On Sat, Oct 4, 2008 at 5:34 AM, Robert <[EMAIL PROTECTED]> wrote: > > I've been banging my head against this problem for several days now, > and have decided to ask for help. I've read the Django DB API and > searched this mailing list. > > Assuming the following models: > > http://dpaste.com/82218/
As a matter of style - here are two suggestions for your agency_docs table: 1) It should be named 'AgencyDoc'. Convention is that classes should be capitalized, with camel case, and singular (since each instance of the class will be an instance of an object). 2) The foreign keys should be agency and document, not agency_id and document_id. Under the hood, the _id suffix will be used in the database, indicating that the column contains an ID, but from a Python/Django API level, you will be dealing with fully fledged objects, not IDs > What I would like to do is access the 'study_level' field. > > When my view is called, I will know the Agency name, and I'd like to > show a list of documents that correspond to that agency. > > Generic views allow me to quite easily do this, but what I can't seem > to do is access the information in agency_docs. When I print out each > document, I'd like to put the 'study_level' information beside each. > > I though that 'select_related' would be my answer, because as far as I > can tell, I need to do a classic JOIN. select_related() won't really help here. select_related() is really just a database optimization; when one object has a foreign key on a second object, you can get the data to both objects using a single SQL call. When performance is critical, this can be a helpful optimization. However, it doesn't provide any additional access to data that wouldn't already be available. > Any help would be greatly appreciated. To solve this problem, you need to look very closely at the data you are trying to display. Are you looking to display a list of agencies? Or are you trying to display a list of the relationships that agencies have? The fact that you want to show study_level suggests that what you should be building is a generic relation on AgencyDoc, not Agency. You can select AgencyDoc objects that point to a specific agency (AgencyDoc.objects.filter(agency_id__name='foo')), and then each object in the list will have access to the study level for the relationship. Each AgencyDoc object has an agency and a document, so you have full access to all the data in the connected objects. Yours, Russ Magee %-) --~--~---------~--~----~------------~-------~--~----~ 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 [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/django-users?hl=en -~----------~----~----~----~------~----~------~--~---
