Django: Set foreign key using integer?

DjangoOptimizationDjango ModelsForeign Keys

Django Problem Overview


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.

Django Solutions


Solution 1 - Django

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

Solution 2 - Django

An alternative that uses [create][1] 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)

[1]: https://docs.djangoproject.com/en/dev/ref/models/querysets/#create "Django create reference"

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Solution 1 - DjangoWill HardyView Answer on Stackoverflow
Solution 2 - DjangoJacindaView Answer on Stackoverflow