Add links to navigation based on INSTALLED_APPS - django

Question

I have a project that has a number of apps. These apps translate into modules that perform different functions for the end user.
Now each deployment of the project may have certain apps enabled or disabled.
What I'd like to achieve is a navigation list of links that only displays links for apps that are in INSTALLED_APPS.
For example deployment 1 has App1 and App2 listed in INSTALLED_APPS so would have a navigation something like this:
Link to App1's view
Link to App2's view
And deployment 2 has App2 and App3 installed so should show a navigation something like this:
Link to App2's view
Link to App3's view
Without having a navigation defined in a base template and editing it for each deployment, I can't see a way of doing this. Even by using { block.super }, this seems to not allow 2 apps to be installed as each child template would append to the parent.

Answer 1

This is strictly from the top of my head, so it may not be the "best" or most appropriate way.
First, if you're going to rely on INSTALLED_APPS you should actually implement something along the lines of:
MY_INSTALLED_APPS = (
'app1',
'app2',
)
INSTALLED_APPS = (
# other installed apps
) + MY_INSTALLED_APPS
Then, instead of having to figure out which of the items in INSTALLED_APPS are yours and which are third-party, you just use MY_INSTALLED_APPS instead for things like building your menus.
Second, in Django, apps aren't tied to views in any meaningful way. There's no concept of a "default" view, and you can't simply link to an app. However, you can some what achieve this idea through the use of namespaces and a convention for view naming, specifically, all your apps will at least have a urlpattern named "index", for example.
Then, in your urls.py you create urlpatterns like:
url(r'^app1/', include('app1.urls', namespace='app1', app_name='app1'),
url(r'^app2/', include('app2.urls', namespace='app2', app_name='app2'),
# etc
In each app, you create a urls.py that has at least one urlpattern:
url(r'^$', some_view, name='index'),
This means that the going to /app1/ in your browser would then load some_view and you can reference this view in your code with a name like: app1:index.
The tricky part is using this in your templates. Django 1.5 will add the ability to use context variables in the {% url %) tag for the view name. As of Django 1.3, you can use this behavior as well via {% load url from future %}. However, even that only gets you part-way to what you need.
In your template, you'll need to loop through the values of MY_INSTALLED_APPS and construct the links. The following should work in Django 1.3-1.4 via {% load url from future %} or Django 1.5:
{% for app in apps %}
{{ app }}
{% endfor %}
The other way to accomplish this is with a template filter, which is your only option in Django <1.3, and may still be preferable in later versions. Something like:
#register.filter
def default_url_for_app(app):
return reverse(app+':index')
And, in your template:
{% for app in apps %}
{{ app }}
{% endfor %}

Answer 2

You can simple import your settings.py file and iterate over the INSTALLED_APPS list.
from django.conf import settings
for installed_app in settings.INSTALLED_APPS:
But I'm also in doubt if this is a smart decision to implement your application as such.

Answer 3

What you really want here is some kind of app registration system. That is, for each app that's loaded (ie included in INSTALLED_APPS), register its homepage with the list of links.
Rather than doing all this automatically, I would follow the example of the admin application and get each app to call a register function on first load. You could do this in the models.py, for example, since you know that will be imported by Django on startup. The register function would take the app name and the url and build a dictionary which could then be passed to each template via a context processor.