CODEWITHSUNDEEP
djangooptimizationdjango-modelsforeign-keys
User
User

Reputation: 65931

Django: Set foreign key using integer?

Is there a way to set foreign key relationship using the integer id of a model? This would be for optimization purposes.

For example, suppose I have an Employee model:

class Employee(models.Model):
  first_name = models.CharField(max_length=100)
  last_name = models.CharField(max_length=100)
  type = models.ForeignKey('EmployeeType')

and 

EmployeeType(models.Model):
  type = models.CharField(max_length=100)

I want the flexibility of having unlimited employee types, but in the deployed application there will likely be only a single type so I'm wondering if there is a way to hardcode the id and set the relationship this way. This way I can avoid a db call to get the EmployeeType object first.

Upvotes: 122

Views: 73745

Answers (3)

Alext
Alext

Reputation: 21

Just to emphasize something on the accepted answer (based on my experience): When you want to add a value in the "id" field of the foreign key (hope that this is clear), you must add "_id" at the end of the name defined by you in that class.

In this example: employee = Employee(first_name="Name", last_name="Name", type_id=4)

Here "type" refers to "type" defined in the "Employee" class.

Note that if you have defined something like "foo_id", you still must add the "_id" suffix and the result will be "foo_id_id"

PS: I can't comment or do other actions (not enough reputation points)

Upvotes: 2

Will Hardy
Will Hardy

Reputation: 14836

Yep:

employee = Employee(first_name="Name", last_name="Name")
employee.type_id = 4
employee.save()

ForeignKey fields store their value in an attribute with _id at the end, which you can access directly to avoid visiting the database.

The _id version of a ForeignKey is a particularly useful aspect of Django, one that everyone should know and use from time to time when appropriate.

caveat: [ < Django 2.1 ]

@RuneKaagaard points out that employee.type is not accurate afterwards in recent Django versions, even after calling employee.save() (it holds its old value). Using it would of course defeat the purpose of the above optimisation, but I would prefer an accidental extra query to being incorrect. So be careful, only use this when you are finished working on your instance (eg employee).

Note: As @humcat points out below, the bug is fixed in Django 2.1

Upvotes: 244

Jacinda
Jacinda

Reputation: 5072

An alternative that uses create to create the object and save it to the database in one line:

employee = Employee.objects.create(first_name='first', last_name='last', type_id=4)

Upvotes: 58

Related Questions