Re: Showing per-user information in admin
Thanks a lot derek! It's just all I was looking for :) On Fri, Aug 12, 2011 at 4:03 PM, Derek wrote: > On Aug 12, 1:29 pm, Isaac wrote: > > Hi dudes, > > > > I'm having trouble at finding a solution for my problem. All I want to > > do is tune django admin to show, for a given user an a given model, only > > instance related to that user. > > > > my pseudocode > > > > class User > > ... > > class A > > foreign_key User > > > > A1 related to User1 > > A2 related to User1 > > A3 related to User2 > > > > then, I want that User1 was only be able to see A1 and A2, and User2 was > > only be able to see A3. > > I know that I can create filters to allow user to filter instances, but > > all I want is not to allow user to make that decision, it will be only > > able to see what its related. > > > > Any clue about how to use admin in this way? > > > > Thanks in advance > > See: > > http://lincolnloop.com/static/slides/2010-djangocon/customizing-the-admin.html#slide45 > for an example of this. Note how the default queryset is now filtered > per user. > > -- > You received this message because you are subscribed to the Google Groups > "Django users" group. > To post to this group, send email to django-users@googlegroups.com. > To unsubscribe from this group, send email to > django-users+unsubscr...@googlegroups.com. > For more options, visit this group at > http://groups.google.com/group/django-users?hl=en. > > -- You received this message because you are subscribed to the Google Groups "Django users" group. To post to this group, send email to django-users@googlegroups.com. To unsubscribe from this group, send email to django-users+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/django-users?hl=en.
Re: Showing per-user information in admin
On Aug 12, 1:29 pm, Isaac wrote: > Hi dudes, > > I'm having trouble at finding a solution for my problem. All I want to > do is tune django admin to show, for a given user an a given model, only > instance related to that user. > > my pseudocode > > class User > ... > class A > foreign_key User > > A1 related to User1 > A2 related to User1 > A3 related to User2 > > then, I want that User1 was only be able to see A1 and A2, and User2 was > only be able to see A3. > I know that I can create filters to allow user to filter instances, but > all I want is not to allow user to make that decision, it will be only > able to see what its related. > > Any clue about how to use admin in this way? > > Thanks in advance See: http://lincolnloop.com/static/slides/2010-djangocon/customizing-the-admin.html#slide45 for an example of this. Note how the default queryset is now filtered per user. -- You received this message because you are subscribed to the Google Groups "Django users" group. To post to this group, send email to django-users@googlegroups.com. To unsubscribe from this group, send email to django-users+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/django-users?hl=en.
Showing per-user information in admin
Hi dudes, I'm having trouble at finding a solution for my problem. All I want to do is tune django admin to show, for a given user an a given model, only instance related to that user. my pseudocode class User ... class A foreign_key User A1 related to User1 A2 related to User1 A3 related to User2 then, I want that User1 was only be able to see A1 and A2, and User2 was only be able to see A3. I know that I can create filters to allow user to filter instances, but all I want is not to allow user to make that decision, it will be only able to see what its related. Any clue about how to use admin in this way? Thanks in advance -- You received this message because you are subscribed to the Google Groups "Django users" group. To post to this group, send email to django-users@googlegroups.com. To unsubscribe from this group, send email to django-users+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/django-users?hl=en.
