Single Django model, multiple tables?
You could, I believe, make a factory function that would return your model with a dynamic db_table.
def getModel(db_table):
class MyClass(models.Model):
# define as usual ...
class Meta:
db_table = db_table
return MyClass
newClass = getModel('29345794_table')
newClass.objects.filter( ...
EDIT: Django does not create a new instance of the class's _meta
attribute each time this function is called. Creating a new instance for _meta
it is dependent upon the name of the class (Django must cache it somewhere). A metaclass can be used to change the name of the class at runtime:
def getModel(db_table):
class MyClassMetaclass(models.base.ModelBase):
def __new__(cls, name, bases, attrs):
name += db_table
return models.base.ModelBase.__new__(cls, name, bases, attrs)
class MyClass(models.Model):
__metaclass__ = MyClassMetaclass
class Meta:
db_table = db_table
return MyClass
not sure if it can be set dynamically on an already-defined class. I haven't done this myself but it might work.
You can set this whenever.
>>> MyModel._meta.db_table = '10293847_table'
>>> MyModel.objects.all()
Create a model for your table dynamically.
from django.db import models
from django.db.models.base import ModelBase
def create_model(db_table):
class CustomMetaClass(ModelBase):
def __new__(cls, name, bases, attrs):
model = super(CustomMetaClass, cls).__new__(cls, name, bases, attrs)
model._meta.db_table = db_table
return model
class CustomModel(models.Model):
__metaclass__ = CustomMetaClass
# define your fileds here
srno = models.IntegerField(db_column='SRNO', primary_key=True)
return CustomModel
and you can start querying the database.
In [6]: t = create_model('trial1')
In [7]: t._meta.db_table
Out[7]: 'trial1'
In [8]: t.objects.all() # default db
Out[8]: [<CustomModel: CustomModel object>, '(remaining elements truncated)...']
In [9]: t.objects.using('test').all() # test db
Out[9]: [<CustomModel: CustomModel object>, '(remaining elements truncated)...']