Develop Reference

Django backwards relation

I am setting up webservices for an application and I have the following models:
class Parent(models.Model):
...
class Child(models.Model):
parent = models.ForeignKey(Course)
...
The relation is One to Many (1 Parent, many Children)
Now, I would like to get all the Parent objects with its particular Child and send it as a JSON Request.
Is it possible to do so without having to first get all the "Childs" and iterate through them looking for the ones related to the particular parent?
I think that would be extremely inefficient for really large databases, plus the "Childs" won't be repeated in other "Parents"
Thank you very much

Every relationship in Django automatically gets its reverse relation added to the model. In the case of a ForeignKey or ManyToManyField that relation contains several objects. In that case, the default attribute name is set to <model>_set, so in this case child_set. This is a manager and can be used as such, so e.g. to iterate over all children:
for child in parent.child_set.all():
do_something()
You can also specify the attribute name used for the reverse relation using the related_name attribute:
class Child(models.Model):
parent = models.ForeignKey(Parent, related_name='children')
for child in parent.children.filter(some_field=True):
do_something()
Read more in the documentation on following relations backwards and how are backward relationships possible.

Why would you need to iterate? Even if Django didn't provide you with a special backwards syntax, you could always do this:
Child.objects.filter(parent=my_parent)
but as a cursory Google for the title of your question would have shown, there is a special syntax for backwards relations:
my_parent.child_set.all()

Yes, in django you can use:
parentInstance.child_set.all()
where parentInstance is one particular parent in your Parent database. That will return all of the Child objects associated with it in an efficient manner. To make it a JSON response, you can try something like this:
import json
from django.http import HttpResponse
response_data = {}
response_data[str(parentInstance)] = parentInstance.child_set.all()
return HttpResponse(json.dumps(response_data), content_type="application/json"
adopted from here.

How do I filter by a ForeignKey's db_column field without going through the foreign object in Django

I have a legacy database with non-django naming conventions. If I have the following (cut down) models:
class Registration(models.Model):
projectId=models.IntegerField(primary_key=True)
class Application(models.Model):
applicationId=models.IntegerField(primary_key=True)
registration=models.ForeignKey(Registration,db_column='projectId')
The ForeignKey instance causes a property to be created on Application called registration_id, but this is neither the correct name for the field (I have a hack to fix this), nor is it able to be used in a QuerySet.
Is there some way of using the id field provided by the ForeignKey on the Application model, rather than having to reference it via Registration?
Ie. I write lots of code like:
Application.objects.get(projectId=1234)
And don't want to have to write it out as:
Application.objects.get(registration__projectId=1234)
or even
Application.objects.get(registration__pk=1234)
I'm slightly surprised that:
Application.objects.get(registration_id=1234)
doesn't work...
Also note, I tried defining the id column as a field as well as the foreignkey which worked for queryset, but inserts complain of trying to insert into the same column twice:
class Application(models.Model):
...
projectId=models.IntegerField()
...

Have you tried this?
Application.objects.get(registration=1234)

I think just doing Application.objects.registration.get(projectId=1234) should do what you want.

Figure out child type with Django MTI or specify type as field?

I'm setting up a data model in django using multiple-table inheritance (MTI) like this:
class Metric(models.Model):
account = models.ForeignKey(Account)
date = models.DateField()
value = models.FloatField()
calculation_in_progress = models.BooleanField()
type = models.CharField( max_length=20, choices= METRIC_TYPES ) # Appropriate?
def calculate(self):
# default calculation...
class WebMetric(Metric):
url = models.URLField()
def calculate(self):
# web-specific calculation...
class TextMetric(Metric):
text = models.TextField()
def calculate(self):
# text-specific calculation...
My instinct is to put a 'type' field in the base class as shown here, so I can tell which sub-class any Metric object belongs to. It would be a bit of a hassle to keep this up to date all the time, but possible. But do I need to do this? Is there some way that django handles this automatically?
When I call Metric.objects.all() every objects returned is an instance of Metric never the subclasses. So if I call .calculate() I never get the sub-class's behavior.
I could write a function on the base class that tests to see if I can cast it to any of the sub-types like:
def determine_subtype(self):
try:
self.webmetric
return WebMetric
except WebMetric.DoesNotExist:
pass
# Repeat for every sub-class
but this seems like a bunch of repetitious code. And it's also not something that can be included in a SELECT filter -- only works in python-space.
What's the best way to handle this?

While it might offend some people's sensibilities, the only practical way to solve this problem is to put either a field or a method in the base class which says what kind of object each record really is. The problem with the method you describe is that it requires a separate database query for every type of subclass, for each object you're dealing with. This could get extremely slow when working with large querysets. A better way is to use a ForeignKey to the django Content Type class.
#Carl Meyer wrote a good solution here: How do I access the child classes of an object in django without knowing the name of the child class?
Single Table Inheritance could help alleviate this issue, depending on how it gets implemented. But for now Django does not support it: Single Table Inheritance in Django so it's not a helpful suggestion.

But do I need to do this?
Never. Never. Never.
Is there some way that django handles this automatically?
Yes. It's called "polymorphism".
You never need to know the subclass. Never.
"What about my WebMetric.url and my TextMetric.text attributes?"
What will you do with these attributes? Define a method function that does something. Implement different versions in WebMetric (that uses url) and TextMetric (that uses text).
That's proper polymorphism.
Please read this: http://docs.djangoproject.com/en/1.2/topics/db/models/#abstract-base-classes
Please make your superclass abstract.
Do NOT do this: http://docs.djangoproject.com/en/1.2/topics/db/models/#multi-table-inheritance
You want "single-table inheritance".

Problem with model inheritance and polymorphism

i came with new django problem. The situtaion: i have a model class UploadItemModel, i subcallss it to create uploadable items, like videos, audio files ...
class UploadItem(UserEntryModel):
category = 'abstract item'
file = models.FileField(upload_to=get_upload_directory)
i subclass it like this:
class Video(UploadItem):
category = 'video'
I need to access category attributes from a custom tag. The problem si that i am getting category='abstract item' even if the class is actually Video.
Any clue?
EDIT: I need to use hierarchy because i have several types of item that user can uplaod(Video, Audio files, PDF text). I need to create a class for each type, but there are lot of things in common between those classes(eg forms).

Any clue?
Yes. AFAIK it doesn't work the way you're hoping. Django Models aren't trivially Python classes. They're more like metaclasses which create instances of a kind of "hidden" class definition. Yes, the expected model class exists, but it isn't quite what you think it is. For one thing, the class you use was built for you from your class definition. That's why some static features of Python classes don't work as you'd expect in Django models.
You can't really make use of class-level items like this.
You might want to create an actual field with a default value or something similar.
class UploadItem(UserEntryModel):
category = models.CharFIeld( default='abstract item' )
file = models.FileField(upload_to=get_upload_directory)
Even after the comments being added to the question, I'm still unclear on why this is being done. There do not seem to be any structural or behavioral differences. These all seem like a single class of objects. Subclasses don't seem to define anything new.
Options.
Simply use the class name instead of this "category" item at the class level. Make the class names good enough that you don't need this "category" item.
Use a property
class UploadItem(UserEntryModel):
file = models.FileField(upload_to=get_upload_directory)
#property
def category( self ):
return self.__class__.__name__

You will need to create an additional field that will be a descriptor for that type.
There is a good tutorial here explaining how to use inheritance in Django models

Can you try overriding the __init__ method of the class to assign a category to each instance? For e.g.
class Video(UploadItem):
def __init__(self, *args, **kwargs):
super(Video, self).__init__(*args, **kwargs)
self.category = 'video'

How do I access the child classes of an object in django without knowing the name of the child class?

In Django, when you have a parent class and multiple child classes that inherit from it you would normally access a child through parentclass.childclass1_set or parentclass.childclass2_set, but what if I don't know the name of the specific child class I want?
Is there a way to get the related objects in the parent->child direction without knowing the child class name?

(Update: For Django 1.2 and newer, which can follow select_related queries across reverse OneToOneField relations (and thus down inheritance hierarchies), there's a better technique available which doesn't require the added real_type field on the parent model. It's available as InheritanceManager in the django-model-utils project.)
The usual way to do this is to add a ForeignKey to ContentType on the Parent model which stores the content type of the proper "leaf" class. Without this, you may have to do quite a number of queries on child tables to find the instance, depending how large your inheritance tree is. Here's how I did it in one project:
from django.contrib.contenttypes.models import ContentType
from django.db import models
class InheritanceCastModel(models.Model):
"""
An abstract base class that provides a ``real_type`` FK to ContentType.
For use in trees of inherited models, to be able to downcast
parent instances to their child types.
"""
real_type = models.ForeignKey(ContentType, editable=False)
def save(self, *args, **kwargs):
if self._state.adding:
self.real_type = self._get_real_type()
super(InheritanceCastModel, self).save(*args, **kwargs)
def _get_real_type(self):
return ContentType.objects.get_for_model(type(self))
def cast(self):
return self.real_type.get_object_for_this_type(pk=self.pk)
class Meta:
abstract = True
This is implemented as an abstract base class to make it reusable; you could also put these methods and the FK directly onto the parent class in your particular inheritance hierarchy.
This solution won't work if you aren't able to modify the parent model. In that case you're pretty much stuck checking all the subclasses manually.

In Python, given a ("new-style") class X, you can get its (direct) subclasses with X.__subclasses__(), which returns a list of class objects. (If you want "further descendants", you'll also have to call __subclasses__ on each of the direct subclasses, etc etc -- if you need help on how to do that effectively in Python, just ask!).
Once you have somehow identified a child class of interest (maybe all of them, if you want instances of all child subclasses, etc), getattr(parentclass,'%s_set' % childclass.__name__) should help (if the child class's name is 'foo', this is just like accessing parentclass.foo_set -- no more, no less). Again, if you need clarification or examples, please ask!

Carl's solution is a good one, here's one way to do it manually if there are multiple related child classes:
def get_children(self):
rel_objs = self._meta.get_all_related_objects()
return [getattr(self, x.get_accessor_name()) for x in rel_objs if x.model != type(self)]
It uses a function out of _meta, which is not guaranteed to be stable as django evolves, but it does the trick and can be used on-the-fly if need be.

It turns out that what I really needed was this:
Model inheritance with content type and inheritance-aware manager
That has worked perfectly for me. Thanks to everyone else, though. I learned a lot just reading your answers!

You can use django-polymorphic for that.
It allows to automatically cast derived classes back to their actual type. It also provides Django admin support, more efficient SQL query handling, and proxy model, inlines and formset support.
The basic principle seems to be reinvented many times (including Wagtail's .specific, or the examples outlined in this post). It takes more effort however, to make sure it doesn't result in an N-query issue, or integrate nicely with the admin, formsets/inlines or third party apps.

Here's my solution, again it uses _meta so isn't guaranteed to be stable.
class Animal(models.model):
name = models.CharField()
number_legs = models.IntegerField()
...
def get_child_animal(self):
child_animal = None
for r in self._meta.get_all_related_objects():
if r.field.name == 'animal_ptr':
child_animal = getattr(self, r.get_accessor_name())
if not child_animal:
raise Exception("No subclass, you shouldn't create Animals directly")
return child_animal
class Dog(Animal):
...
for a in Animal.objects.all():
a.get_child_animal() # returns the dog (or whatever) instance

You can achieve this looking for all the fields in the parent that are an instance of django.db.models.fields.related.RelatedManager. From your example it seems that the child classes you are talking about are not subclasses. Right?

An alternative approach using proxies can be found in this blog post. Like the other solutions, it has its benefits and liabilities, which are very well put in the end of the post.