Django ORM: latest order per customer without N+1 queries

Backend Live Coding November 13, 2026 python

Get the latest order of each customer, without N+1.

python 
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
# ======================================================================
# Q: Get the latest order of each customer, without N+1.
# ======================================================================

latest = Order.objects.filter(customer=OuterRef('pk')).order_by('-created_at')
customers = Customer.objects.annotate(
    last_order_at=Subquery(latest.values('created_at')[:1]),
    last_order_id=Subquery(latest.values('id')[:1]),
)

# Alternatively in PostgreSQL: DISTINCT ON
Order.objects.order_by('customer_id', '-created_at').distinct('customer_id')

Related Posts

Python shallow copy vs deep copy
Python shallow copy vs deep copy
Python closures and late binding in lambdas
Python closures and late binding in lambdas
Storing heartbeats from 250,000 screens with Redis write coalescing
Storing heartbeats from 250,000 screens with Redis write coalescing
Writing a retry decorator with exponential backoff in Python
Writing a retry decorator with exponential backoff in Python
SQL: what a CTE is and how to find duplicate records
SQL: what a CTE is and how to find duplicate records
Threads vs processes vs asyncio: when to use each
Threads vs processes vs asyncio: when to use each