> Hmm, I guess I'm missing the "real world use case" for this. Most > importantly, I'm missing how this slow model relates to the fast model > and what kind of query it is executing. I'm especially curious about the > "as this statement is executed as an instance of the model" bit.
Ok, I'll try to explain this better. The fast model is coupled directly to a table. The Slow model is not, it is activated by a specific sql query that depends on a specific condition. AFAIK to get data from the slow model, I do: SLOMO = SlowModel.objects.raw( SQLstatement ,[CONDITION]) However, if I set the primary_key of SlowModel to OneToOne (pointing to FastModel), as suggested above, and try to retreive SlowModel data from FastModel (FastModelInstance.slow) I get an error that data is missing. My guess was that this was because I did not manage to tell the instance of Fastmodel that it first needs to pass the SQL to its SlowModel connection before I can ask it questions. #------------------------------- class FastModel(models.Model): idgroup = models.IntegerField(primary_key=True) field1 = models.IntegerField() field2 = models.CharField(max_length=250, blank=True) field3 = models.CharField(max_length=250, blank=True) class Meta: db_table = u'my_db_table' class SlowModel(models.Model): rid_fastmodel = models.IntegerField(primary_key=True) calcfield1 = models.TextField(blank=True) calcfield2= models.TextField(blank=True) SQLstatement = "SELECT rid_fastmodel, calcfield1, calcfield2 from (SELECT ....) JOIN (...) JOIN (...) WHERE some_condition = %s" Is this better? thanks joris -- You received this message because you are subscribed to the Google Groups "Django users" group. To view this discussion on the web visit https://groups.google.com/d/msg/django-users/-/TPkzlZJ7NfkJ. 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.