Django Introduce a foreign key from a model CharField - django

Question

I have migrated over 10,000 records from my old mySQL database to Django/sqlite. In my old mysql schema's Song table, the artist field was not a 1 to many field but was just a mysql varchar field. In my new Django model, I converted the artist field to a ForeignKey and used temp_artist to temporarily store the artist's name from the old database.
How do I create each Song instance's artist foreignkey based on the temp_artist field? I'm assuming I should use the manager's get_or_create method but where and how do I write the code?
my model below:
class Artist (models.Model):
name = models.CharField(max_length=100)
class Song (models.Model):
artist = models.ForeignKey(Artist, blank=True, null=True, on_delete=models.CASCADE, verbose_name="Artist")
temp_artist = models.CharField(null=True, blank=True, max_length=100)
title = models.CharField(max_length=100, verbose_name="Title")
duration = models.DurationField(null=True, blank=True, verbose_name="Duration")

Answer 1

You can write a custom management command that performs this logic for you. The docs provide good instructions on how to set it up. Your command code would look something like this:
# e.g., migrateauthors.py
from django.core.management.base import BaseCommand
from myapp import models
class Command(BaseCommand):
help = 'Migrate authors from old schema'
def handle(self, *args, **options):
for song in myapp.models.Song.objects.all():
song.artist, _ = models.Artist.objects.get_or_create(name=song.temp_artist)
song.save()
Then you simply run the management command with manage.py migrateauthors. Once this is done and verified you can remove the temporary field from your model.

Answer 2

Since you don't have a usable foreign key at the moment you would have to dig down to raw_sql. If you were still on mysql you could have used the UPDATE JOIN syntax. But unfortunately Sqlite does not support UPDATE JOIN.
Luckily for you you have only a few thousand rows and that makes it possible to iterate through them and update each row individually.
raw_query = '''SELECT s.*, a.id as fkid
FROM myapp_song s
INNER JOIN myapp_artist a on s.temp_artist = a.name'''
for song in Song.objects.raw(raw_query)
song.artist_id = s.fkid
song.save()
This might take a few minutes to complete because you don't have an index on temp_artist and name. Take care to replace myapp with the actual name of your app.
Edit1:
Though Sqlite doesn't have update JOIN, it does allow you to SET a value with a subquery. So this will also work.
UPDATE myapp_song set artist_id =
(SELECT id from myapp_artist WHERE name = myapp_song.temp_artist)
type it in the sqlite console or GUI. Make sure to replace myapp with your own app name. This will be very quick because it's a single query. All other solutions including my alternative solution in this answer involve 10,000 queries.
Edit 2
If your Artist table is empty at the moment, before you do all this you will have to populate it, here is an easy query that does it
INSERT INTO stackoverflow_artist(name)
SELECT distinct temp_artist from stackoverflow_song
note that you should have a unique index on Artist.name