On Thursday 18 February 2010 15:41:40 Sander wrote: > let's say the following (pure hypothetical): > - an Manager can read questions at example.com/manager/read > - an Employee can post questions at example.com/employee/add > > Let me try myself: > In this case, following your guidelines, I think I have to create an > app called 'question' whith the functionality of adding and listing > the questions > Also the Employee and Manager should be different apps because they > have functionality of changing their information. > The apps DO depent on eachoter because a question is placed by a > 'Employee' but thats fine > > Am I on the right track? First of all you need to make sure whether you need an Employer (and a seperate Manager) model. Perhaps you can do with a UserProfile (see auth docs). And then there can be a role column to define Employer/Manager/etc or you can make use of Group's (again, documented). If this works out for you; you have an app named `profile`.
Now it shouldn't be too complicated if you write a Question app which has `add` and `read` functionality (BTW they seem related to Question to me). If you think `question` app is getting bloated (you judgement call) you can refactor administrative stuff into a, say, `manager` app. I hope all this is making sense. But the single most useful advice I can give is; get your feet wet. Get out of the land of purely hypothetical and implement something. You wont regret the time invested in Django, trust me. Also you can take a look at existing open source projects based on Django. I'd like to name a few but they all seem unnecessarily complex to me. :/ -- Saygılarımla, Atamert Ölçgen -+- --+ +++ www.muhuk.com mu...@jabber.org -- You received this message because you are subscribed to the Google Groups "Django users" group. To post to this group, send email to django-us...@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.
