Add unique embed_key to List model

This commit is contained in:
Joachim 2021-12-04 16:06:07 +01:00
parent 749c902070
commit 3bd28afe93
2 changed files with 38 additions and 0 deletions

View file

@ -0,0 +1,29 @@
# Generated by Django 3.2.5 on 2021-12-04 10:55
from django.db import migrations, models
import uuid
def gen_uuid(apps, schema_editor):
"""sets an unique UUID for embed_key"""
book_lists = apps.get_model("bookwyrm", "List")
db_alias = schema_editor.connection.alias
for book_list in book_lists.objects.using(db_alias).all():
book_list.embed_key = uuid.uuid4()
book_list.save(broadcast=False)
class Migration(migrations.Migration):
dependencies = [
("bookwyrm", "0119_user_feed_status_types"),
]
operations = [
migrations.AddField(
model_name="list",
name="embed_key",
field=models.UUIDField(editable=False, null=True, unique=True),
),
migrations.RunPython(gen_uuid, reverse_code=migrations.RunPython.noop),
]

View file

@ -1,4 +1,6 @@
""" make a list of books!! """
import uuid
from django.apps import apps
from django.db import models
from django.db.models import Q
@ -43,6 +45,7 @@ class List(OrderedCollectionMixin, BookWyrmModel):
through="ListItem",
through_fields=("book_list", "book"),
)
embed_key = models.UUIDField(unique=True, null=True, editable=False)
activity_serializer = activitypub.BookList
def get_remote_id(self):
@ -105,6 +108,12 @@ class List(OrderedCollectionMixin, BookWyrmModel):
group=None, curation="closed"
)
def save(self, *args, **kwargs):
"""on save, update embed_key and avoid clash with existing code"""
if not self.embed_key:
self.embed_key = uuid.uuid4()
return super(List, self).save(*args, **kwargs)
class ListItem(CollectionItemMixin, BookWyrmModel):
"""ok"""