Change type of Django model field from CharField to ForeignKey

Adding on top of Joey's answer, detailed steps for Django 2.2.11.

Here are the models from my use case, that consists of a Company and Employee model. We have to convert designation to a foreign key field. The app name is called core

class Company(CommonFields):
    name = models.CharField(max_length=255, blank=True, null=True

class Employee(CommonFields):
    company = models.ForeignKey("Company", on_delete=models.CASCADE, blank=True, null=True)
    designation = models.CharField(max_length=100, blank=True, null=True)

Step 1

Create a foreign key designation_link in Employee and mark it as null=True

class Designation(CommonFields):
    name = models.CharField(max_length=255)
    company = models.ForeignKey("Company", on_delete=models.CASCADE, blank=True, null=True)

class Employee(CommonFields):
    company = models.ForeignKey("Company", on_delete=models.CASCADE, blank=True, null=True)
    designation = models.CharField(max_length=100, blank=True, null=True)
    designation_link = models.ForeignKey("Designation", on_delete=models.CASCADE, blank=True, null=True)

Step 2

Create empty migration. Using the command:

python app_code/manage.py makemigrations --empty --name transfer_designations core

This will create a following file in migrations directory.

# Generated by Django 2.2.11 on 2020-04-02 05:56

from django.db import migrations


class Migration(migrations.Migration):

    dependencies = [
        ('core', '0006_auto_20200402_1119'),
    ]

    operations = [
    ]

Step 3

Populate the empty migration with a function that loops over all Employees, creates a Designation and links it to the Employee.

In my use case each Designation is also linked to a Company. Which means that Designation may contain two rows for "managers", one for company A, another for company B.

Final migration would look something like this:

# core/migrations/0007_transfer_designations.py

# Generated by Django 2.2.11 on 2020-04-02 05:56

from django.db import migrations

def link_designation(apps, schema_editor):
    Employee = apps.get_model('core', 'Employee')
    Designation = apps.get_model('core', 'Designation')
    for emp in Employee.objects.all():
        if(emp.designation is not None and emp.company is not None):
            desig, created = Designation.objects.get_or_create(name=emp.designation, company=emp.company)
            emp.designation_link = desig
            emp.save()

class Migration(migrations.Migration):

    dependencies = [
        ('core', '0006_auto_20200402_1119'),
    ]

    operations = [
        migrations.RunPython(link_designation),
    ]

Step 4

Finally run this migration using:

python app_code/manage.py migrate core 0007

That's a continuation of the great answer by Joey. How to rename the new field to the original name?

If the field has data, it probably means that you are using it elsewhere in your project, therefore this solution will leave you with a field named differently, and you have to either refactor the project to use the new field or delete the old field and rename the new one.

Be aware that this process is not going to prevent you to refactor code. If you where using a CharField with CHOICES, you were accessing its content with get_filename_display(), for example.

If you try to delete the field to make a migration, for then renaming the other field and make another migration, you'll see Django complaining because you cannot delete a field that you are using in the project.

Just create an empty migration as Joey explained, and put this in operations:

operations = [
    migrations.RemoveField(
        model_name='app_name',
        name='old_field_name',
    ),
    migrations.RenameField(
        model_name='app_name',
        old_name='old_field_name_link',
        new_name='old_field_name',
    ),
]

Then run migrate and you'll have the changes made in your database, but obviously not in your model, it's time now to delete the old field and to rename new ForeignKey field to the original name.

I don't think that doing this is particularly hacky, but still, only do this kind of things if you are fully understanding what are you messing with.