I'm trying to create a Django ORM mapping that's compatible with an existing data model, so I'm trying to work with an existing set of table and column names.
I've got a multi-table inheritance situation where a class InformationObject derives from class Object. I'd like to let Django handle this the usual way:
class Object(models.Model):
class Meta:
db_table = "object"
class InformationObject(Object):
class Meta:
db_table = "information_object"
In this case Django would automatically create a one-to-one field on the inheriting model called object_ptr_id. However, on the schema I'm constrained to use, the reference to the Object is simply called "id". So:
Is there a way to somehow specify the name of the column Django auto-magically uses for multi-table inheritance?
The alternative, which I'll have to use otherwise, is to use an explicit one-to-one field, but then I won't be able to inherit non-开发者_Python百科database methods from the Object model:
class Object(models.Model):
class Meta:
db_table = "object"
class InformationObject(models.Model):
class Meta:
db_table = "information_object"
id = models.OneToOneField(Object, primary_key=True, db_column="id")
Any ideas? Maybe I could create a common base class for both of them and put non-db methods there...?
From the django docs (development version):
As mentioned, Django will automatically create a OneToOneField linking your child class back any non-abstract parent models. If you want to control the name of the attribute linking back to the parent, you can create your own OneToOneField and set parent_link=True to indicate that your field is the link back to the parent class.
As mentioned by @fusion quoting from the docs, you will have to create a OneToOneField if you want to specify the column, while using model inheritance. The inherited fields will be available in the child class in both self scope AND the one-to-one field.
class Object(models.Model):
class Meta:
db_table = "object"
column_1 = models.CharField()
class InformationObject(Object):
class Meta:
db_table = "information_object"
# arbitrary property name (parent_link)
parent_link = models.OneToOneField(Object, primary_key=True, db_column="id", parent_link=True)
In this example:
>>> inf_obj = InformationObject.objects.get(pk=1)
>>> print inf_obj.column_1 == inf_obj.parent_link.column_1
True
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