it will be included in all templates.
If you really find that too much work, you could use the (new in 1.3) TemplateResponse class.
Or simply create a context processor. See
http://docs.djangoproject.com/en/dev/ref/templates/api/#writing-your-own-context-processors
Put this in context_processor.py
def root_categories(request):
return {
'user': request.user,
}
in settings.py add the context processor.
now in your template try: {{ user }}