i would not call this a real versioning as this will not keep track of
which object has been edited.
if you have 10 different ponies and edited pony 3 to 3', you can not
tell that 3' was 3 before. neither can you tell which are the current
ponies, you will only get the last pony with current_pony.
or did i miss anything?
david
On 28 Okt., 17:01, hcarvalhoalves <hcarvalhoal...@gmail.com> wrote:
> On Oct 27, 4:54 pm, Todd Blanchard <tblanch...@mac.com> wrote:
>
>
>
> > Total django noob here. Rails/PHP/WebObjects refugee here.
>
> > I'm starting a project where some models need to be fully versioned.
>
> > IOW, record update is forbidden - every save to a changed model should
> > result in a new record with a new version number or timestamp along
> > with identity pulled from the current authenticated user's session.
> > Queries/relationships should be specified as "fetch newest" or "fetch
> > history". IOW sometimes I want to traverse a relationship and just
> > get the "current" record. Sometimes I want to get the history of that
> > relationship.
>
> > Anybody done this? Got any tips?
>
> > Thanks,
> > -Todd Blanchard
>
> You can do all that without much magic.
>
> class MyVersionedModel(models.Model):
> user = models.ForeignKey('User', editable=False)
> timestamp = models.DateTimeField(auto_now_add=True,
> editable=False)
>
> class Meta:
> ordering = ['-timestamp']
> get_latest_by = 'timestamp'
> abstract = True
>
> def save(self, *args, **kwargs):
> self.pk = None # Forces insert
> return super(MyVersionedObject, self).save(*args, **kwargs)
>
> Now you just need to inherit this mixin on all models you want
> versioned.
>
> class MyVersionedPony(MyVersionedObject):
> ... # Whatever other fields
>
> In your views you can:
>
> def my_pony_view(request):
> ...
> my_pony.user = request.user
> my_pony.save() # Always inserts instead of updating
>
> current_pony = MyVersionedPony.objects.latest() # Last pony
> all_ponies = MyVersionedPony.objects.all() # Ordered descending
>
> If you need to encapsulate logic for more complex queries, you can
> create a model manager. [1]
>
> http://docs.djangoproject.com/en/dev/topics/db/managers/
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