Django - Efficiently bulk create inherited models

Django - Efficiently bulk create inherited models

By : user2954764
Date : November 22 2020, 10:31 AM
Hope this helps The workaround I settled on was wrapping all of my collected create()s in a with transaction.atomic():. This greatly reduced running time by not opening any database connections or executing any queries until all of the Python had returned.
A downside could be that if any errors at all are encountered all changes are rolled back and the database is untouched. This could be remedied by chunking the create()s into batches and opening and closing a transaction around each one. (In my case this was not the desired behavior because I wanted all of the data or none of it.)
code :

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In Django, how could one use Django's update_object generic view to edit forms of inherited models?

In Django, how could one use Django's update_object generic view to edit forms of inherited models?

By : pramit
Date : March 29 2020, 07:55 AM
this will help Alright, here's what I've done, and it seems to work and be a sensible design (though I stand to be corrected!).
In a core library (e.g. mysite.core.views.create_update), I've written a decorator:
code :
from django.contrib.contenttypes.models import ContentType
from django.views.generic import create_update

def update_object_as_child(parent_model_class):
   Given a base models.Model class, decorate a function to return  
   create_update.update_object, on the child class.

   def update_object(request, object_id):

  kwargs should have an object_id defined.

  def decorator(function):
      def wrapper(request, **kwargs):
          # may raise KeyError
          id = kwargs['object_id']

          parent_obj = parent_model_class.objects.get( pk=id )

          # following http://www.djangosnippets.org/snippets/1031/
          child_class = parent_obj.content_type.model_class()

          kwargs['model'] = child_class

          # rely on the generic code for testing/validation/404
          return create_update.update_object(request, **kwargs)
      return wrapper

  return decorator
from mysite.core.views.create_update import update_object_as_child

def edit_animal(request, object_id):
urlpatterns += patterns('animals.views',
  url(r'^edit/(?P<object_id>\d+)$', 'edit_animal', name="edit_animal"),
Django forms: most DRY way to organize create / update forms for inherited models

Django forms: most DRY way to organize create / update forms for inherited models

By : living free
Date : March 29 2020, 07:55 AM
around this issue You're right that the class-based view documentation is rather primitive. No doubt it will improve, but for the moment you need to be prepared to experiment and to read the source.
You are trying to do two things:
code :
from django.views import generic

class PostCreateView(generic.CreateView):
    form_class = PostForm
    model = Post

class PostUpdateView(generic.UpdateView):
    form_class = PostForm
    model = Post

class BlogPostCreateView(generic.CreateView):
    form_class = BlogPostForm
    model = BlogPost

class BlogPostUpdateView(generic.UpdateView):
    form_class = BlogPostForm
    model = BlogPost
from django.views import generic

class PostView(generic.FormView):
    form_class = PostForm
    model = Post

class PostCreateView(PostView, generic.CreateView): pass
class PostUpdateView(PostView, generic.UpdateView): pass

class BlogPostView(PostView):
    form_class = BlogPostForm
    model = BlogPost

class BlogPostCreateView(BlogPostView, generic.CreateView): pass
class BlogPostUpdateView(BlogPostView, generic.UpdateView): pass
Django AutoSlugField across inherited models

Django AutoSlugField across inherited models

By : jane doe
Date : March 29 2020, 07:55 AM
should help you out OK so after digging through the source code it turns out the docs have updated since I lasted looked.
So, if you add objects = models.Manager() to your BaseModel and pass that to the AutoSlugField. This will check the slug against BaseModel rather than the sub class.
code :
class BaseModel(models.Model):
    objects = models.Manager()
    name = models.CharField(max_length=255)
    slug = AutoSlugField(populate_from='name', unique=True, db_index=True, managers=objects)
    # ... other fields
Django: How to save inherited models?

Django: How to save inherited models?

By : Tommy Karlsson
Date : March 29 2020, 07:55 AM
this will help Because you haven't declared your Conversation model to be abstract. You're using multi-table inheritance. Have a look at the docs.
If you want all the data stored in your child then you should do something like -
code :
class ConversationBase(models.Model):
    contact             = models.ForeignKey(Contact)
    conversation_datetime = models.DateTimeField()    
    notes               = models.TextField(_(u'Notes'),        blank=True)  

    class Meta:
        absract = True

class Conversation(ConversationBase):

class ConversationHistory(ConversationBase):
    log_date_time = CreationDateTimeField()
Django bulk create for models with multiple required fields

Django bulk create for models with multiple required fields

By : Egor Babosiuk
Date : March 29 2020, 07:55 AM
I hope this helps . I have a Person model, a Demographic model, and a PersonDemographic model which contains detailed methods using data from both models, and saves some very expensive calculations as a field for specific person-demographic instances. How do I use bulk create to facilitate entries into my PersonDemographic model? , You can do it like this:
code :
demographic = Demographic.objects.get(id=7)
    [PersonDemographic(person=person, demographic=demographic) for person in Person.objects.some_query_set()]
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