Check permission inside a template in Django - django

Can I use the Auth application's permission checking inside a template in Django? (I want to display a simple form at the end of the template for privileged users)
And more importantly, should I do it at all or is this no the "Django way"?

If you are looking to check for permissions in templates, the following code would suffice:
{% if perms.app_label.can_do_something %}
<form here>
{% endif %}
Where model refers to the model that the user need permissions to see the form for.
Refer to https://docs.djangoproject.com/en/stable/topics/auth/default/#permissions for more examples.
The currently logged-in user's permissions are stored in the template variable {{ perms }}
(This requires the following context processor to be enabled: django.contrib.auth.context_processors.auth)

Tested on Django 2.0 +
If you want to see all the permissions the logged in user has, on your template (.html),
print :
{{ perms.app_name }}
Or
{{ perms }}
In order to check if user has permission , use:
{% if perms.app_name.change_model_name_lower_cased %}
E.g :
{% if perms.Utilization.change_invoice %}
Here: Utilization is my App name. Invoice is a model name.
Note that in general, there will be 4 kinds of permissions:
change [E.g Utilization.change_projectemail]
view [E.g Utilization.view_invoice]
delete [E.g Utilization.delete_invoicetype]
add [E.g Utilization.add_invoicetype]
Also , if you want to see all permissions a user has due to the groups he belongs to, launch Django shell...
user = User.objects.get(username='somename')
user.get_group_permissions()
Here, all permissions listed, are due to the groups he belongs to.

One more Unique way to do this is:
{% if 'app_label.permission' in perms %}
<form here>
{% endif %}
Example:
{% if 'auth.view_group' in perms %}
<p> Hello World! </p>
{% endif %}
This comes handy when you want to use your default/custom authentication permissions whether you've created an app for your model or not because this method don't need an app name. It just need the permission name from your permissions table.
You can put multiple checks also using and/or commands:
{% if 'auth.view_group' in perms and 'auth.add_group' in perms %}
<form here>
{% endif %}

If you need more granularity in checking perms (on a particular object for example), check out this extension: http://django-authority.readthedocs.org/en/latest/check_templates/

And for those using Jinja templates and struggling with this question, like I did for a few hours/days...
Use an extended Jinja Environment, and add the request object to it:
# in settings.py
TEMPLATES = [
{
"BACKEND": "django.template.backends.jinja2.Jinja2",
'DIRS': ['jinja2'],
"APP_DIRS": True,
"OPTIONS": {
'environment': 'main.jinjaconfig.env.environment',
}
},
# Other template backends...
]
# main/jinjaconfig/env.py
from django.template.context_processors import request
from jinja2 import Environment
# ...
def environment(**options):
env = Environment(**options)
# Update globals with the functions and objects you need, here 'request'
env.globals.update({
'request': request,
# Other globals like 'static', 'url', ...
})
return env
Then you can access the request object in Jinja templates, and also request.user and request.user.has_perm() and all user related functions:
{% if request.user.has_perm('app_label.can_do_something') %}
{# Stuff .. #}
{% endif %}

Related

Django - add link with custom admin page href

In my Django project, I have created a custom admin page for an app via the get_urls() method. I'd like to add a link to the app's main model index view that will take users to this custom page - however, I'm having some trouble creating this link element correctly and I don't seem to be able to piece together the right way to do it - I'm just left with a Reverse for 'export' not found. 'export' is not a valid view function or pattern name. error.
I've set up the admin for the app like so:
# my_project/observations/admin.py
from django.template.response import TemplateResponse
from django.urls import path
class ObservationAdmin(SimpleHistoryAdmin, SoftDeletionModelAdmin):
change_list_template = 'export_link.html'
def get_urls(self):
urls = super().get_urls()
custom_urls = [
path('export/', self.admin_site.admin_view(self.export_view), name='export')
]
return custom_urls + urls
def export_view(self, request):
context = dict(
self.admin_site.each_context(request),
)
return TemplateResponse(request, 'export.html', context)
and the two templates that are referenced:
# my_project/observations/templates/export.html
{% extends "admin/base_site.html" %}
{% block content %}
<div>
Some custom content
</div>
{% endblock %}
# my_project/observations/templates/export_link.html
{% extends 'admin/change_list.html' %}
{% block object-tools-items %}
<li>
Export
</li>
{{ block.super }}
{% endblock %}
Navigating directly to http://localhost:8000/admin/observations/observation/export/ works perfectly, I see the custom content page exactly as I want it... so the issue I'm striking is with the link template - I get the Reverse... error when I navigate to the model index page.
Perhaps the argument I'm passing to url is incorrect, or I need to register that URL elsewhere - but I don't quite know. The other examples of link elements like this that I've been able to find don't reference URLs created via the admin class' get_urls() method - so any guidance on this would be greatly appreciated.
Thanks very much, let me know if there's any other info that I can provide to help sort this out.
I think the problems is in missing namespace in your export_link.html template. Instead of:
Export
try:
Export

How get current user in a template tag?

How can i get the current user in a django template tags? (request object is not accessible)
Or how can i access to request object?
If you want to access the current user in a template tag, you must pass it as a parameter in the templates, like so:
{% my_template_tag user %}
Then make sure your template tag accepts this extra parameter. Check out the documentation on this topic. You should also check out simple tags.
The user is always attached to the request, in your templates you can do the following:
{% if user.is_authenticated %}
{% endif %}
You don't have to specify "request" to access its content
UPDATE:
Be aware: is_authenticated() always return True for logged user (User objects), but returns False for AnonymousUser (guest users). Read here: https://docs.djangoproject.com/en/1.7/ref/contrib/auth/
This question was already answered here:
{% if user.is_authenticated %}
<p> Welcome '{{ user.username }}'</p>
{% else %}
Login
{% endif %}
and make sure you have the request template context processor installed in your settings.py:
TEMPLATE_CONTEXT_PROCESSORS = (
...
'django.core.context_processors.request',
...
)
Note:
Use request.user.get_username() in views & user.get_username in
templates. Preferred over referring username attribute directly.
Source
This template context variable is available if a RequestContext is used.
django.contrib.auth.context_processors.auth is enabled by default & contains the variable user
You do NOT need to enable django.core.context_processors.request template context processor.
Source : https://docs.djangoproject.com/en/dev/topics/auth/default/#authentication-data-in-templates
Suppose you have a profile page of every registered user, and you only want to show the edit link to the owner of the profile page (i.e., if the current user is accessing his/her profile page, the user can see the edit button, but the user can't see the edit button on other user's profile page.
In your html file:
<h2>Profile of {{ object.username }}</h2>
{% if object.username == user.username %}
Edit
{% endif %}
Then your urls.py file should contain:
from django.urls import path
from .views import ProfileUpdateView
urlpatterns = [
...
path('<int:pk>/profile/update', ProfileUpdateView.as_view(), name = 'profile_update'),
...
]
considering you have appropriate ProfileUpdateView and appropriate model

How to reset user password from the admin interface

In my website, I want to let the admins reset the password of any user.
With reset I mean exactly what the password_reset view does (under contrib.auth): Send a confirmation link to that user email.
How would be the best way of doing that? Is there an already app/snippet that does that?
Edit:
Let's suppose user john is an admin. What I want is to let john reset any user's password through the admin interface. For example, to reset max password, he will just go to the max user, and click on any link to reset his password.
What I finally did was to add a custom ModelAdmin:
from django.contrib.auth.forms import PasswordResetForm
from django.contrib.auth.admin import UserAdmin
class CustomUserAdmin(UserAdmin):
...
def reset_password(self, request, user_id):
if not self.has_change_permission(request):
raise PermissionDenied
user = get_object_or_404(self.model, pk=user_id)
form = PasswordResetForm(data={'email': user.email})
form.is_valid()
form.save(email_template_name='my_template.html')
return HttpResponseRedirect('..')
def get_urls(self):
urls = super(UserAdmin, self).get_urls()
my_urls = patterns('',
(r'^(\d+)/reset-password/$',
self.admin_site.admin_view(self.reset_password)
),
)
return my_urls + urls
and I also had to override the change_form.html template, like this:
{% extends "admin/change_form.html" %}
{% load i18n %}
{% block object-tools %}
{% if change %}{% if not is_popup %}
<ul class="object-tools">
{# You can also give a name to that pattern and refer to it below using 'url' #}
<li>Reset password</li>
<li>{% trans "History" %}</li>
{% if has_absolute_url %}
<li><a href="../../../r/{{ content_type_id }}/{{ object_id }}/" class="viewsitelink">
{% trans "View on site" %}</a>
</li>
{% endif%}
</ul>
{% endif %}{% endif %}
{% endblock %}
The result looks like this:
If you want a more detailed explanation, I blogged about it.
The passreset app just exposes the django views via urls.py, and adjusts the login template to show a "Forgot my password" link.
The built-in django password reset views and templates are meant for self-reset. I guess the reset form could be prepopulated with a different user's email address (in the query string) but you'd still need to make adjustments such as changing the email template - "You're receiving this e-mail because you requested a password reset for your user account" is probably not what you want:
https://code.djangoproject.com/browser/django/trunk/django/contrib/admin/templates/registration/password_reset_email.html
Therefore you should expose the views at different URLs if you want to include self-reset as well.
Hook the django views into urls.py like so:
urlpatterns += patterns('django.contrib.auth.views',
url(r'^accounts/password/reset/$',
'password_reset',
name='password-reset'),
url(r'^accounts/password/reset/done/$',
'password_reset_done',
name='password-reset-done'),
url(r'^accounts/password/reset/confirm/(?P<uidb36>[-\w]+)/(?P<token>[-\w]+)/$',
'password_reset_confirm',
name='password-reset-confirm'),
url(r'^accounts/password/reset/complete/$',
'views.password_reset_complete',
name='password-reset-complete')
)
and where you want to make adjustments, pass in e.g. your own email template:
url(r'^/accounts/password/reset/$',
'password_reset',
{'email_template_name': 'my_templates/password_reset_email.html'}
name='password-reset'),
The "password_reset" view has more parameters you can tweak:
https://docs.djangoproject.com/en/dev/topics/auth/#module-django.contrib.auth.views
("post_reset_redirect" comes to mind as another one for your purposes)
To show a corresponding link you'd either change the User admin (careful, already registered - unregister then register your own, subclassed plus additional link field) or the change_form template itself.
I'm unaware of an app that provides this out-of-the-box, so I upvoted the question :-).
Yep, there is an app for that. Check here:
https://github.com/bendavis78/django-passreset

if..else custom template tag

I'm implementing a custom permissions application in my Django project, and I'm lost as to how to implement a custom template tag that checks a logged in user's permissions for a specific object instance and shows a piece of HTML based on the outcome of the check.
What I have now is (pseudocode):
{% check_permission request.user "can_edit" on article %}
<form>...</form>
{% endcheck %}
('check_permission' is my custom template tag).
The templatetag takes in the user, the permission and the object instance and returns the enclosed HTML (the form). This currently works fine.
What I would like to do however, is something like:
{% if check_permission request.user "can_edit" on article %}
<form>...</form>
{% else %}
{{ article }}
{% endif %}
I've read about the assignment tag, but my fear is that I would pollute the context variable space with this (meaning I might overwrite previous permission context variables). In other words, as the context variables are being defined on different levels (the view, middleware in my case, and now this assignment template tag), I'm worried about maintainability.
You can use template filters inside if statements. So you could rewrite your tag as a filter:
{% if request.user|check_can_edit:article %}
Note that it's tricky to pass multiple arguments of different types to a filter, so you'll probably want to use one filter per permission, above I've used check_can_edit.
You can definitely do that if you're willing to write some more lines of python code to improve your template readability! :)
You need to parse the tag content yourself, even the parameters it takes and then resolve them, if you want to use variables on them.
The tag implemented below can be used like this:
{% load mytag %}
{% mytag True %}Hi{% else %}Hey{% endmytag %} Bro
Or with a variable:
{% mytag myobject.myflag %}Hi{% else %}Hey{% endmytag %} Bro
So, here's the way I did it:
from django.template import Library, Node, TemplateSyntaxError
register = Library()
#register.tag
def mytag(parser, token):
# Separating the tag name from the "test" parameter.
try:
tag, test = token.contents.split()
except (ValueError, TypeError):
raise TemplateSyntaxError(
"'%s' tag takes two parameters" % tag)
default_states = ['mytag', 'else']
end_tag = 'endmytag'
# Place to store the states and their values
states = {}
# Let's iterate over our context and find our tokens
while token.contents != end_tag:
current = token.contents
states[current.split()[0]] = parser.parse(default_states + [end_tag])
token = parser.next_token()
test_var = parser.compile_filter(test)
return MyNode(states, test_var)
class MyNode(Node):
def __init__(self, states, test_var):
self.states = states
self.test_var = test_var
def render(self, context):
# Resolving variables passed by the user
test_var = self.test_name.resolve(context, True)
# Rendering the right state. You can add a function call, use a
# library or whatever here to decide if the value is true or false.
is_true = bool(test_var)
return self.states[is_true and 'myvar' or 'else'].render(context)
And that's it. HTH.
In Django 2 the assignment tag was replaced by simple_tag() but you could store the custom tag result as a template variable:
# I'm assuming that check_permission receives user and article,
# checks if the user can edit the article and return True or False
{% check_permission user article as permission_cleared %}
{% if permission_cleared %}
<form>...</form>
{% else %}
{{ article }}
{% endif %}
Check the current doc about custom template tags: https://docs.djangoproject.com/en/2.1/howto/custom-template-tags/#simple-tags
inside my_tags.py
from django import template
register = template.Library()
#register.simple_tag(takes_context=True)
def make_my_variable_true(context):
context['my_variable'] = True
return '' # without this you'll get a "None" in your html
inside my_template.html
{% load my_tags %}
{% make_my_variable_true %}
{% if my_variable %}foo{% endif %}
In this case best solution is to use custom filter. If you don't want write long code for custom tag. Also if you don't want to copy/paste others code.
Here is an example
Inside templatetag
register = template.Library()
def exam_available(user, skill):
skill = get_object_or_404(Skill, id=skill)
return skill.exam_available(user)
register.filter('exam_available', exam_available)
Inside template
{{ request.user|exam:skill.id }}
or
{% if request.user|exam:skill.id %}
Since one of the main common of it is to use request.user or any specific object(id) inside model's custom method, so filtering that individual object or user is the easiest way to make it done. :)

Get user group in a template

I want to display a menu that changes according to the user group of the currently logged in user, with this logic being inside of my view, and then set a variable to check in the template to determine which menu items to show....I'd asked this question before, but my logic was being done in the template. So now I want it in my view...The menu looks as below
<ul class="sidemenu">
<li>General List </li>
<li>Sales List </li>
<li>Add a New Record </li>
<li>Edit Existing Record </li>
<li>Filter Records </li>
<li>Logout </li>
</ul>
Suppossing the user is management, they'll see everything...But assuming the user is in the group sales, they'll only see the first two and the last two items...and so on. I also want a dynamic redirect after login based on the user's group. Any ideas?
The standard Django way of checking permissions is by the individual permission flags rather than testing for the group name.
If you must check group names, being aware that Users to Groups is a many-to-many relationship, you can get the first group in the list of groups in your template with something like this:
{{ user.groups.all.0 }}
or using it like this in a conditional (untested but should work):
{% ifequal user.groups.all.0 'Sales' %}
...
{% endif %}
If you go with the preferred permission model you would do something like the following.
...
{% if perms.vehicle.can_add_vehicle %}
<li>Add a New Record </li>
{% endif %}
{% if perms.vehicle.can_change_vehicle %}
<li>Edit Existing Record </li>
{% endif %}
...
These are the permissions automatically created for you by syncdb assuming your app is called vehicle and the model is called Vehicle.
If the user is a superuser they automatically have all permissions.
If the user is in a Sales group they won't have those vehicle permissions (unless you've added those to the group of course).
If the user is in a Management group they can have those permissions, but you need to add them to the group in the Django admin site.
For your other question, redirect on login based on user group: Users to Groups is a many-to-many relationship so it's not really a good idea to use it like a one-to-many.
user.groups.all.0.name == "groupname"
Create a user_tags.py in your app/templatetags follow above:
# -*- coding:utf-8 -*-
from __future__ import unicode_literals
# Stdlib imports
# Core Django imports
from django import template
# Third-party app imports
# Realative imports of the 'app-name' package
register = template.Library()
#register.filter('has_group')
def has_group(user, group_name):
"""
Verifica se este usuário pertence a um grupo
"""
groups = user.groups.all().values_list('name', flat=True)
return True if group_name in groups else False
And finally in template use it:
{% if request.user|has_group:"Administradores"%}
<div> Admins can see everything </div>
{% endif %}
If you are working with Custom User Model (best practice with Django), you can create a method:
CustomUser(AbstractUser):
# Your user stuff
def is_manager(self):
return self.groups.filter(name='Management').exists()
Then inside your template you just call it this way:
{% if user.is_manager %}
{# Do your thing #}
{% endif %}
That method will be also useful validating permission in other parts of your code (views, etc.)