I have a model named Lal in Django. I have successfully registered it in my admin.py file.
Now , what I want is if a particular url is hit then, I should be able to directly redirect using reverse_lazy to the page displaying the model contents that I had registered in admin.py file.
The url that gets generated when I access the model displaying template page using my admin-login directly is:
http://127.0.0.1:8000/admin/test1_app/lal
With the help of this line of code,
url(r'yahoo/$', RedirectView.as_view(url = reverse_lazy('admin:app_list',kwargs={'app_label': 'test1_app'})), name="yahoo")
I am able to successfully generate the following url:
http://127.0.0.1:8000/admin/test1_app/
What should I add more to generate the model's url i.e. this one:
http://127.0.0.1:8000/admin/test1_app/lal
The url name for a model's changelist is admin:{{ app_label }}_{{ model_name }}_changelist. So for your app/model you want:
reverse_lazy('admin:test1_app_lal_changelist')
See the docs on reversing admin urls for more info.
Related
I have a django project in which the the form records the name and the email addresses of the users and I was able to put that data using forms.py and models.py.
What I want to do nextis to create an action through which I can download that in csv file.
My admin page looks like this now and I want to add action right above.
in order to add an action to a model in the admin page you have to create a new class like this and register it with your model:
admin.py
from youSite.views import downloadCSV
from yourSite.models import Info
class infoAdmin(admin.ModelAdmin):
actions =[downloadCSV]
admin.site.register(infoObject, infoAdmin)
You have to create the function in your views and import it into the admin page. It create a new action in that model.
Hope it helps
I need to get from which admin page(app) current request is from in django admin.
Currently request.resolver_match.app_name only returns "admin" which is not what I want.
I have noticed that my app name is in 'view_name' and 'url_name' but is it reliable to parse these variables to access current app name?
Django 1.11 LTS
EDIT: For example, when a user enters admin page for my course app with the above method I still only get 'admin' in my request which should be 'course' not 'admin'. My ultimate goal is to hide some of my app model fields in admin page based on user group.
Thanks
From a ModelAdmin you have access to the model via self.model. In a ModelAdmin method you can thus get the app name using self.model._meta.app_label.
I you need to access it from the template rather than the ModelAdmin, self.model._meta is passed to the context as opts. You can thus access it via {% if opts.app_label == "some_app" %}.
you can simply do this by tracking requested.user i.e who is logged in currently.
something like this
if(requested.user=='admin'):
(show all fields)
else:
(show mentioned fields)
here you are restricting access of your's models fields based on currently logged in user
so I have set up Django allauth on my Django project and connected to Instagram,
when doing so I have now on my admin site Social accounts category with my account registers, all good so far
on the lower page, I can see a field called extra data,
how can I put it inside the normal Users database so I can use it to take how many followers I got out of the extra data?
can I request the followers with the Token i have maybe?
You can simply access the SocialAccount model like any other django model:
from allauth.socialaccount.models import SocialAccount
def instagram(request):
data = SocialAccount.objects.get(user=request.user).extra_data
follows = data.get('counts')
return render(request, 'Path.to.html', {"follows": follows})
After I login, I need to redirect to another page while adding URL parameters to the URL of the next page. I get the value of these parameters after the user is authenticated because they need to be accessed from the user database table. I heard about using the next parameter but I don't know how I would use it since I need to access the database table and I can't do that from urls.py. This is my url.py line for login right now:
url(r'^$',auth_views.login, name='login',kwargs={
'authentication_form':loginPlaceHolderForm,
}),
I'm not really sure what other info you need so just ask for it in the comments and I'll be sure to add it.
Also I'm using Django 1.11
EDIT:
For more clarification: What I want is something like this /colors?team=blue
And let's say the team can be red, blue or green and you get this value from the team column in the given row that you get when the user logs in.
You could try to override djangos class-based view LoginView.
In views.py
from django.contrib.auth.views import LoginView
class MyLoginView(LoginView):
authentication_form = loginPlaceHolderForm
def get_redirect_url(self):
print(self.request.user)
# get data for user here
user_data_query_string = ''
url = '{}?{}'.format(
reverse('app:some-name')
user_data_query_string)
return url
In urls.py
url(r'^$', MyLoginView.as_view(), name='login'),
See also this question about adding GET querystring parameters to djangos HttpRedirect.
I have a custom user model (it's actually named User as I didn't see any need to name it otherwise) in my Django 1.5c1 project (currently running on the latest from the Django 1.5 branch on github). AUTH_USER_MODEL is defined in my settings properly, so the auth module works correctly and I can log in etc. fine.
However, with the custom user module enabled, the admin site doesn't work. When I add admin.autodiscover() to my urls.py, every page on the site (not just admin pages) throws a NotRegistered exception and says The model User is not registered. The traceback shows that admin.autodiscover() is trying to call admin.site.unregister(User), apparently before it has registered that model.
I tried renaming my user model to something other than User, but it didn't seem to work. I also tried creating my own admin.py for that app, and then I tried manually registering my custom User model with the custom UserAdmin model specified in admin.py before admin.autodiscover() ran, but that actually caused a separate exception saying that User was already registered.
What should I try next in order to get admin.autodiscover() working?
It looks like you need to jump through a few extra hoops if you want your custom User model to work with the admin. From the documentation:
...your User model must define some additional attributes and methods.
These methods allow the admin to control access of the User to admin
content:
class models.CustomUser
is_staff True if the user is allowed to have access to the admin site.
is_active True if the user account is currently active.
has_perm(perm, obj=None) True if the user has the named
permission.
has_module_perms(app_label) True if the user has perm
to access models in the given app.
I set up a brand new empty project with a custom user model and attempted to recreate the situation, which led to a diagnosis: we had added the django-usertools package to the project, which has not been updated for Django 1.5 and apparently conflicts with custom user models. Removing that package from the installed apps list in settings resolved the issue.