User permissions for Django module

Question

I'm having a small issue with my permissions in my Django template.

I'm trying to, based on permissions, show an icon in the menu bar for my project. I want to have it so that if the user has the permissions to add a new follow-up to the project, they can see the icon, if they don't have that permission, then do not display the link.

My permission syntax is follow.add_followup, which I got from printing user.get_all_permissions().

I have tried this code in my template:

...
{% if user.has_perm('followup.add_followup') %}
<li><a href="{% url followup-new p.id %}">Log</a></li>
{% endif %}
...

But when I display the template, I am presented with this error:

TemplateSyntaxError at /project/232/view/

Could not parse the remainder: '(followup.add_followup)' from 'user.has_perm(followup.add_followup)'

Any thoughts? This has been giving me a headache! :)

score 11 · Answer 1 · edited Feb 06 '11 at 07:30

Since you are using the Django permission system, it's better you use the followingg template syntax...

{%if perms.followup.add_followup%}your URL here{%endif%}

EDIT: Django automatically creates 3 permissions for each model, 'add', 'change' and 'delete'. If there exists no model for adding a link, then you must add the permission from related model, in the model class Meta... Likewise:

somemodels.py

class SomeModel(Model):
    ...
    class Meta:
    permissions = (('add_followup','Can see add urls'),(...))

In the Django auth user admin page, you can see your permission. In the template layer, permission is presented with the basic Django style,

<app_label>.<codename>

which, in this case, will be like:

{%if perms.somemodels.add_followup%}your URL here{%endif%}

If there is no model, related to the job you wish to do, the add the permission to a model...

In your template, you can write

{{perms.somemodels}}

to seal available permissions to that user, where somemodel is the name of the applicaton that you add your permission to one of its models.

Liberathor · Answer 2 · 2012-06-08T16:36:31.773

2

This is my very simple solution, in your template add this:

for example:

.......

{% if 'user.can_drink' in user.get_all_permissions %}
   {{ user }} can drink.
   .......
{% else %}
   {{ user }} can´t drink.
    ........
{% endif %}

.......

edited Jun 08 '12 at 16:36

answered Jun 08 '12 at 16:26

Liberathor

29
6

score 2 · Answer 3 · answered Feb 28 '12 at 21:38

2

Django documentation detailing answer #2: https://docs.djangoproject.com/en/dev/topics/auth/#id9

The currently logged-in user's permissions are stored in the template variable {{ perms }}. This is an instance of django.contrib.auth.context_processors.PermWrapper, which is a template-friendly proxy of permissions.

answered Feb 28 '12 at 21:38

Kevin Audleman

334
2
4

Thank you... this is by far the most useful answer! – Greg Feb 21 '13 at 02:25
Updated link: https://docs.djangoproject.com/en/dev/topics/auth/default/#permissions - works for Django 1.5 to 1.7 (which is latest at time of writing). – Rob Grant Aug 21 '14 at 08:13

score 1 · Accepted Answer · edited May 23 '17 at 12:33

1

I have tried this code in my template:

This kind of complex decision-making goes in the view functions.

Or it goes into the context which is then presented to the template.

https://stackoverflow.com/search?q=%5Bdjango%5D+context

When to use context processor

Do this in your view

def my_view( request ):
    followup= user.has_perm('followup.add_followup')
    # etc.
    return render_to_response( template, {'followup':followup,... )

Then your template is simply

{% if followup %}
<li><a href="{% url followup-new p.id %}">Log</a></li>
{% endif %}

edited May 23 '17 at 12:33

Community

1
1

answered Feb 03 '11 at 16:29

S.Lott

384,516
81
508
779

The reason I was trying it in the template, was because I didn't want to show icons that the logged in user, couldn't use. – TheLifeOfSteve Feb 03 '11 at 16:31
1

You can create a boolean value in the view function and use a simple {%if%} in the template to display or not display. – S.Lott Feb 03 '11 at 16:34
Not really true or helpful, see answers below. – Greg Feb 21 '13 at 02:26

User permissions for Django module

4 Answers4