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Refs #31051 -- Optimized serialize_db_to_string() by avoiding creation of models list.

This commit is contained in:
Matthijs Kooijman 2020-04-02 11:38:23 +02:00 committed by Mariusz Felisiak
parent 289d0ec6fd
commit 75520e1767
1 changed files with 18 additions and 18 deletions
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20
django/db/backends/base/creation.py
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@ -97,24 +97,24 @@ class BaseDatabaseCreation:
Designed only for test runner usage; will not handle large
amounts of data.
"""
# Build list of all models to serialize.
# Iteratively return every object for all models to serialize.
def get_objects():
from django.db.migrations.loader import MigrationLoader
loader = MigrationLoader(self.connection)
model_list = []
for app_config in apps.get_app_configs():
if (
app_config.models_module is not None and
app_config.label in loader.migrated_apps and
app_config.name not in settings.TEST_NON_SERIALIZED_APPS
):
model_list.extend(app_config.get_models())
# Make a function to iteratively return every object
def get_objects():
for model in model_list:
if (model._meta.can_migrate(self.connection) and
router.allow_migrate_model(self.connection.alias, model)):
queryset = model._default_manager.using(self.connection.alias).order_by(model._meta.pk.name)
for model in app_config.get_models():
if (
model._meta.can_migrate(self.connection) and
router.allow_migrate_model(self.connection.alias, model)
):
queryset = model._default_manager.using(
self.connection.alias,
).order_by(model._meta.pk.name)
yield from queryset.iterator()
# Serialize to a string
out = StringIO()