python - What's the correct way to use transaction.on_commit with Celery tasks in Django? - Stack Overflow

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Approach 1: Without transaction.atomic

Source: /

def order_created(order):
    order.status = 'created'
    order.save()
    transaction.on_commit(
        partial(send_order_email.delay, order_id=order.id)
    )

Approach 2: With transaction.atomic

def order_created(order):
    with transaction.atomic():
        order.status = 'created'
        order.save()
        transaction.on_commit(
            partial(send_order_email.delay, order_id=order.id)
        )

Which approach is correct and why? The save() operation makes database changes, but examples differ on whether transaction.atomic is needed. (ATOMIC_REQUESTS=False in settings)

Approach 1: Without transaction.atomic

Source: https://adamj.eu/tech/2022/08/22/use-partial-with-djangos-transaction-on-commit/

def order_created(order):
    order.status = 'created'
    order.save()
    transaction.on_commit(
        partial(send_order_email.delay, order_id=order.id)
    )

Approach 2: With transaction.atomic

def order_created(order):
    with transaction.atomic():
        order.status = 'created'
        order.save()
        transaction.on_commit(
            partial(send_order_email.delay, order_id=order.id)
        )

Which approach is correct and why? The save() operation makes database changes, but examples differ on whether transaction.atomic is needed. (ATOMIC_REQUESTS=False in settings)

• python
• django
• celery

Share Improve this question Follow asked Jan 24 at 7:09 swaranswaran 11599 bronze badges

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1 Answer 1

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In the first example transaction.on_commit will result in error if there is no transaction open (related docs), plus save() will be executed right away so there is no need for on_commit at all.

In second example, Celery task will be queued at the end of successful transaction. As I understand your question, this is expected behaviour

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