Re: [Django] #21699: Provide a way to define a model without being registered into AppConfig

2014-01-05 Thread Django 
#21699: Provide a way to define a model without being registered into AppConfig
--+-
 Reporter:  mitar |Owner:  nobody
 Type:  New feature   |   Status:  new
Component:  Core (Other)  |  Version:  master
 Severity:  Normal|   Resolution:
 Keywords:  app-loading   | Triage Stage:  Someday/Maybe
Has patch:  0 |  Needs documentation:  0
  Needs tests:  0 |  Patch needs improvement:  0
Easy pickings:  0 |UI/UX:  0
--+-

Comment (by aaugustin):

 `ModelBase.__new__` cannot return a class previously registered in the app
 registry any more.

 `get_registered_model` is still used in `ModelSignal.connect` and in
 `add_lazy_relation`. In both cases it triggers different code paths
 depending on whether the target model was already created and its
 `class_prepared` signal was sent.

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Re: [Django] #21699: Provide a way to define a model without being registered into AppConfig

2013-12-30 Thread Django 
#21699: Provide a way to define a model without being registered into AppConfig
--+-
 Reporter:  mitar |Owner:  nobody
 Type:  New feature   |   Status:  new
Component:  Core (Other)  |  Version:  master
 Severity:  Normal|   Resolution:
 Keywords:  app-loading   | Triage Stage:  Someday/Maybe
Has patch:  0 |  Needs documentation:  0
  Needs tests:  0 |  Patch needs improvement:  0
Easy pickings:  0 |UI/UX:  0
--+-

Comment (by mitar):

 > So we're talking of hacking the internals heavily here ;-)

 Yes. :-) But from such hacking new use cases emerge. :-)

 > Currently ModelBase.new always returns a class registered in the app
 registry.

 Yes. This is why we register, and then try to unregister. So maybe then
 #21698 is a better approach. The issue there is that you still send a
 signal when a new model is registered, despite then removing it almost
 immediately. Maybe an alternative would be that API could provide a way to
 register a model without sending a signal, and then to unregister. But
 this is just a hack for current approach with `get_registered_model`.

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Re: [Django] #21699: Provide a way to define a model without being registered into AppConfig

2013-12-30 Thread Django 
#21699: Provide a way to define a model without being registered into AppConfig
--+-
 Reporter:  mitar |Owner:  nobody
 Type:  New feature   |   Status:  new
Component:  Core (Other)  |  Version:  master
 Severity:  Normal|   Resolution:
 Keywords:  app-loading   | Triage Stage:  Someday/Maybe
Has patch:  0 |  Needs documentation:  0
  Needs tests:  0 |  Patch needs improvement:  0
Easy pickings:  0 |UI/UX:  0
--+-
Changes (by aaugustin):

 * stage:  Unreviewed => Someday/Maybe


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Re: [Django] #21699: Provide a way to define a model without being registered into AppConfig

2013-12-30 Thread Django 
#21699: Provide a way to define a model without being registered into AppConfig
--+--
 Reporter:  mitar |Owner:  nobody
 Type:  New feature   |   Status:  new
Component:  Core (Other)  |  Version:  master
 Severity:  Normal|   Resolution:
 Keywords:  app-loading   | Triage Stage:  Unreviewed
Has patch:  0 |  Needs documentation:  0
  Needs tests:  0 |  Patch needs improvement:  0
Easy pickings:  0 |UI/UX:  0
--+--

Comment (by aaugustin):

 So we're talking of hacking the internals heavily here ;-)

 The implementation of Apps has become very straightforward if you ignore
 all the deprecated code. In the short term I'd just go for `del
 apps.all_models[app_label][model_name]` or `del
 apps.get_app_config(app_label).models[model_name]`, followed by
 `apps.clear_cache()`. The first one is marginally faster, the second one
 is marginally less likely to change. Make sure `model_name` is lowercase.

 Yesterday I tried to stop storing deferred models with the app registry.
 It would have solved your use case nicely -- what you're doing looks a lot
 like deferred models. Sadly it didn't work.

 This request has larger ramifications. Currently ModelBase.__new__ always
 returns a class registered in the app registry. It's quite a hack, and
 it's done for bad reasons that no longer exist since Django 1.4 (new
 project layout). I'm going to keep the ticket open with the following
 scope: get rid of get_registered_model, and then figure out what the
 registration API becomes.

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Re: [Django] #21699: Provide a way to define a model without being registered into AppConfig

2013-12-30 Thread Django 
#21699: Provide a way to define a model without being registered into AppConfig
--+--
 Reporter:  mitar |Owner:  nobody
 Type:  New feature   |   Status:  new
Component:  Core (Other)  |  Version:  master
 Severity:  Normal|   Resolution:
 Keywords:  app-loading   | Triage Stage:  Unreviewed
Has patch:  0 |  Needs documentation:  0
  Needs tests:  0 |  Patch needs improvement:  0
Easy pickings:  0 |UI/UX:  0
--+--

Comment (by mitar):

 I am just copying from #21698, to make things more clear here. So my
 comment how I see that unregistering could help:
 > So there are two ways to allow making custom subclasses of models which
 should not be registered. One is to somehow allow some configuration in
 Meta (#21682), another is to call current metaclass which adds it, but be
 able to remove it from registry immediately afterwards (a bad side-effect
 is that registration signal has been already made).

 I didn't know about "no-op subclass" approach. This seems interesting
 idea.

 > Maybe the goal is to avoid leaking memory when creating a large number
 of such models?

 Yes. So what we are doing is developing support for dynamic schema. Users
 can add dynamically fields from the many hop away models. In a way we make
 proxy fields to things you would need multiple hops and we do one big join
 instead in the background to not have to fetch them again and again as you
 are hopping around. (Those hops can cross content type relations as well.)
 (Which mean you can change concrete implementation of a model, but the
 rest of your code can still work as it is referencing them through
 registered names.) We add those fields as virtual fields into base model
 and create a dummy proxy model to keep them, when we return a new queryset
 for them. So the syntax is something like:

 {{{
 queryset = models.Node.objects.regpoint('config').registry_fields(
 name='core.general#name',
 type='core.type#type',
 router_id='core.routerid#router_id',
 project='core.project#project.name',
 ).regpoint('monitoring').registry_fields(
 last_seen='core.general#last_seen',
 network_status='core.status#network',
 monitored_status='core.status#monitored',
 health_status='core.status#health',
 peers='network.routing.topology#link_count',
 ).order_by('type')
 }}}

 So, after you fetch those additional fields, returned queryset contains a
 dynamically created proxy model to `models.Node`, with `name` and others
 added as virtual fields. As this proxy model is useful only for this
 particular queryset, we do not want to register it globally. `core.status`
 is a registered name for a particular registry point, which can be
 implemented by any model which announces that it is implementing it. So a
 concrete related model can change (be extended by an app), but your
 queryset does not have to change.

 All this is a bit involved and we do not yet have good documentation, but
 you can check [https://dev.wlan-si.net/wiki/Nodewatcher/Registry this
 description] and our
 
[https://github.com/wlanslovenija/nodewatcher/blob/77ec018255fa0546b362376976db2f7b874968c1/nodewatcher/core/registry/lookup.py#L138
 current implantation of creating a dynamic proxy model and unregistered
 immediately afterwards is here]. We do not want that such models pile up.

 I know this is not public API, but it would still be good to have some
 clear way to: or not register a model when created, or be able to remove
 it from the registry. As you are planing on rewriting AppConfig, I just
 wanted to raise a point that some internal API (which I understand can
 change at any time) would be useful here.

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Re: [Django] #21699: Provide a way to define a model without being registered into AppConfig

2013-12-29 Thread Django 
#21699: Provide a way to define a model without being registered into AppConfig
--+--
 Reporter:  mitar |Owner:  nobody
 Type:  New feature   |   Status:  new
Component:  Core (Other)  |  Version:  master
 Severity:  Normal|   Resolution:
 Keywords:  app-loading   | Triage Stage:  Unreviewed
Has patch:  0 |  Needs documentation:  0
  Needs tests:  0 |  Patch needs improvement:  0
Easy pickings:  0 |UI/UX:  0
--+--
Changes (by aaugustin):

 * version:  1.6 => master


Comment:

 Maybe the goal is to avoid leaking memory when creating a large number of
 such models?

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Re: [Django] #21699: Provide a way to define a model without being registered into AppConfig

2013-12-29 Thread Django 
#21699: Provide a way to define a model without being registered into AppConfig
--+--
 Reporter:  mitar |Owner:  nobody
 Type:  New feature   |   Status:  new
Component:  Core (Other)  |  Version:  1.6
 Severity:  Normal|   Resolution:
 Keywords:  app-loading   | Triage Stage:  Unreviewed
Has patch:  0 |  Needs documentation:  0
  Needs tests:  0 |  Patch needs improvement:  0
Easy pickings:  0 |UI/UX:  0
--+--

Comment (by aaugustin):

 `Model._deferred` might be even more suitable in that case, since it's
 what Django does for its own dynamic models.

 Of course it would be better to provide a formal API.

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Re: [Django] #21699: Provide a way to define a model without being registered into AppConfig

2013-12-29 Thread Django 
#21699: Provide a way to define a model without being registered into AppConfig
--+--
 Reporter:  mitar |Owner:  nobody
 Type:  New feature   |   Status:  new
Component:  Core (Other)  |  Version:  1.6
 Severity:  Normal|   Resolution:
 Keywords:  app-loading   | Triage Stage:  Unreviewed
Has patch:  0 |  Needs documentation:  0
  Needs tests:  0 |  Patch needs improvement:  0
Easy pickings:  0 |UI/UX:  0
--+--

Comment (by aaugustin):

 Setting Model._meta.auto_created most likely does exactly what the OP
 wants, but since they didn't specify, I can't be sure.

 https://github.com/django/django/blob/98b52ae2/django/apps/registry.py#L222

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Re: [Django] #21699: Provide a way to define a model without being registered into AppConfig

2013-12-29 Thread Django 
#21699: Provide a way to define a model without being registered into AppConfig
--+--
 Reporter:  mitar |Owner:  nobody
 Type:  New feature   |   Status:  new
Component:  Core (Other)  |  Version:  1.6
 Severity:  Normal|   Resolution:
 Keywords:  app-loading   | Triage Stage:  Unreviewed
Has patch:  0 |  Needs documentation:  0
  Needs tests:  0 |  Patch needs improvement:  0
Easy pickings:  0 |UI/UX:  0
--+--

Comment (by charettes):

 I'm also curious to know why you'd need such a feature? There was no way
 of doing this with the legacy `django.db.models.loading` AFAIK.

 Wouldn't making a `django.apps.registry.Apps` no-op subclass and
 specifying it as your `DynamicModel.Meta.apps` work?

 {{{
 #!python
 class Void(django.apps.registry.Apps):
 def register_model(self, app_label, model):
 pass

 void = Void()

 Meta = type('Meta', (), {'apps': void})
 MyDynamicModel = type('MyDynamicModel', (django.db.models.Model,),
 {'Meta': Meta})
 }}}

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Re: [Django] #21699: Provide a way to define a model without being registered into AppConfig

2013-12-28 Thread Django 
#21699: Provide a way to define a model without being registered into AppConfig
--+--
 Reporter:  mitar |Owner:  nobody
 Type:  New feature   |   Status:  new
Component:  Core (Other)  |  Version:  1.6
 Severity:  Normal|   Resolution:
 Keywords:  app-loading   | Triage Stage:  Unreviewed
Has patch:  0 |  Needs documentation:  0
  Needs tests:  0 |  Patch needs improvement:  0
Easy pickings:  0 |UI/UX:  0
--+--
Changes (by aaugustin):

 * needs_better_patch:   => 0
 * needs_tests:   => 0
 * needs_docs:   => 0


Comment:

 I know I'm insisting heavily, but can you formulate this feature request
 in terms of effects on the public API of the app registry, or any other
 public API of Django?

 For example:

 - "I would like app_config.get_model(xxx) to return None when I've created
 a model in this way."
 - "I would like Django not to crash miserably with this stack trace when
 I've created a model in that way."

 A request on the internal implementation of the app registry doesn't make
 sense because only public APIs are guaranteed to keep working as
 documented across versions. Why would you care that the model is or isn't
 referenced somewhere in the app registry, provided it's never returned or
 used by any API? In fact, if this feature gets implemented, it's likely
 that Django will register this model with some sort of "hidden" flag.

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