#35442: N+1 queries from RelatedManager + only("pk")
-------------------------------------+-------------------------------------
Reporter: REGNIER Guillaume | Owner: nobody
Type: Uncategorized | Status: new
Component: Database layer | Version: 4.2
(models, ORM) |
Severity: Normal | Resolution:
Keywords: | Triage Stage:
| Unreviewed
Has patch: 0 | Needs documentation: 0
Needs tests: 0 | Patch needs improvement: 0
Easy pickings: 0 | UI/UX: 0
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Comment (by Natalia Bidart):
Hello Guillaume!
I'm having a hard time understanding the goal of your code. I assume you
have provided a simplified version to be used as a small example, which I
appreciate, but with the information given, the proper way to fetch
records from the database without generating N+1 queries would be:
{{{#!python
>>> from django.db import connection, reset_queries
>>> reset_queries(); connection.queries
[]
>>> company = Company.objects.prefetch_related("employees").last()
>>> for e in company.employees.all():
>>> print(e.pk)
7
8
9
10
>>> connection.queries
[{'sql': 'SELECT "ticket_35442_company"."id",
"ticket_35442_company"."name" FROM "ticket_35442_company" ORDER BY
"ticket_35442_company"."id" DESC LIMIT 1',
'time': '0.002'},
{'sql': 'SELECT "ticket_35442_employee"."id",
"ticket_35442_employee"."company_id" FROM "ticket_35442_employee" WHERE
"ticket_35442_employee"."company_id" IN (5)',
'time': '0.001'}]
}}}
Or, my preferred (a single query):
{{{#!python
>>> for e in Employee.objects.filter(company=company).only("pk"):
print(e.pk)
7
8
9
10
>>> connection.queries
[{'sql': 'SELECT "ticket_35442_employee"."id" FROM "ticket_35442_employee"
WHERE "ticket_35442_employee"."company_id" = 5',
'time': '0.000'}]
}}}
--
Ticket URL: <https://code.djangoproject.com/ticket/35442#comment:2>
Django <https://code.djangoproject.com/>
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