How does django decide which transaction to choose on transaction.on_commit()

Question

I had to use transaction.on_commit() for synchronous behaviour in one of the signals of my project. Though it works fine, I couldn't understand how does transaction.on_commit() decide which transaction to take. I mean there can be multiple transactions at the same time. But how does django know which transaction to take by using transaction.on_commit()

Answer 1

According to the docs

You can also wrap your function in a lambda:

transaction.on_commit(lambda: some_celery_task.delay('arg1')) The function you pass in will be called immediately after a hypothetical database write made where on_commit() is called would be successfully committed.

If you call on_commit() while there isn’t an active transaction, the callback will be executed immediately.

If that hypothetical database write is instead rolled back (typically when an unhandled exception is raised in an atomic() block), your function will be discarded and never called.

If you are using it on post_save method with sender=SomeModel. Probably the on_commit is executed each time a SomeModel object is saved. Without the proper code we would not be able to tell the exact case.

Answer 2

If I understand the question correctly, I think the docs on Savepoints explains this.

Essentially, you can nest any number of transactions, but on_commit() is only called after the top most one commits. However, on_commit() that's nested within a savepoint will only be called if that savepoint was committed and all the ones above it are committed. So, it's tied to which ever one is currently open at the point it's called.