Thanks for the suggestions. The "convert to list" approach sounded
appealing, but I had to get back a queryset since this was for an RSS
feed.
After reading up more, finally decided it might actually make more
sense to go with a bit of denormalization and create a "combined_date"
field on the
If you know how to do it in SQL, just make your own manager.
http://docs.djangoproject.com/en/1.1/topics/db/managers/
Yay for Django!
Shawn
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I think you'll have to sort the list yourself by creating a calculated
field, since you're only returning a small number, I don't see this as
a big issue.
So, get the records back that you want via the ORM, and then copy them
over to a list.
HTH
John
On Feb 20, 6:26 pm, shacker
Given a model like:
class Item(models.Model):
title = models.CharField(max_length=140)
created_date =
models.DateTimeField(default=datetime.datetime.now)
completed = models.BooleanField(default=False)
completed_date = models.DateTimeField(blank=True,null=True)
...
I want to
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